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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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implement-magic-dictionary | [C++] Clean Code | c-clean-code-by-alexander-rjx5 | \nclass MagicDictionary {\npublic:\n /** Initialize your data structure here. */\n MagicDictionary() {\n \n }\n \n /** Build a dictionary | alexander | NORMAL | 2017-09-10T03:19:54.684000+00:00 | 2017-09-10T03:19:54.684000+00:00 | 1,205 | false | ```\nclass MagicDictionary {\npublic:\n /** Initialize your data structure here. */\n MagicDictionary() {\n \n }\n \n /** Build a dictionary through a list of words */\n void buildDict(vector<string> dict) {\n for (string w : dict) {\n words.insert(w);\n }\n }\n \... | 7 | 1 | [] | 0 |
implement-magic-dictionary | Efficient Trie and Java 8 w/ Explanation | efficient-trie-and-java-8-w-explanation-xor6s | The implementation is a simple Trie, with the method relaxedSearch.\n\nrelaxedSearch searches for a word, with one deviation from a normal trie.\n\nIf there is | Cubicon | NORMAL | 2017-09-10T03:03:22.317000+00:00 | 2017-09-10T03:03:22.317000+00:00 | 3,974 | false | The implementation is a simple Trie, with the method relaxedSearch.\n\nrelaxedSearch searches for a word, with one deviation from a normal trie.\n\nIf there is a match with the current character, it proceeds as usual in that branch.\nBut for all the non matched characters, it still continues searching, by incrementing ... | 7 | 2 | [] | 4 |
implement-magic-dictionary | 2 Solution Beginner Friendly Set and simple array using Zip (Python3) | 2-solution-beginner-friendly-set-and-sim-8daq | Approach\n Describe your approach to solving the problem. \n- 1. Array and Zip \n- 2. Set method\n\n# Complexity\n\n- Array and Zip\n - Time complexity: O(n | amitpandit03 | NORMAL | 2023-04-16T19:27:25.061918+00:00 | 2023-04-16T19:27:42.551740+00:00 | 342 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n- 1. Array and Zip \n- 2. Set method\n\n# Complexity\n\n- Array and Zip\n - Time complexity: O(n * m)\n\n - Space complexity: O(n + m)\n- Set \n - Time complexity: O(26M) \n\n - Space complexity: O(n * m)\n\n# 1. Simple Array and zip meth... | 6 | 0 | ['Array', 'Hash Table', 'String', 'Design', 'Python3'] | 0 |
implement-magic-dictionary | My Java Solution with Trie and Backtracking Search | my-java-solution-with-trie-and-backtrack-vztk | Implement magic dictionary with two operations: buildDict and search. It\'s easy to come up with Trie to build a dict. The difficulty here for me is to implemen | xiaoxiao7 | NORMAL | 2020-02-29T09:45:55.200783+00:00 | 2020-02-29T11:03:57.496699+00:00 | 539 | false | Implement magic dictionary with two operations: `buildDict` and `search`. It\'s easy to come up with `Trie` to build a dict. The difficulty here for me is to implement the `search` operation.\nI use the backtracking way to find if there exists one word in the dict which has only one character different from the word I ... | 6 | 0 | ['Backtracking', 'Trie', 'Java'] | 1 |
implement-magic-dictionary | C++ 0ms solution with Hashtable | c-0ms-solution-with-hashtable-by-jasperj-gega | \tclass MagicDictionary {\n\tpublic:\n\t\t/ Initialize your data structure here. */\n\t\tunordered_map> m;\n\t\tMagicDictionary() {\n\n\t\t}\n\n\t\t/ Build a di | jasperjoe | NORMAL | 2020-01-07T07:10:18.578013+00:00 | 2020-01-07T07:10:18.578045+00:00 | 416 | false | \tclass MagicDictionary {\n\tpublic:\n\t\t/** Initialize your data structure here. */\n\t\tunordered_map<int,vector<string>> m;\n\t\tMagicDictionary() {\n\n\t\t}\n\n\t\t/** Build a dictionary through a list of words */\n\t\tvoid buildDict(vector<string> dict) {\n\t\t\tfor(auto x:dict){\n\t\t\t\tm[x.size()].push_back(x)... | 6 | 0 | ['Hash Table', 'C', 'C++'] | 2 |
implement-magic-dictionary | Python Simple Solution | python-simple-solution-by-yangshun-otxu | Since the input set is small and the character set is finite, it is acceptable to enumerate through all possible combinations and see if the new word exists in | yangshun | NORMAL | 2017-09-10T03:09:43.455000+00:00 | 2017-09-10T03:09:43.455000+00:00 | 1,042 | false | Since the input set is small and the character set is finite, it is acceptable to enumerate through all possible combinations and see if the new word exists in the dictionary and the time complexity for search is O(k^2) where k is the length of the word.\n\nFunnily enough, the comment in the skeleton for the `search` m... | 5 | 4 | [] | 1 |
implement-magic-dictionary | Two C++ solutions, Hash Table / Trie | two-c-solutions-hash-table-trie-by-zefen-htpl | Solution 1. Hash Table\n\nclass MagicDictionary {\npublic:\n /** Initialize your data structure here. */\n MagicDictionary() {}\n \n /** Build a dic | zefengsong | NORMAL | 2017-10-14T22:07:11.531000+00:00 | 2017-10-14T22:07:11.531000+00:00 | 669 | false | **Solution 1.** Hash Table\n```\nclass MagicDictionary {\npublic:\n /** Initialize your data structure here. */\n MagicDictionary() {}\n \n /** Build a dictionary through a list of words */\n void buildDict(vector<string> dict) {\n for(auto x: dict) m[x.size()].push_back(x);\n }\n \n /** ... | 5 | 0 | ['Hash Table', 'Trie', 'C++'] | 2 |
implement-magic-dictionary | Very simple python 🐍 solution using set | very-simple-python-solution-using-set-by-1xhi | \nclass MagicDictionary:\n\n def __init__(self):\n self.myDict=set()\n\n def buildDict(self, dictionary: List[str]) -> None:\n for word in d | injysarhan | NORMAL | 2020-11-03T19:00:23.081720+00:00 | 2020-11-03T19:00:46.654215+00:00 | 292 | false | ```\nclass MagicDictionary:\n\n def __init__(self):\n self.myDict=set()\n\n def buildDict(self, dictionary: List[str]) -> None:\n for word in dictionary:\n self.myDict.add(word)\n\n def search(self, searchWord: str) -> bool:\n \n for word in self.myDict:\n if l... | 4 | 0 | ['Python', 'Python3'] | 1 |
implement-magic-dictionary | [Python] Solve this question in real interview with video explanation | python-solve-this-question-in-real-inter-qijq | My step by step approach to solve this question like in real interview with video explanation:\nhttps://youtu.be/zjnb7QUZMPQ\n\nIntuition: Hahsmap / Looping\n\n | iverson52000 | NORMAL | 2020-10-18T23:10:57.697302+00:00 | 2020-10-18T23:10:57.697333+00:00 | 322 | false | My step by step approach to solve this question like in real interview with video explanation:\nhttps://youtu.be/zjnb7QUZMPQ\n\nIntuition: Hahsmap / Looping\n\n**Code**\n```\nclass MagicDictionary:\n\n def __init__(self):\n """\n Initialize your data structure here.\n """\n self.m = colle... | 4 | 0 | ['Python'] | 0 |
implement-magic-dictionary | [C++] simple easy solution using Trie | c-simple-easy-solution-using-trie-by-sah-i08d | \nclass TrieNode {\npublic:\n TrieNode *children[26];\n bool is_terminal;\n \n TrieNode() {\n for(int i=0;i<26;++i) {\n children[i | sahilgoyals | NORMAL | 2020-07-09T09:51:12.165039+00:00 | 2020-07-09T09:51:12.165070+00:00 | 396 | false | ```\nclass TrieNode {\npublic:\n TrieNode *children[26];\n bool is_terminal;\n \n TrieNode() {\n for(int i=0;i<26;++i) {\n children[i]=NULL;\n }\n is_terminal=false;\n }\n};\n\nclass MagicDictionary {\npublic:\n /** Initialize your data structure here. */\n TrieNode ... | 4 | 0 | ['Backtracking', 'Trie', 'C++'] | 2 |
implement-magic-dictionary | Java Trie | java-trie-by-hobiter-uudf | \nclass MagicDictionary {\n public class Node{\n boolean isW;\n Node[] nodes;\n public Node() {\n nodes = new Node[26];\n | hobiter | NORMAL | 2020-05-24T02:24:38.240643+00:00 | 2020-05-24T02:24:38.240676+00:00 | 264 | false | ```\nclass MagicDictionary {\n public class Node{\n boolean isW;\n Node[] nodes;\n public Node() {\n nodes = new Node[26];\n isW = false;\n }\n }\n \n Node root;\n\n /** Initialize your data structure here. */\n public MagicDictionary() {\n root... | 4 | 0 | [] | 1 |
implement-magic-dictionary | Simple Iterative Java Solution | simple-iterative-java-solution-by-soviet-lr7u | In order for a string to exist in a magic dictionary one character in the string must be modified. This means we can compare the incoming string against strings | sovietaced | NORMAL | 2017-09-10T06:28:43.158000+00:00 | 2017-09-10T06:28:43.158000+00:00 | 982 | false | In order for a string to exist in a magic dictionary one character in the string must be modified. This means we can compare the incoming string against strings in the dictionary and count how many characters do not match between the two. If only one character does not match the word exists in the magic dictionary. \n\... | 4 | 1 | [] | 2 |
implement-magic-dictionary | Trie DFS Solution C++ || Easy Solution | trie-dfs-solution-c-easy-solution-by-jar-fm8g | Code\ncpp []\n// TrieNode class\nclass TrieNode{\npublic:\n TrieNode* child[26];\n bool endWord;\n \n TrieNode(){\n endWord = false;\n\n | JArrazola | NORMAL | 2024-10-05T16:46:26.943688+00:00 | 2024-10-05T16:48:50.467273+00:00 | 224 | false | # Code\n```cpp []\n// TrieNode class\nclass TrieNode{\npublic:\n TrieNode* child[26];\n bool endWord;\n \n TrieNode(){\n endWord = false;\n\n for (size_t i = 0; i < 26; i++)\n child[i] = nullptr;\n }\n};\n\nclass MagicDictionary {\nprivate: \n TrieNode *root; // Trie root\n... | 3 | 0 | ['String', 'Depth-First Search', 'Trie', 'C++'] | 1 |
implement-magic-dictionary | [Python 3] - Trie - Simple Solution | python-3-trie-simple-solution-by-dolong2-d640 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | dolong2110 | NORMAL | 2023-07-20T16:26:39.734164+00:00 | 2024-03-12T02:14:18.525027+00:00 | 729 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(N*K)$$ with N is the number of word and K is length of each word\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Spa... | 3 | 0 | ['String', 'Trie', 'Python3'] | 1 |
implement-magic-dictionary | 676: Solution with step by step explanation | 676-solution-with-step-by-step-explanati-fx6f | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. init method\n 1. Create an empty dictionary to hold the words and t | Marlen09 | NORMAL | 2023-03-20T06:33:03.511296+00:00 | 2023-03-20T06:33:03.511331+00:00 | 726 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. __init__ method\n 1. Create an empty dictionary to hold the words and their replacements.\n 2. Store the dictionary in the class attribute self.dict.\n2. buildDict method\n 1. For each word in the dictionary argu... | 3 | 0 | ['Hash Table', 'String', 'Trie', 'Python', 'Python3'] | 0 |
implement-magic-dictionary | My fast Python solution | my-fast-python-solution-by-obose-d729 | Intuition\n Describe your first thoughts on how to solve this problem. \nMy first thoughts on how to solve this problem is to create a data structure that store | Obose | NORMAL | 2023-01-15T00:34:06.553925+00:00 | 2023-01-15T00:34:06.553967+00:00 | 469 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nMy first thoughts on how to solve this problem is to create a data structure that stores all the words in the dictionary. Then, when searching for a word, I will iterate through each word in the data structure and compare it to the search... | 3 | 0 | ['Python3'] | 0 |
implement-magic-dictionary | [Java] | Trie | | java-trie-by-nepo_ian-szqr | \nclass MagicDictionary {\n\n private class Node {\n Node[] links;\n boolean isEnd;\n Node() {\n links = new Node[26];\n | nepo_ian | NORMAL | 2022-10-12T10:51:12.376371+00:00 | 2022-10-12T10:51:12.376415+00:00 | 409 | false | ```\nclass MagicDictionary {\n\n private class Node {\n Node[] links;\n boolean isEnd;\n Node() {\n links = new Node[26];\n isEnd = false;\n }\n }\n \n private Node root;\n public MagicDictionary() {\n root = new Node();\n }\n \n public vo... | 3 | 0 | ['Trie', 'Java'] | 0 |
implement-magic-dictionary | Easiest and understandable code | easiest-and-understandable-code-by-asppa-hyho | just change each word and check in trie\nTIME COMPLEXITY(n26logm)\n\n\nclass MagicDictionary {\n\n class Trie\n {\n boolean end;\n Trie node | asppanda | NORMAL | 2022-02-01T06:31:09.330423+00:00 | 2022-02-01T06:31:45.495128+00:00 | 400 | false | just change each word and check in trie\nTIME COMPLEXITY(n*26*logm)\n\n```\nclass MagicDictionary {\n\n class Trie\n {\n boolean end;\n Trie node[]=new Trie[26];\n Trie()\n {\n end=false;\n for(int i=0;i<26;i++)\n {\n node[i]=null;\n ... | 3 | 0 | ['Depth-First Search', 'Trie', 'Java'] | 1 |
implement-magic-dictionary | [Python3] Solution with using trie | python3-solution-with-using-trie-by-maos-onuh | \nclass MagicDictionary:\n def __init__(self):\n self.trie = {}\n self.end = \'end\'\n\n def buildDict(self, dictionary: List[str]) -> None: | maosipov11 | NORMAL | 2021-12-08T05:55:36.998771+00:00 | 2021-12-08T05:55:36.998797+00:00 | 251 | false | ```\nclass MagicDictionary:\n def __init__(self):\n self.trie = {}\n self.end = \'end\'\n\n def buildDict(self, dictionary: List[str]) -> None:\n for word in dictionary:\n node = self.trie\n for c in word:\n if c not in node:\n node[c] =... | 3 | 0 | ['Array', 'Trie', 'Python3'] | 1 |
implement-magic-dictionary | C++ Easy to Understand Trie Solution | c-easy-to-understand-trie-solution-by-us-eh7c | \tstruct node {\n\t\t\tint count;\n\t\t\tint ew;\n\t\t\tnode ar[26];\n\t\t};\n\t\tnode getNode() {\n\t\t\tnode n=new node();\n\t\t\tn->ew=0; \n\t\t\tfor(int i=0 | user3408lm | NORMAL | 2021-07-29T13:03:25.855033+00:00 | 2021-07-29T13:03:25.855073+00:00 | 122 | false | \tstruct node {\n\t\t\tint count;\n\t\t\tint ew;\n\t\t\tnode* ar[26];\n\t\t};\n\t\tnode* getNode() {\n\t\t\tnode* n=new node();\n\t\t\tn->ew=0; \n\t\t\tfor(int i=0;i<26;i++) n->ar[i]=NULL;\n\t\t\tn->count=0;\n\t\t\treturn n;\n\t\t}\n\t\tvoid insert(node* root,string s) {\n\t\t\tnode* temp=root;\n\t\t\tfor(int i=0;i<s.s... | 3 | 0 | [] | 0 |
implement-magic-dictionary | Java || Trie || buildDict - O(dictionary.length*dictionary[i].length) || search - O(26*word.length) | java-trie-builddict-odictionarylengthdic-4x4d | public class Trie {\n\n\tprivate class Node {\n\n\t\tchar data;\n\t\tNode[] children;\n\t\tboolean isTerminal;\n\n\t\tpublic Node(char data, boolean isTerminal) | LegendaryCoder | NORMAL | 2021-02-04T08:48:58.835283+00:00 | 2021-02-04T08:48:58.835309+00:00 | 150 | false | public class Trie {\n\n\tprivate class Node {\n\n\t\tchar data;\n\t\tNode[] children;\n\t\tboolean isTerminal;\n\n\t\tpublic Node(char data, boolean isTerminal) {\n\t\t\tthis.data = data;\n\t\t\tthis.isTerminal = isTerminal;\n\t\t\tthis.children = new Node[26];\n\t\t}\n\n\t}\n\n\tprivate Node root;\n\n\t// O(1)\n\tpubl... | 3 | 0 | [] | 1 |
implement-magic-dictionary | Python Trie Solution | python-trie-solution-by-realslimshady-i8cf | \nclass TrieNode:\n def __init__(self):\n self.children = collections.defaultdict(TrieNode)\n self.isWord = False\n\nclass Trie:\n def __ini | realslimshady | NORMAL | 2020-08-01T11:53:25.448640+00:00 | 2020-08-01T11:53:25.448688+00:00 | 121 | false | ```\nclass TrieNode:\n def __init__(self):\n self.children = collections.defaultdict(TrieNode)\n self.isWord = False\n\nclass Trie:\n def __init__(self):\n self.root = TrieNode()\n \n def insert(self,word):\n node = self.root\n for w in word:\n node = node.c... | 3 | 0 | [] | 0 |
implement-magic-dictionary | Python, modified search in Trie, O(AL) search | python-modified-search-in-trie-oal-searc-iohl | python\nclass TrieNode(object):\n def __init__(self):\n self.nxts = dict()\n self.isword = False\n \n def insert(self, word):\n | hungwei-andy | NORMAL | 2019-08-12T08:00:43.649031+00:00 | 2019-08-12T08:00:43.649061+00:00 | 398 | false | ```python\nclass TrieNode(object):\n def __init__(self):\n self.nxts = dict()\n self.isword = False\n \n def insert(self, word):\n cur = self\n for c in word:\n if c not in cur.nxts:\n cur.nxts[c] = TrieNode()\n cur = cur.nxts[c]\n cur... | 3 | 0 | [] | 0 |
implement-magic-dictionary | EASY TO UNDERSTAND || C++ | easy-to-understand-c-by-krish_kakadiya-gmb9 | C++\n\n# Code\n\nclass MagicDictionary {\npublic:\n vector<string> a;\n MagicDictionary() {\n \n }\n \n void buildDict(vector<string> dict | krish_kakadiya | NORMAL | 2023-09-08T16:20:55.778795+00:00 | 2023-09-08T16:20:55.778815+00:00 | 175 | false | C++\n\n# Code\n```\nclass MagicDictionary {\npublic:\n vector<string> a;\n MagicDictionary() {\n \n }\n \n void buildDict(vector<string> dictionary) {\n for(int i=0;i<dictionary.size();i++){\n a.push_back(dictionary[i]);\n }\n }\n \n bool search(string searchWord)... | 2 | 0 | ['C++'] | 0 |
implement-magic-dictionary | Solution | solution-by-deleted_user-du0s | C++ []\nclass MagicDictionary {\n public:\n void buildDict(vector<string> dictionary) {\n for (const string& word : dictionary)\n for (int i = 0; i < w | deleted_user | NORMAL | 2023-04-18T12:41:04.514127+00:00 | 2023-04-18T12:47:45.627641+00:00 | 682 | false | ```C++ []\nclass MagicDictionary {\n public:\n void buildDict(vector<string> dictionary) {\n for (const string& word : dictionary)\n for (int i = 0; i < word.length(); ++i) {\n const string replaced = getReplaced(word, i);\n dict[replaced] = dict.count(replaced) ? \'*\' : word[i];\n }\n }\n... | 2 | 0 | ['C++', 'Java', 'Python3'] | 0 |
implement-magic-dictionary | Easy c++ code using Trie Data Structure | easy-c-code-using-trie-data-structure-by-fn7v | \n\n# Code\n\nclass Trie{\n public:\n Trie* child[26];\n bool isEnd;\n Trie(){\n for(int i=0;i<26;i++){\n child[i]=NULL;\n | Shristha | NORMAL | 2023-01-13T16:58:22.559360+00:00 | 2023-01-13T16:58:22.559403+00:00 | 530 | false | \n\n# Code\n```\nclass Trie{\n public:\n Trie* child[26];\n bool isEnd;\n Trie(){\n for(int i=0;i<26;i++){\n child[i]=NULL;\n }\n isEnd=false; \n }\n};\nclass MagicDictionary {\npublic:\n Trie* root=new Trie();\n MagicDictionary() {\n \n } \n \n ... | 2 | 0 | ['C++'] | 1 |
implement-magic-dictionary | EASY TRIE SOLUTION C++ | easy-trie-solution-c-by-amangupta12207-r75e | Intuition\ninsert all the string in the trie \nIn search operation traverse the whole seachString, \nif trie does not contains that char then skip the curr char | Amangupta12207 | NORMAL | 2022-12-17T14:26:56.958133+00:00 | 2022-12-17T14:26:56.958160+00:00 | 1,042 | false | # Intuition\ninsert all the string in the trie \nIn search operation traverse the whole seachString, \nif trie does not contains that char then skip the curr character and check if u can find the remaining substring (implemented in helper function) return true else return false and if trie does contain that char then w... | 2 | 0 | ['Trie', 'C++'] | 0 |
implement-magic-dictionary | [C++] Trie || DFS || O(N) || Easy to Read | c-trie-dfs-on-easy-to-read-by-atman_2407-d64c | Explanation:\n\nHow to Implement Trie ?\n\nCheck this problem: 208 Implement Trie (Prefix Tree)Link\n\n\nContinue....\n\nPattern = Variation of Trie.\n\n- Inser | atman_2407 | NORMAL | 2022-03-04T05:44:27.290335+00:00 | 2022-03-04T05:44:27.290369+00:00 | 264 | false | **Explanation:**\n\nHow to Implement Trie ?\n\nCheck this problem: 208 Implement Trie (Prefix Tree)[Link](https://leetcode.com/problems/implement-trie-prefix-tree/)\n\n\nContinue....\n\nPattern = Variation of Trie.\n\n- Insert is same.\n\n- In search, we must perform one mismatch. Same word can not return true.\n\n ... | 2 | 0 | ['Depth-First Search', 'Trie', 'C'] | 0 |
implement-magic-dictionary | [C++] Trie Solution | c-trie-solution-by-logan-47-43ol | \nclass TrieNode {\npublic:\n TrieNode* children[26];\n bool isEnd;\n \n TrieNode() {\n for(int i=0; i < 26; i++) {\n children[i] | logan-47 | NORMAL | 2022-01-29T11:24:07.504756+00:00 | 2022-01-29T11:24:07.504781+00:00 | 305 | false | ```\nclass TrieNode {\npublic:\n TrieNode* children[26];\n bool isEnd;\n \n TrieNode() {\n for(int i=0; i < 26; i++) {\n children[i] = NULL;\n }\n isEnd = false;\n }\n};\n\n\nclass MagicDictionary {\npublic:\n \n TrieNode* head;\n MagicDictionary() {\n head... | 2 | 0 | ['Trie', 'C++'] | 2 |
implement-magic-dictionary | You guys know that the naive solution beats 100% here? | you-guys-know-that-the-naive-solution-be-6us3 | Just wanted to leave it here that for this question the absolute best solution (for the test cases presented) is a naive linear search:\n\n1. Save the dictionar | aalmos | NORMAL | 2021-11-29T08:56:09.390534+00:00 | 2021-11-29T08:56:09.390563+00:00 | 108 | false | Just wanted to leave it here that for this question the absolute best solution (for the test cases presented) is a naive linear search:\n\n1. Save the dictionary as is.\n2. Upon `Search` iterate over the dictionary and find any string that differs with a single character.\n\nThis is faster than any dictionary or trie a... | 2 | 0 | [] | 0 |
implement-magic-dictionary | [Python] Simple Trie + DFS Recursive Solution with Explanation | python-simple-trie-dfs-recursive-solutio-a0gg | For the Trie implementation I am not creating another TrieNode class instead using a dictionary as node and for wordEnd I am using "~" == True, if "~" is presen | Saksham003 | NORMAL | 2021-11-14T09:03:54.520376+00:00 | 2021-11-14T09:16:55.302877+00:00 | 237 | false | For the Trie implementation I am not creating another TrieNode class instead using a dictionary as node and for wordEnd I am using "~" == True, if "~" is present in dictionary then the word ends here, since dictionary only contains lower case english letters so it will not be a problem\n```\nclass MagicDictionary:\n ... | 2 | 0 | ['Depth-First Search', 'Trie', 'Python', 'Python3'] | 0 |
implement-magic-dictionary | Python implementation using dictionary | python-implementation-using-dictionary-b-dn7r | Simple Python implementation using dictionary\n ```\n class MagicDictionary:\n def init(self):\n self.dict = {}\n self.word = set()\n def bu | maflint7 | NORMAL | 2021-08-31T19:47:51.593063+00:00 | 2021-08-31T19:50:20.641223+00:00 | 292 | false | Simple Python implementation using dictionary\n ```\n class MagicDictionary:\n def __init__(self):\n self.dict = {}\n self.word = set()\n def buildDict(self, dictionary: List[str]) -> None:\n # SC: O(n) # TC: O(nm)\n for word in dictionary:\n for i in range(len(word)):\n\t\t... | 2 | 0 | ['Python', 'Python3'] | 0 |
implement-magic-dictionary | Java | Easy Solution | Set | Without Trie | java-easy-solution-set-without-trie-by-u-g8pb | \nclass MagicDictionary {\n\n /** Initialize your data structure here. */\n Set<String> set;\n public MagicDictionary() {\n set=new HashSet<>(); | user8540kj | NORMAL | 2021-05-04T12:59:51.582371+00:00 | 2021-05-04T12:59:51.582417+00:00 | 120 | false | ```\nclass MagicDictionary {\n\n /** Initialize your data structure here. */\n Set<String> set;\n public MagicDictionary() {\n set=new HashSet<>();\n }\n \n public void buildDict(String[] dictionary) {\n for(String s:dictionary) set.add(s);\n }\n \n public boolean search(String ... | 2 | 0 | [] | 0 |
implement-magic-dictionary | Java || HashMap || buildDict - O(dictionary.length*dictionary[i].length) || search - O(word.length) | java-hashmap-builddict-odictionarylength-teca | class MagicDictionary {\n\n HashMap map;\n\n\t// O(1)\n\tpublic MagicDictionary() {\n\t\tmap = new HashMap();\n\t}\n\n\t\t// O(dictionary.length * dictionary | LegendaryCoder | NORMAL | 2021-02-04T09:12:00.560850+00:00 | 2021-02-04T09:12:00.560898+00:00 | 100 | false | class MagicDictionary {\n\n HashMap<String, Integer> map;\n\n\t// O(1)\n\tpublic MagicDictionary() {\n\t\tmap = new HashMap<String, Integer>();\n\t}\n\n\t\t// O(dictionary.length * dictionary[i].length)\n\t\tpublic void buildDict(String[] dictionary) {\n\t\t\tfor (String word : dictionary) {\n\t\t\t\tchar[] str = wo... | 2 | 0 | [] | 0 |
implement-magic-dictionary | JAVA || Trie || Clean code | java-trie-clean-code-by-igloo11-774c | \nclass MagicDictionary {\n class TrieNode {\n Map<Character, TrieNode> children;\n boolean endOfWord;\n public TrieNode() {\n | Igloo11 | NORMAL | 2021-01-12T04:29:12.596682+00:00 | 2021-01-12T04:29:12.596726+00:00 | 281 | false | ```\nclass MagicDictionary {\n class TrieNode {\n Map<Character, TrieNode> children;\n boolean endOfWord;\n public TrieNode() {\n children = new HashMap<>();\n endOfWord = false;\n }\n }\n \n //Insert by iteration\n public void insert(TrieNode root, Strin... | 2 | 0 | ['Trie', 'Java'] | 0 |
implement-magic-dictionary | Javascript Trie+DFS approach | javascript-triedfs-approach-by-yestrup-o2gi | \nfunction TrieNode(){\n this.children = new Map()\n this.endOfWord = false;\n}\n\nvar MagicDictionary = function() {\n this.root = new TrieNode()\n};\ | ratiboi | NORMAL | 2020-10-13T19:06:32.914851+00:00 | 2020-10-13T19:06:32.914895+00:00 | 160 | false | ```\nfunction TrieNode(){\n this.children = new Map()\n this.endOfWord = false;\n}\n\nvar MagicDictionary = function() {\n this.root = new TrieNode()\n};\n\nMagicDictionary.prototype.buildDict = function(dictionary) {\n for(let word of dictionary){\n let curr = this.root\n for(let letter of wo... | 2 | 0 | ['Depth-First Search', 'JavaScript'] | 0 |
implement-magic-dictionary | Java - beats 100% - No HashMap/Trie - Easy to understand | java-beats-100-no-hashmaptrie-easy-to-un-givq | This solution doesn\'t use HashMap or Trie. \nbuildDict(String[] dict) complexity is O(1). \nsearch(String word) complexity is O(NK) where N is the total number | chere | NORMAL | 2020-07-02T03:45:23.909851+00:00 | 2020-07-02T03:45:23.909878+00:00 | 95 | false | This solution doesn\'t use HashMap or Trie. \n```buildDict(String[] dict)``` complexity is O(1). \n```search(String word)``` complexity is O(NK) where N is the total number of words in dictionary and K is the length of word to be searched.\n\n\n```\nclass MagicDictionary {\n\n String[] dictionary;\n /** Initializ... | 2 | 0 | [] | 0 |
implement-magic-dictionary | Python3 trie solution - Implement Magic Dictionary | python3-trie-solution-implement-magic-di-y2im | \nfrom collections import defaultdict\n\nclass Node:\n def __init__(self):\n self.child = defaultdict(Node)\n self.isEnd = False\n\nclass Magic | r0bertz | NORMAL | 2020-05-12T04:39:39.121163+00:00 | 2020-05-13T08:31:30.536050+00:00 | 276 | false | ```\nfrom collections import defaultdict\n\nclass Node:\n def __init__(self):\n self.child = defaultdict(Node)\n self.isEnd = False\n\nclass MagicDictionary:\n\n def __init__(self):\n self.trie = Node()\n \n def buildDict(self, words: List[str]) -> None:\n for w in words:\n ... | 2 | 0 | ['Trie', 'Python', 'Python3'] | 0 |
implement-magic-dictionary | Python, DFS solution, beats 96% Time and 100% Space | python-dfs-solution-beats-96-time-and-10-qifx | The intuition behind this approach is to create a tree with nodes that hold characters. A path from the root node to a leaf node defines a string in the diction | andrewqho | NORMAL | 2019-08-12T05:40:30.792179+00:00 | 2019-08-12T05:43:15.117687+00:00 | 283 | false | The intuition behind this approach is to create a tree with nodes that hold characters. A path from the root node to a leaf node defines a string in the dictionary. \n\nWe define a structure called a letterNode that holds the current node\'s character value, the possible characters it can lead to, and whether it is a l... | 2 | 0 | [] | 2 |
implement-magic-dictionary | Python - really easy to understand (Explanations) | python-really-easy-to-understand-explana-ywnh | Instead of using a Trie, we can make a dictionary with the lengths of the words as the key and the list of all the words with the length as values, something li | crazyphotonzz | NORMAL | 2018-07-18T17:10:45.290275+00:00 | 2018-07-18T17:10:45.290275+00:00 | 207 | false | Instead of using a Trie, we can make a dictionary with the lengths of the words as the key and the list of all the words with the length as values, something like this:\n```defaultdict(<type \'list\'>, {8: [u\'leetcode\'], 5: [u\'hello\', u\'hallo\']})```\n\nNow, we simply have to go through each word in the dictionary... | 2 | 0 | [] | 0 |
implement-magic-dictionary | C++ Trie Solution avoid memory leak | c-trie-solution-avoid-memory-leak-by-qin-jvkb | Many have posted C++/Java Trie solution, for Java, we do not need to consider possible memory leak, but for C++, a lot of people use pointers without releasing | qinzhou | NORMAL | 2018-01-12T22:15:46.265000+00:00 | 2018-01-12T22:15:46.265000+00:00 | 352 | false | Many have posted C++/Java Trie solution, for Java, we do not need to consider possible memory leak, but for C++, a lot of people use pointers without releasing them (or not completely release them after terminates), here I post my Trie solution with *shared_ptr*, a smart pointer which guarantees completely memory recyc... | 2 | 0 | [] | 0 |
implement-magic-dictionary | Java implementation using Trie | java-implementation-using-trie-by-marvin-sjul | The basic idea is:\n\nFirst, insert the dictionary list into the Trie;\n\nSecond, for each given word, scan character by character. For each character, try modi | marvinbai | NORMAL | 2017-09-20T19:01:55.406000+00:00 | 2017-09-20T19:01:55.406000+00:00 | 709 | false | The basic idea is:\n\nFirst, insert the dictionary list into the Trie;\n\nSecond, for each given word, scan character by character. For each character, try modify it and then call the search method in Trie. If trie.search() returns true then return true. If not then see if the Trie has exactly this character, if yes th... | 2 | 0 | [] | 1 |
implement-magic-dictionary | Simple JavaScript Solution | simple-javascript-solution-by-dnshi-63gz | js\n/**\n * Initialize your data structure here.\n */\nvar MagicDictionary = function() {\n this.table = {}\n};\n\n/**\n * Build a dictionary through a list | dnshi | NORMAL | 2017-09-25T00:04:36.977000+00:00 | 2017-09-25T00:04:36.977000+00:00 | 306 | false | ```js\n/**\n * Initialize your data structure here.\n */\nvar MagicDictionary = function() {\n this.table = {}\n};\n\n/**\n * Build a dictionary through a list of words \n * @param {string[]} dict\n * @return {void}\n */\nMagicDictionary.prototype.buildDict = function(dict) {\n for (let word of dict) {\n t... | 2 | 0 | [] | 0 |
implement-magic-dictionary | ✅ Trie + DFS | ✅ Easy Code | trie-dfs-easy-code-by-sajal0701-4v6x | IntuitionThe problem requires us to implement a dictionary where we can check if modifying exactly one character in a search word can match any word in the dict | Sajal0701 | NORMAL | 2025-02-26T14:17:53.012629+00:00 | 2025-02-26T14:17:53.012629+00:00 | 121 | false | # Intuition
The problem requires us to implement a dictionary where we can check if modifying exactly one character in a search word can match any word in the dictionary. A **Trie** (Prefix Tree) is an ideal data structure for this because it allows us to efficiently store words and perform character-based searches.
#... | 1 | 0 | ['Depth-First Search', 'Trie', 'C++'] | 0 |
implement-magic-dictionary | ✅Easy and Simple Solution With Explanation ✅ Clean Code ✅ Best Solution | easy-and-simple-solution-with-explanatio-fzed | \nGuy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F\nFo | ayushluthra62 | NORMAL | 2024-08-11T18:29:08.445758+00:00 | 2024-08-11T18:29:08.445792+00:00 | 198 | false | \n***Guy\'s if you find this solution helpful \uD83D\uDE0A, PLEASE do UPVOTE. By doing that it motivate\'s me to create more better post like this \u270D\uFE0F***<br>\n**Follow me on LinkeDin [[click Here](https://www.linkedin.com/in/ayushluthra62/)]**\n\n# Complexity\n- Time complexity:\nbuildDict Function : O(N)\n Se... | 1 | 0 | ['Hash Table', 'String', 'Depth-First Search', 'Design', 'Trie', 'C++'] | 0 |
implement-magic-dictionary | Beats 98%, easy to understand... simple approach | beats-98-easy-to-understand-simple-appro-zfw0 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | manyaagargg | NORMAL | 2024-07-26T02:03:51.607121+00:00 | 2024-07-26T02:03:51.607145+00:00 | 116 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Array', 'Hash Table', 'String', 'Depth-First Search', 'Design', 'Trie', 'C++'] | 0 |
implement-magic-dictionary | Java 8 solution | using Map | Easy to understand | java-8-solution-using-map-easy-to-unders-i5rb | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | bhargava2123 | NORMAL | 2024-07-06T10:07:24.485264+00:00 | 2024-07-06T10:07:24.485297+00:00 | 32 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
implement-magic-dictionary | Just a dictionary | just-a-dictionary-by-pawan481992-374i | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\nM is nu | pawan481992 | NORMAL | 2023-12-29T14:01:15.982187+00:00 | 2023-12-29T14:01:15.982216+00:00 | 353 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\nM is number of strings and N is the length of String\nBuild time\n- Time complexity:\n $$O(m*n)$$\n\n\n- search time\n$$O(n*n)$$\n\n- Space complexi... | 1 | 0 | ['Java'] | 0 |
implement-magic-dictionary | C# Solution using Trie | c-solution-using-trie-by-mohamedabdelety-hof2 | \npublic class Node{\n public bool endWord;\n public Dictionary<char,Node> childs;\n public Node(bool _endWord = false){\n childs = new Dictiona | mohamedAbdelety | NORMAL | 2023-11-05T08:07:16.585774+00:00 | 2023-11-05T08:07:16.585796+00:00 | 30 | false | ```\npublic class Node{\n public bool endWord;\n public Dictionary<char,Node> childs;\n public Node(bool _endWord = false){\n childs = new Dictionary<char,Node>();\n endWord = _endWord;\n }\n}\n\npublic class MagicDictionary {\n Node head;\n\n public MagicDictionary() {\n head = n... | 1 | 0 | ['Depth-First Search', 'Trie', 'C#'] | 0 |
implement-magic-dictionary | c++ using hashmap; | c-using-hashmap-by-sanket33245-ywdk | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | sanket33245 | NORMAL | 2023-08-25T21:42:33.108697+00:00 | 2023-08-25T21:42:33.108725+00:00 | 117 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Hash Table', 'C++'] | 0 |
implement-magic-dictionary | [C++] Intuitive & Simple Beginner Friendly TRIE Solution | c-intuitive-simple-beginner-friendly-tri-3myx | Intuitive Trie Solution, the insert operation is same as the basic trie insert operation. For searching just replace each character of searchWord one-by-one and | shrut_23 | NORMAL | 2023-08-23T10:22:22.030077+00:00 | 2023-08-23T16:13:53.505949+00:00 | 19 | false | Intuitive Trie Solution, the insert operation is same as the basic trie insert operation. For searching just replace each character of `searchWord` one-by-one and check whether we are not relacing the original character (since we need to change 1 character).\n\n# Time Complexity\n- Building the Trie: `O(n * m)`, where ... | 1 | 0 | ['String', 'Design', 'Trie', 'C++'] | 0 |
implement-magic-dictionary | TRIE + DEPTH FIRST SEARCH || BEATS 50% || JAVA || EASY CODE | trie-depth-first-search-beats-50-java-ea-mb1t | Code\n\nclass TrieNode {\n Map<Character, TrieNode> children;\n boolean isEnd;\n TrieNode() {\n children = new HashMap<>();\n isEnd = fal | youssef1998 | NORMAL | 2023-06-25T09:23:14.108917+00:00 | 2023-06-25T09:23:14.108945+00:00 | 191 | false | # Code\n```\nclass TrieNode {\n Map<Character, TrieNode> children;\n boolean isEnd;\n TrieNode() {\n children = new HashMap<>();\n isEnd = false;\n }\n}\n\nclass Trie {\n private final TrieNode root = new TrieNode();\n Trie(){}\n \n public void insert(String str) {\n TrieNod... | 1 | 0 | ['String', 'Depth-First Search', 'Design', 'Trie', 'Java'] | 0 |
implement-magic-dictionary | ✨ JavaScript - Clean Solution Using Trie | javascript-clean-solution-using-trie-by-rnnwr | \nvar MagicDictionary = function() {\n this.trie = {\n children: {}, \n }\n};\n\n/** \n * @param {string[]} dictionary\n * @return {void}\n */\nMagicDictio | ngseke | NORMAL | 2023-06-23T17:21:34.019853+00:00 | 2023-06-23T17:30:21.610616+00:00 | 51 | false | ```\nvar MagicDictionary = function() {\n this.trie = {\n children: {}, \n }\n};\n\n/** \n * @param {string[]} dictionary\n * @return {void}\n */\nMagicDictionary.prototype.buildDict = function(dictionary) {\n dictionary.forEach(word => {\n let current = this.trie\n\n ;[...word].forEach((char, index, { leng... | 1 | 0 | ['Tree', 'Depth-First Search', 'Trie', 'Recursion', 'JavaScript'] | 0 |
implement-magic-dictionary | [Java] Simple Trie solution | java-simple-trie-solution-by-ytchouar-j1ad | java\nclass MagicDictionary {\n private final TrieNode root;\n\n public MagicDictionary() {\n this.root = new TrieNode();\n }\n \n public | YTchouar | NORMAL | 2023-05-26T05:54:10.669398+00:00 | 2023-05-28T06:37:21.885952+00:00 | 1,398 | false | ```java\nclass MagicDictionary {\n private final TrieNode root;\n\n public MagicDictionary() {\n this.root = new TrieNode();\n }\n \n public void buildDict(final String[] dictionary) {\n for(String word : dictionary)\n this.buildDictHelper(this.root, word);\n }\n\n public v... | 1 | 0 | ['Java'] | 0 |
implement-magic-dictionary | C++ | Map and Bitmask | c-map-and-bitmask-by-pikachuu-p3is | Code\n\nclass MagicDictionary {\n map<pair<string, string>, int> Map;\npublic:\n MagicDictionary() {\n \n }\n \n void buildDict(vector<str | pikachuu | NORMAL | 2022-12-28T13:50:13.607129+00:00 | 2022-12-28T13:50:13.607172+00:00 | 27 | false | # Code\n```\nclass MagicDictionary {\n map<pair<string, string>, int> Map;\npublic:\n MagicDictionary() {\n \n }\n \n void buildDict(vector<string> dictionary) {\n for(string word: dictionary) {\n for(int i = 0; i < word.length(); i++) {\n Map[{word.substr(0, i), w... | 1 | 0 | ['Hash Table', 'String', 'Bit Manipulation', 'Design', 'C++'] | 0 |
implement-magic-dictionary | Trie simple approach | trie-simple-approach-by-sajnani-9djo | Intuition\nTrie approach\n\n# Approach\n1. Trie\n2. Recurse to find if there is a difference of one character in the most efficient way\n\n\n\n# Code\n\nclass M | sajnani | NORMAL | 2022-12-22T02:38:18.400174+00:00 | 2022-12-22T02:38:18.400214+00:00 | 693 | false | # Intuition\nTrie approach\n\n# Approach\n1. Trie\n2. Recurse to find if there is a difference of one character in the most efficient way\n\n\n\n# Code\n```\nclass MagicDictionary {\n\n Trie trie=new Trie();\n static class Trie{\n Map<Character,Trie> trieMap=new HashMap<Character,Trie>(); \n boolean eo... | 1 | 0 | ['Java'] | 0 |
implement-magic-dictionary | C++ Easy hashing || In search_word function replace each char with all of 26 chars || Explained | c-easy-hashing-in-search_word-function-r-z369 | Complexity\n- Time complexity:\nO(n * m)-> for build_dictionary function,n is the number of words to be inserted into the data structure and m is the size of th | dee_stroyer | NORMAL | 2022-11-26T12:40:34.506669+00:00 | 2022-11-26T12:40:34.506702+00:00 | 31 | false | # Complexity\n- Time complexity:\nO(n * m)-> for **build_dictionary** function,n is the number of words to be inserted into the data structure and m is the size of that word. But as it is given that the maximum size of any word can be 100 so m can be neglected and it can be considered as O(n * 100) ~= O(N).\n \nO(n)-> ... | 1 | 0 | ['Hash Table', 'String', 'C++'] | 0 |
implement-magic-dictionary | [C++] Try all combinations || very easy to understand | c-try-all-combinations-very-easy-to-unde-wg2m | \nclass MagicDictionary {\npublic:\n unordered_set<string>st;\n MagicDictionary() {\n st.clear();\n }\n \n void buildDict(vector<string> a | NehaGupta_09 | NORMAL | 2022-10-02T16:56:23.907802+00:00 | 2022-10-02T16:56:23.907843+00:00 | 850 | false | ```\nclass MagicDictionary {\npublic:\n unordered_set<string>st;\n MagicDictionary() {\n st.clear();\n }\n \n void buildDict(vector<string> arr) {\n for(int i=0;i<arr.size();i++)\n {\n st.insert(arr[i]);\n }\n }\n \n bool search(string s) {\n for(int... | 1 | 0 | ['C', 'C++'] | 0 |
implement-magic-dictionary | [C++] Try all combinations || very easy to understand | c-try-all-combinations-very-easy-to-unde-m9zy | \nclass MagicDictionary {\npublic:\n unordered_set<string>st;\n MagicDictionary() \n\t{\n st.clear(); \n }\n \n void buildDict(vector<str | akshat0610 | NORMAL | 2022-10-02T16:54:40.019958+00:00 | 2022-10-02T16:54:40.019999+00:00 | 402 | false | ```\nclass MagicDictionary {\npublic:\n unordered_set<string>st;\n MagicDictionary() \n\t{\n st.clear(); \n }\n \n void buildDict(vector<string> dict) \n\t{\n for(int i=0;i<dict.size();i++)\n {\n \t st.insert(dict[i]);\n\t }\n }\n \n bool search(string word) \n\t{\n ... | 1 | 0 | ['C', 'C++'] | 0 |
implement-magic-dictionary | Java Solution | Using Trie | java-solution-using-trie-by-nitwmanish-0o77 | \nclass MagicDictionary {\n private static class Node {\n\t\tprivate char data;\n\t\tprivate boolean isEnd;\n\t\tprivate Node[] children;\n\n\t\tpublic Node( | nitwmanish | NORMAL | 2022-09-29T20:39:57.115374+00:00 | 2022-11-05T02:32:31.713329+00:00 | 89 | false | ```\nclass MagicDictionary {\n private static class Node {\n\t\tprivate char data;\n\t\tprivate boolean isEnd;\n\t\tprivate Node[] children;\n\n\t\tpublic Node(char data) {\n\t\t\tthis.data = data;\n\t\t\tthis.isEnd = false;\n\t\t\tthis.children = new Node[26];\n\t\t}\n\t}\n \n private Node root;\n \n pu... | 1 | 0 | ['Tree', 'Trie', 'Java'] | 0 |
implement-magic-dictionary | C++ | Trie | Easy Solution | c-trie-easy-solution-by-sankalp_sharma_2-fmbr | \nclass MagicDictionary {\npublic:\n \n class Trie{\n public:\n Trie *child[26];\n bool isEnd;\n Trie()\n {\n | Sankalp_Sharma_29 | NORMAL | 2022-09-26T18:31:59.453195+00:00 | 2022-09-26T18:31:59.453243+00:00 | 14 | false | ```\nclass MagicDictionary {\npublic:\n \n class Trie{\n public:\n Trie *child[26];\n bool isEnd;\n Trie()\n {\n isEnd=false;\n for(int i=0;i<26;i++)\n child[i]=NULL;\n }\n };\n Trie *root=new Trie();\n MagicDictionary() {\n ... | 1 | 0 | ['Trie', 'C'] | 0 |
implement-magic-dictionary | [C++] Trie + DFS | Try all combinations | c-trie-dfs-try-all-combinations-by-nilab-w8ls | ```\nclass MagicDictionary {\nprivate:\n struct Node{\n Node arr[26];\n bool end = false;\n };\n Node root;\npublic:\n MagicDictionary | nilabjodey | NORMAL | 2022-09-26T17:31:33.723371+00:00 | 2022-09-26T17:31:33.723414+00:00 | 458 | false | ```\nclass MagicDictionary {\nprivate:\n struct Node{\n Node* arr[26];\n bool end = false;\n };\n Node* root;\npublic:\n MagicDictionary() {\n root = new Node();\n }\n \n void trie_insert(string &word){\n Node* node = root;\n for(int i = 0; i < word.size(); i++){\... | 1 | 0 | ['Depth-First Search', 'Trie'] | 0 |
implement-magic-dictionary | Cpp Solution using trie | cpp-solution-using-trie-by-sanket_jadhav-yma7 | ```\nclass node{\n public:\n node v[26];\n int end=0;\n \n bool ispresent(char c){\n return this->v[c-\'a\']!=NULL;\n }\n \n void c | Sanket_Jadhav | NORMAL | 2022-09-23T06:04:30.236012+00:00 | 2022-09-23T06:04:30.236048+00:00 | 388 | false | ```\nclass node{\n public:\n node* v[26];\n int end=0;\n \n bool ispresent(char c){\n return this->v[c-\'a\']!=NULL;\n }\n \n void create(char c,node* n){\n this->v[c-\'a\']=n;\n }\n \n void setend(){\n this->end=1;\n }\n};\n\nclass trie{\n public:\n node* ro... | 1 | 0 | ['Trie', 'C++'] | 0 |
implement-magic-dictionary | c++ | using trie | low T.C. | c-using-trie-low-tc-by-venomhighs7-6rm7 | ```\nclass node{\npublic:\n node links[26];\n bool flag=false;\n};\n\nclass MagicDictionary {\n node root;\npublic:\n MagicDictionary() {\n r | venomhighs7 | NORMAL | 2022-09-18T10:36:34.207470+00:00 | 2022-09-18T10:36:34.207507+00:00 | 187 | false | ```\nclass node{\npublic:\n node* links[26];\n bool flag=false;\n};\n\nclass MagicDictionary {\n node* root;\npublic:\n MagicDictionary() {\n root=new node();\n }\n \n void addKey(string s) {\n node* temp=root;\n \n for(int i=0; i<s.length(); i++) {\n if(temp-... | 1 | 0 | [] | 0 |
implement-magic-dictionary | Python | DFS on a Trie | Notes | python-dfs-on-a-trie-notes-by-rktayal-i8hs | \n/*\nwe can approach this problem using trie. Problem is very similar to \nAdd and search word data structures.\nWe can create a standard trie and search for w | rktayal | NORMAL | 2022-08-18T01:12:21.834663+00:00 | 2022-08-18T01:12:21.834716+00:00 | 29 | false | ```\n/*\nwe can approach this problem using trie. Problem is very similar to \nAdd and search word data structures.\nWe can create a standard trie and search for words in that.\nif at all, we come across a character which is not present down the path,\nwe still continue to see if word is still searchable. If we are abl... | 1 | 0 | ['Depth-First Search', 'Trie', 'Python'] | 0 |
implement-magic-dictionary | EASY TO UNDERSTAND || C++ || RECURSION || TRIE | easy-to-understand-c-recursion-trie-by-a-7zm8 | struct Trie{\n Trie child[26];\n bool isEnd;\n};\n\nclass MagicDictionary {\npublic:\n \n Trie root;\n \n MagicDictionary() {\n \n | aniketm | NORMAL | 2022-08-02T17:41:20.796800+00:00 | 2022-08-02T17:41:20.796847+00:00 | 136 | false | struct Trie{\n Trie *child[26];\n bool isEnd;\n};\n\nclass MagicDictionary {\npublic:\n \n Trie *root;\n \n MagicDictionary() {\n \n root = new Trie();\n for(int i=0;i<26;i++)\n root->child[i] = NULL;\n root->isEnd = false;\n }\n \n void insert(string &s... | 1 | 0 | ['Depth-First Search', 'Trie', 'C'] | 0 |
implement-magic-dictionary | 72% TC and 63% SC easy python solution | 72-tc-and-63-sc-easy-python-solution-by-gkyho | \ndef __init__(self):\n\tself.root = dict()\n\tself.root[\'end\'] = 0\n\ndef buildDict(self, dictionary: List[str]) -> None:\n\tfor word in dictionary:\n\t\tc = | nitanshritulon | NORMAL | 2022-07-16T09:45:57.430928+00:00 | 2022-07-16T09:45:57.430974+00:00 | 50 | false | ```\ndef __init__(self):\n\tself.root = dict()\n\tself.root[\'end\'] = 0\n\ndef buildDict(self, dictionary: List[str]) -> None:\n\tfor word in dictionary:\n\t\tc = self.root\n\t\tfor char in word:\n\t\t\tif(char not in c):\n\t\t\t\tc[char] = dict()\n\t\t\t\tc[char][\'end\'] = 0\n\t\t\tc = c[char]\n\t\tc[\'end\'] = 1\n\... | 1 | 1 | ['Trie', 'Python', 'Python3'] | 0 |
implement-magic-dictionary | [Java] Trie | java-trie-by-vcamarggo-s3mw | ```\nstatic class TrieTree {\n\n TrieNode root = new TrieNode(\' \');\n\n public void insert(String s) {\n TrieNode current = root;\n | vcamarggo | NORMAL | 2022-07-10T13:48:46.280418+00:00 | 2022-07-10T13:48:46.280462+00:00 | 67 | false | ```\nstatic class TrieTree {\n\n TrieNode root = new TrieNode(\' \');\n\n public void insert(String s) {\n TrieNode current = root;\n for (char c : s.toCharArray()) {\n current = current.children.computeIfAbsent(c, k -> new TrieNode(c));\n }\n cur... | 1 | 0 | ['Trie'] | 0 |
implement-magic-dictionary | Python: Trie+DFS | python-triedfs-by-vokasik-7k0b | The idea:\n\n1. Create a trie from words. O(|words| \ max_word_len)\n2. Check each combination in the trie of the search_word with 1 char skipped. O(25\max_word | vokasik | NORMAL | 2022-06-25T02:40:01.834065+00:00 | 2022-06-25T02:41:57.337466+00:00 | 171 | false | The idea:\n\n1. Create a trie from words. O(|words| \\* max_word_len)\n2. Check each combination in the trie of the search_word with 1 char skipped. O(25\\*max_word_len^2) \nExclude from the search the exact search_word if it is present in the original dict\n\ne.g.\n```\nsearch_word=hello, dict: hello, hullo\n\n_ello -... | 1 | 0 | ['Depth-First Search', 'Trie', 'Python'] | 0 |
implement-magic-dictionary | #Accepted solution in C++ using Tries data structure (Very easy & intuitive) | accepted-solution-in-c-using-tries-data-a7w2x | \nclass Node {\n private:\n Node* links[26];\n bool flag = false;\n public:\n bool containsKey(char ch) {\n return links[ch-\'a\'] != null | yashrajyash | NORMAL | 2022-05-22T18:27:22.622464+00:00 | 2022-05-22T18:27:22.622506+00:00 | 62 | false | ```\nclass Node {\n private:\n Node* links[26];\n bool flag = false;\n public:\n bool containsKey(char ch) {\n return links[ch-\'a\'] != nullptr;\n }\n void put(char ch, Node* node) {\n links[ch-\'a\'] = node;\n }\n Node* get(char ch) {\n return links[ch-\'a\'];\n }\n ... | 1 | 0 | ['Depth-First Search', 'C'] | 0 |
implement-magic-dictionary | trie | trie-by-biswajit2329-i1f1 | class MagicDictionary {\npublic:\n struct node\n {\n char val;\n nodearr[26]={NULL};\n bool isend;\n node(char ch)\n {\ | biswajit2329 | NORMAL | 2022-05-18T03:05:19.994024+00:00 | 2022-05-18T03:05:19.994061+00:00 | 91 | false | class MagicDictionary {\npublic:\n struct node\n {\n char val;\n node*arr[26]={NULL};\n bool isend;\n node(char ch)\n {\n val=ch;\n isend=false;\n }\n };\n node*root=new node(\'\\0\');\n void insert(string word)\n {\n node*temp=roo... | 1 | 0 | ['Trie', 'C', 'C++'] | 0 |
implement-magic-dictionary | [C++] Trie solution with complexity analysis | c-trie-solution-with-complexity-analysis-oqa3 | \nclass MagicDictionary {\nprivate:\n struct Trie {\n private:\n bool isEnd = false;\n Trie *children[26];\n public:\n Trie () {\n | tinfu330 | NORMAL | 2022-05-10T17:58:23.072187+00:00 | 2022-05-10T17:58:23.072212+00:00 | 142 | false | ```\nclass MagicDictionary {\nprivate:\n struct Trie {\n private:\n bool isEnd = false;\n Trie *children[26];\n public:\n Trie () {\n for (int i = 0; i < 26; i++) {\n children[i] = NULL;\n }\n }\n \n void insert(string &s) {\n ... | 1 | 0 | ['Backtracking', 'Trie', 'C', 'C++'] | 0 |
implement-magic-dictionary | python solution || hash with dict || faster than 60% || time: O(nk) || space: O(nk) | python-solution-hash-with-dict-faster-th-3y9b | \tclass MagicDictionary:\n\n\t\tdef init(self):\n\t\t\tself.magic_dict = dict() \n\n\t\tdef buildDict(self, dictionary: List[str]) -> None:\n\t\t\tfor word in d | longyunhust | NORMAL | 2022-05-02T16:23:24.977617+00:00 | 2022-05-03T03:40:23.559383+00:00 | 45 | false | \tclass MagicDictionary:\n\n\t\tdef __init__(self):\n\t\t\tself.magic_dict = dict() \n\n\t\tdef buildDict(self, dictionary: List[str]) -> None:\n\t\t\tfor word in dictionary: \n\t\t\t\tword_len = len(word)\n\t\t\t\tfor i in range(word_len): \n\t\t\t\t\tkey = word[:i] + \'*\' + word[i+1:]\n\t\t\t\t\tif key not in self.m... | 1 | 0 | [] | 0 |
implement-magic-dictionary | [Java]Trie || Recursion | javatrie-recursion-by-javacoder911-v2hr | ```\nclass MagicDictionary {\n TrieNode root;\n public MagicDictionary() {\n root = new TrieNode();\n }\n \n public void buildDict(String[ | javaCoder911 | NORMAL | 2022-03-20T05:50:21.941165+00:00 | 2022-03-20T05:50:21.941208+00:00 | 132 | false | ```\nclass MagicDictionary {\n TrieNode root;\n public MagicDictionary() {\n root = new TrieNode();\n }\n \n public void buildDict(String[] dictionary) {\n for(String s: dictionary){\n insertWord(s);\n }\n }\n \n public boolean search(String searchWord) {\n ... | 1 | 0 | ['Trie', 'Java'] | 0 |
implement-magic-dictionary | Java solution | java-solution-by-user9210fp-6ezz | ```class MagicDictionary {\n List myList;\n\n public MagicDictionary() {\n myList = new LinkedList();\n }\n \n public void buildDict(Strin | user9210Fp | NORMAL | 2022-03-06T03:49:40.733785+00:00 | 2022-03-06T03:49:40.733812+00:00 | 111 | false | ```class MagicDictionary {\n List<String> myList;\n\n public MagicDictionary() {\n myList = new LinkedList<String>();\n }\n \n public void buildDict(String[] dictionary) {\n for(int i =0; i < dictionary.length; i++)\n {\n myList.add(dictionary[i]);\n }\n }\n \... | 1 | 0 | [] | 0 |
implement-magic-dictionary | [C++] Trie + Backtracking solution | c-trie-backtracking-solution-by-tinfu330-6tbn | \nclass Trie {\nprivate:\n vector<Trie *> child;\n bool isEnd;\npublic:\n Trie () : child(26, NULL) {\n isEnd = false;\n }\n \n void in | tinfu330 | NORMAL | 2022-01-11T07:27:12.271799+00:00 | 2022-01-11T07:29:32.927794+00:00 | 117 | false | ```\nclass Trie {\nprivate:\n vector<Trie *> child;\n bool isEnd;\npublic:\n Trie () : child(26, NULL) {\n isEnd = false;\n }\n \n void insert(string &s) {\n Trie *curr = this;\n for (char &c : s) {\n int cIndex = c - \'a\';\n if (curr->child[cIndex] == NULL)... | 1 | 0 | ['Backtracking', 'Trie', 'C', 'C++'] | 0 |
implement-magic-dictionary | ⌘[ Java ] 3 Solutions | Observations | Trie | HashMap | java-3-solutions-observations-trie-hashm-aiky | Solution 1 : \n TC : \n\t buildDict : O(n), where n is the number of words in the dictionary\n\t search : O(n*k), where k is the length of word we are checkin | Kaustubh_22 | NORMAL | 2022-01-05T04:00:01.222089+00:00 | 2022-01-18T17:57:22.983598+00:00 | 206 | false | # Solution 1 : \n* TC : \n\t* buildDict : `O(n) `, where `n` is the number of words in the dictionary\n\t* search : `O(n*k)`, where `k` is the length of word we are checking\n \n* SC : `O(n)`\n* Drawback : `Kills memory as the dictionary increases`\n```\nclass MagicDictionary {\n \n HashMap<Integer, HashSet<... | 1 | 0 | ['Trie', 'Java'] | 0 |
implement-magic-dictionary | Golang Hashmap | golang-hashmap-by-archaeoraptor-9cxt | golang\ntype MagicDictionary struct {\n\tneightbour map[string]int\n\torigin map[string]int\n}\n\nfunc Constructor() MagicDictionary {\n\tm := MagicDictiona | archaeoraptor | NORMAL | 2021-12-30T06:55:33.423569+00:00 | 2021-12-30T06:55:33.423670+00:00 | 86 | false | ```golang\ntype MagicDictionary struct {\n\tneightbour map[string]int\n\torigin map[string]int\n}\n\nfunc Constructor() MagicDictionary {\n\tm := MagicDictionary{map[string]int{}, map[string]int{}}\n\treturn m\n}\n\nfunc (this *MagicDictionary) BuildDict(dictionary []string) {\n\tfor _, word := range dictionary {\n... | 1 | 0 | ['Go'] | 0 |
implement-magic-dictionary | JAVA SOLUTION | | 100% LESS MEMORY USE | | ~78.64% FASTER | java-solution-100-less-memory-use-7864-f-c7mc | \nclass MagicDictionary {\n HashSet<String> hs; \n public MagicDictionary() {\n hs = new HashSet<>();\n }\n \n public void buildDict(Stri | Rajnish_vedi | NORMAL | 2021-12-23T09:37:07.956512+00:00 | 2021-12-23T09:37:41.031712+00:00 | 137 | false | ```\nclass MagicDictionary {\n HashSet<String> hs; \n public MagicDictionary() {\n hs = new HashSet<>();\n }\n \n public void buildDict(String[] dictionary) {\n for(String s:dictionary){\n hs.add(s);\n }\n }\n \n public boolean search(String searchWord) {\n f... | 1 | 0 | ['Java'] | 0 |
implement-magic-dictionary | Easy Java Solution No extra space | easy-java-solution-no-extra-space-by-rag-8p7z | class MagicDictionary {\n \n\tString[] magicDictionary;\n public MagicDictionary() {}\n \n public void buildDict(String[] dictionary) {\n mag | raghu6306766 | NORMAL | 2021-10-30T15:27:20.259832+00:00 | 2021-10-30T15:27:20.259857+00:00 | 98 | false | class MagicDictionary {\n \n\tString[] magicDictionary;\n public MagicDictionary() {}\n \n public void buildDict(String[] dictionary) {\n magicDictionary = dictionary;\n }\n \n public boolean canMatch(String searchWord , String dictWord){\n if(dictWord.length() == searchWord.length() ... | 1 | 0 | ['Java'] | 0 |
maximize-win-from-two-segments | [Java/C++/Python] DP + Sliding Segment O(n) | javacpython-dp-sliding-segment-on-by-lee-bum4 | Intuition\nMaximize Win From One Segments,\ncan be solved by sliding window.\n\nNow we can slide the second segment,\nand get calculate the result from dp.\n\n\ | lee215 | NORMAL | 2023-02-04T16:15:42.255814+00:00 | 2023-02-05T07:56:09.276218+00:00 | 13,250 | false | # **Intuition**\nMaximize Win From **One** Segments,\ncan be solved by sliding window.\n\nNow we can slide the second segment,\nand get calculate the result from `dp`.\n<br>\n\n# **Explanation**\nMaintain a sliding segment(sliding window),\nwhere `A[i] - A[j] <= k`.\n\n`dp[k]` means the the maximum number\nwe can cove... | 184 | 0 | ['C', 'Python', 'Java'] | 23 |
maximize-win-from-two-segments | Sliding Window +DP O(n) | sliding-window-dp-on-by-priyanshu_chaudh-2pj8 | Intuition\n what if we were allowed to choose only one segment(sliding window)\n what if i save the last segment with maximum prizes that I can attain from 0<= | Priyanshu_Chaudhary_ | NORMAL | 2023-02-04T16:19:06.438147+00:00 | 2023-02-07T08:52:14.659295+00:00 | 5,504 | false | # Intuition\n* what if we were allowed to choose only one segment(sliding window)\n* what if i save the last segment with maximum prizes that I can attain from 0<=j<=i(dynamic programming)\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n* if I am currently at ith position, the maximum prizes I ... | 41 | 0 | ['Dynamic Programming', 'Sliding Window', 'C++'] | 5 |
maximize-win-from-two-segments | DP+Binary Search||C++||Recursive Solution | dpbinary-searchcrecursive-solution-by-ba-t62f | \nclass Solution {\npublic:\n vector<vector<int>>dp;\n int fn(int ind,vector<int>&arr,int segment,int k) \n {\n if(segment==0)\n retu | baibhavkr143 | NORMAL | 2023-02-04T16:19:04.619062+00:00 | 2023-02-13T06:48:06.899999+00:00 | 3,052 | false | ```\nclass Solution {\npublic:\n vector<vector<int>>dp;\n int fn(int ind,vector<int>&arr,int segment,int k) \n {\n if(segment==0)\n return 0;\n if(ind>=arr.size())\n return 0;\n if(dp[ind][segment]!=-1)\n return dp[ind][segment];\n \n int non_... | 30 | 1 | ['Binary Search', 'Dynamic Programming', 'Recursion', 'C'] | 4 |
maximize-win-from-two-segments | ✅ Easy Sliding Window with DP + Visual Example ✅ | easy-sliding-window-with-dp-visual-examp-tx74 | Idea\n- Use sliding window along with 1D dynamic programming. \n- Sliding window will find all possible segments while dp array stores the maximum segment up to | L30XL1U | NORMAL | 2023-02-07T08:57:18.332855+00:00 | 2023-02-09T05:23:29.880759+00:00 | 1,040 | false | # Idea\n- Use sliding window along with 1D dynamic programming. \n- Sliding window will find all possible segments while `dp` array stores the maximum segment up to each index. \n- Combining these two concepts, sliding window is used for current segment length while `dp` finds a previous maximum segment. \n - Notice... | 27 | 0 | ['C++'] | 1 |
maximize-win-from-two-segments | Prefix Count || O(n) || sliding window || with Image | prefix-count-on-sliding-window-with-imag-9d58 | Intuition\nSince we only need to choose two segement , why not consider one to left and one to right for all index \'i\'\n\n\n\n# Approach\nUsing sliding window | pkpawan | NORMAL | 2023-02-04T17:09:04.518119+00:00 | 2023-02-04T17:12:00.517717+00:00 | 3,008 | false | # Intuition\nSince we only need to choose two segement , why not consider one to left and one to right for all index \'i\'\n\n\n\n# Approach\nUsing sliding window we can find the max count for [0,i] for ... | 24 | 0 | ['Sliding Window', 'Prefix Sum', 'C++'] | 3 |
maximize-win-from-two-segments | Why intuitive approach didn't work out (greedy) | why-intuitive-approach-didnt-work-out-gr-so34 | For all those who are thinking why the intuitive greedy approach didn\'t work out, here is a conversation image\n\n | Dr_Mashoor_Gulati | NORMAL | 2023-02-04T20:44:32.982628+00:00 | 2023-02-04T20:44:32.982669+00:00 | 517 | false | For all those who are thinking why the intuitive greedy approach didn\'t work out, here is a conversation image\n\n | 17 | 0 | ['Greedy'] | 1 |
maximize-win-from-two-segments | DP: leftMax + rightMax | dp-leftmax-rightmax-by-hardcracker-0ov1 | Intuition\nFor every index i, if we can get leftMax[i-1] and rightMax[i], we can iterate through i and get the max result.\n\nleftMax[i]: max on the left of ind | hardcracker | NORMAL | 2023-02-04T16:47:05.894500+00:00 | 2023-02-04T16:58:58.722739+00:00 | 1,431 | false | # Intuition\nFor every index ```i```, if we can get ```leftMax[i-1]``` and ```rightMax[i]```, we can iterate through ```i``` and get the max result.\n\n```leftMax[i]```: max on the left of index ```i``` (including ```i```) \n```rightMax[i]```: max on the right of index ```i``` (including ```i```)\n\nTo get ```leftMax[i... | 17 | 0 | ['Java'] | 1 |
maximize-win-from-two-segments | Catch Up | catch-up-by-votrubac-wueo | We can run the first [l1, r1] and the second [l2, r2] sliding windows at the same time.\n\nAs r2 < l1, the second window catches up as the first one moves.\n\nW | votrubac | NORMAL | 2023-02-07T06:50:47.563104+00:00 | 2023-02-07T06:50:47.563146+00:00 | 796 | false | We can run the first `[l1, r1]` and the second `[l2, r2]` sliding windows at the same time.\n\nAs `r2 < l1`, the second window catches up as the first one moves.\n\nWe track the maximum win we can get from the second window (`max_w2`) so far.\n\nThe win from the first window is `w1`, and we return the best result we ca... | 12 | 0 | ['C'] | 1 |
maximize-win-from-two-segments | [Python] Binary Search + DP | python-binary-search-dp-by-himanshu3889-mxbs | \n def maximizeWin(self, arr: List[int], k: int) -> int:\n n = len(arr)\n best = [0]*(n+1) # best segment after >= i \n | Himanshu3889 | NORMAL | 2023-02-04T16:16:25.537039+00:00 | 2023-02-08T13:42:04.409787+00:00 | 1,063 | false | \n def maximizeWin(self, arr: List[int], k: int) -> int:\n n = len(arr)\n best = [0]*(n+1) # best segment after >= i \n res = 0\n for i in range(n-1,-1,-1): # curr seg start at ith\n e = bisect.bisect_right(arr,arr[i]+k) # take maximum as possible\n... | 12 | 1 | ['Dynamic Programming', 'Binary Tree'] | 1 |
maximize-win-from-two-segments | c++ solution dp memoization | c-solution-dp-memoization-by-dilipsuthar-tc3d | \nclass Solution {\npublic:\n long long range=0;\n long long n;\n long long dp[100500][3];\n int find(vector<int>&nums,int index,int count)\n {\n | dilipsuthar17 | NORMAL | 2023-02-04T17:35:12.076352+00:00 | 2023-02-04T17:35:12.076386+00:00 | 1,483 | false | ```\nclass Solution {\npublic:\n long long range=0;\n long long n;\n long long dp[100500][3];\n int find(vector<int>&nums,int index,int count)\n {\n if(count==0||index>=n)\n {\n return 0;\n }\n if(dp[index][count]!=-1)\n {\n return dp[index][count]... | 10 | 0 | ['Dynamic Programming', 'Memoization', 'C', 'C++'] | 0 |
maximize-win-from-two-segments | C++ | | DP+BS | | Memoization | | Well Explained Code | c-dpbs-memoization-well-explained-code-b-ue24 | Intuition\nWe will be going through all the possible combinations of windows and return the max sum.The code will be making sure that the windows don\'t overlap | H_Dabas02 | NORMAL | 2023-02-05T05:23:33.947690+00:00 | 2023-02-05T05:25:01.032953+00:00 | 745 | false | # Intuition\nWe will be going through all the possible combinations of windows and return the max sum.The code will be making sure that the windows don\'t overlap.We will be traversing from right to left as it\'ll allow us to apply binary search to find the start of the window whose end is \'Arr[i]\'\n\n# Approach\n->S... | 7 | 0 | ['Binary Search', 'Dynamic Programming', 'Memoization', 'C++'] | 2 |
maximize-win-from-two-segments | Simple dp solution with explanation | simple-dp-solution-with-explanation-by-s-dd8b | Intuition\nLet\'s start with a simple problem with one segment. This problem could be solved with a sliding window. To solve the problem with 2 segments we can | svolkovichs | NORMAL | 2023-02-05T23:45:01.906452+00:00 | 2023-02-05T23:45:01.906484+00:00 | 350 | false | # Intuition\nLet\'s start with a simple problem with one segment. This problem could be solved with a sliding window. To solve the problem with 2 segments we can use dynamic programming.\n\n# Approach\nLet\'s define a dp[k] as the max segment in the first k elements. We can iterate an array and for every, i build a max... | 6 | 0 | ['Dynamic Programming', 'JavaScript'] | 0 |
maximize-win-from-two-segments | C++ O(N) || Sliding Window || DP | c-on-sliding-window-dp-by-abhay5349singh-ewqb | Connect with me on LinkedIn: https://www.linkedin.com/in/abhay5349singh/\n\n\nclass Solution {\npublic:\n int maximizeWin(vector<int>& prizePositions, int k) | abhay5349singh | NORMAL | 2023-08-01T11:19:23.150725+00:00 | 2023-08-01T11:19:23.150749+00:00 | 284 | false | **Connect with me on LinkedIn**: https://www.linkedin.com/in/abhay5349singh/\n\n```\nclass Solution {\npublic:\n int maximizeWin(vector<int>& prizePositions, int k) {\n int n=prizePositions.size();\n \n int i=0, j=0;\n \n // maximum no. of prizes which can be selected for segment [... | 5 | 0 | ['Dynamic Programming', 'C++'] | 0 |
maximize-win-from-two-segments | Please help me find error in this code | please-help-me-find-error-in-this-code-b-wb3u | My thought process was- first find the subarray in which the count of prizes is max for a given k.. then remove those elements already picked and repeat the sam | rUdy_c0des | NORMAL | 2023-02-04T17:23:39.936536+00:00 | 2023-02-04T17:23:39.936573+00:00 | 370 | false | My thought process was- first find the subarray in which the count of prizes is max for a given k.. then remove those elements already picked and repeat the same process for 1 more time on the new array since there should be exactly 2 segments.. the code I wrote is as follows- (It passed 61 out of 65 test cases during... | 5 | 0 | ['Java'] | 3 |
maximize-win-from-two-segments | Using Binary Search || Easy To understand 🌟 | using-binary-search-easy-to-understand-b-htfe | Approach\nFind the Most appropriate value from all index\'s and pick that value and maximum of remaining values as answer.\nFind same thing for all 1<=i<=n.\n\n | Sriyansh2001 | NORMAL | 2023-02-04T18:15:45.063255+00:00 | 2023-02-04T18:15:45.063282+00:00 | 1,511 | false | # Approach\nFind the Most appropriate value from all index\'s and pick that value and maximum of remaining values as answer.\nFind same thing for all 1<=i<=n.\n\n# Complexity\n- Time complexity:\n O(N LogN)\n\n- Space complexity:\n O(2N)\n\n# Code\n```\nclass Solution {\npublic:\n int lower(vector<int> &v,int val)... | 4 | 0 | ['Binary Search', 'Dynamic Programming', 'C++'] | 1 |
maximize-win-from-two-segments | Beats 100% by speed and memory, Python simple solution. | beats-100-by-speed-and-memory-python-sim-ta5p | Intuition\nLet\'s iterate through the array and for each position, calculate the count of prizes in the interval if the interval ends at the current position an | stivsh | NORMAL | 2023-02-05T14:19:27.176393+00:00 | 2023-02-05T14:48:44.179967+00:00 | 793 | false | # Intuition\nLet\'s iterate through the array and for each position, calculate the count of prizes in the interval if the interval ends at the current position and store the count and the position in the "intervals" list.\n```\nintervals = [(count of prizes, end pos), (count of prizes, end pos)]\n```\n\nLet\'s also sto... | 3 | 0 | ['Python3'] | 0 |
maximize-win-from-two-segments | C++ || Maintain 2 arrays, front and back, to calculate the answer | c-maintain-2-arrays-front-and-back-to-ca-nwwf | In the solution, what I have tried to do is to have 2 arrays which store, for each index i, for the front array, the value stored in array tells us about the ma | pranjal_gaur | NORMAL | 2023-02-05T04:51:19.242543+00:00 | 2023-02-05T04:51:19.242569+00:00 | 163 | false | In the solution, what I have tried to do is to have 2 arrays which store, for each index i, for the front array, the value stored in array tells us about the maximum window that can satisfy the given condition of k, and similarily for the back row, the values tell us about the maximum window size that satisfies the giv... | 3 | 0 | ['C'] | 0 |
maximize-win-from-two-segments | python3 Solution | python3-solution-by-motaharozzaman1996-m3xj | \n\nclass Solution:\n def maximizeWin(self, prizePositions: List[int], k: int) -> int:\n dp=[0]*(len(prizePositions)+1)\n res=0\n j=0\n | Motaharozzaman1996 | NORMAL | 2023-02-04T16:35:25.866459+00:00 | 2023-02-04T16:35:25.866507+00:00 | 415 | false | \n```\nclass Solution:\n def maximizeWin(self, prizePositions: List[int], k: int) -> int:\n dp=[0]*(len(prizePositions)+1)\n res=0\n j=0\n for i,a in enumerate(prizePositions):\n while prizePositions[j]<prizePositions[i]-k:\n j+=1\n dp[i+1]=max(dp[i],i... | 3 | 1 | ['Python', 'Python3'] | 0 |
maximize-win-from-two-segments | Map Intuitive Approach!! | map-intuitive-approach-by-the_arc_knight-4qqd | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | the_arc_knight_24 | NORMAL | 2023-02-04T16:18:26.788505+00:00 | 2023-02-04T16:18:26.788537+00:00 | 1,205 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 1 | ['C++'] | 2 |
maximize-win-from-two-segments | Python Solution Using Binary Search | python-solution-using-binary-search-by-n-arrb | For More DSA concepts and problem solution subscribe my channel\nhttps://www.youtube.com/channel/UCPbJzXhXAPB3aQnrBcbnqgQ\n# Code\n\nfrom bisect import bisect_r | ng2203 | NORMAL | 2023-02-25T04:51:43.618135+00:00 | 2024-04-13T19:52:43.070543+00:00 | 136 | false | ### For More DSA concepts and problem solution subscribe my channel\nhttps://www.youtube.com/channel/UCPbJzXhXAPB3aQnrBcbnqgQ\n# Code\n```\nfrom bisect import bisect_right,bisect_left\nclass Solution:\n def maximizeWin(self, prizePositions: List[int], k: int) -> int:\n ans=prv=0\n for i,item in enumera... | 2 | 0 | ['Python3'] | 1 |
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