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calculate-digit-sum-of-a-string | python3 simple solution | python3-simple-solution-by-mxmb-c2ds | python\ndef digitSum(self, s: str, k: int) -> str:\n while len(s) > k:\n arr = [s[i:i+k] for i in range(0,len(s),k)] # divide \n ar | mxmb | NORMAL | 2022-04-23T18:02:35.890528+00:00 | 2022-04-23T18:03:17.623646+00:00 | 156 | false | ```python\ndef digitSum(self, s: str, k: int) -> str:\n while len(s) > k:\n arr = [s[i:i+k] for i in range(0,len(s),k)] # divide \n arr = [str(sum(int(i) for i in s)) for s in arr] # replace\n s = \'\'.join(arr) # merge\n return s\n``` | 1 | 0 | ['Python', 'Python3'] | 0 |
calculate-digit-sum-of-a-string | C++ | 0ms Solution | Recursion with comments | c-0ms-solution-recursion-with-comments-b-ewr2 | \nclass Solution {\npublic:\n string digitSum(string s, int k) {\n // if length is smaller or equal to k, no more rounds left\n if (s.length() | rohit2706 | NORMAL | 2022-04-21T13:28:47.861191+00:00 | 2022-04-21T13:28:47.861234+00:00 | 38 | false | ```\nclass Solution {\npublic:\n string digitSum(string s, int k) {\n // if length is smaller or equal to k, no more rounds left\n if (s.length() <= k)\n return s;\n \n // find string after this round\n string round_str;\n for (int i=0; i<s.length(); i+=k) {\n ... | 1 | 0 | [] | 0 |
calculate-digit-sum-of-a-string | [Python] | Simple Simulation solution | python-simple-simulation-solution-by-gre-29di | This is my first post, so let\'s start with an easy one.\nWhat we want to do here is to simulate the rounds by iterating until our string s is shorter than k, a | gregalletti | NORMAL | 2022-04-20T10:55:39.471003+00:00 | 2022-04-20T16:49:40.797272+00:00 | 118 | false | *This is my first post, so let\'s start with an easy one.*\nWhat we want to do here is to **simulate the rounds** by iterating until our string `s` is shorter than `k`, and in the loop we:\n* split `s` in groups of length `k` and store the results in a list `groups`\n* calculate the sum of the digits of each group `g` ... | 1 | 0 | ['Python', 'Python3'] | 0 |
calculate-digit-sum-of-a-string | Easy C++ Solution | easy-c-solution-by-shreya_2506-aaxx | class Solution {\npublic:\n string digitSum(string s, int k) {\n \n while(s.size()>k){\n \n int a=0,cnt=0;\n s | Shreya_2506 | NORMAL | 2022-04-20T06:32:15.375269+00:00 | 2022-04-20T06:32:15.375300+00:00 | 83 | false | class Solution {\npublic:\n string digitSum(string s, int k) {\n \n while(s.size()>k){\n \n int a=0,cnt=0;\n string t="";\n \n for(int i=0;i<s.size();i++){\n \n a+=(s[i]-\'0\');\n cnt++;\n if(... | 1 | 0 | ['String', 'C'] | 0 |
calculate-digit-sum-of-a-string | a few solutions | a-few-solutions-by-claytonjwong-l5lk | Recursively reduce the input string s into k-sized chunks.\n\n---\n\nKotlin\n\nclass Solution {\n fun digitSum(s: String, k: Int): String {\n var N = | claytonjwong | NORMAL | 2022-04-19T15:45:21.174902+00:00 | 2022-04-19T15:48:03.044563+00:00 | 47 | false | Recursively reduce the input string `s` into `k`-sized chunks.\n\n---\n\n*Kotlin*\n```\nclass Solution {\n fun digitSum(s: String, k: Int): String {\n var N = s.length\n if (N <= k)\n return s\n var f = { s: String -> s.toCharArray().map{ it.toInt() - \'0\'.toInt() }.sum()!! }\n ... | 1 | 0 | [] | 0 |
calculate-digit-sum-of-a-string | Python || Easy Solution | python-easy-solution-by-nashvenn-7ix9 | Please upvote if it helps. Thank you!\n\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n while len(s) > k:\n new_s = \'\'\n | nashvenn | NORMAL | 2022-04-18T20:26:48.458671+00:00 | 2022-04-18T20:26:48.458712+00:00 | 66 | false | **Please upvote if it helps. Thank you!**\n```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n while len(s) > k:\n new_s = \'\'\n for i in range(0, len(s), k):\n new_s += str(sum([int(c) for c in s[i:i+k]]))\n s = new_s[:]\n return s\n``` | 1 | 0 | ['Python'] | 0 |
calculate-digit-sum-of-a-string | Python beats 100% memory 85% space | With Comments | python-beats-100-memory-85-space-with-co-3z5u | \nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n # Requirement 3; Merge groups when string s > k\n while len(s) > k:\ | krazynukz | NORMAL | 2022-04-17T18:41:51.443134+00:00 | 2022-04-17T18:41:51.443182+00:00 | 88 | false | ```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n # Requirement 3; Merge groups when string s > k\n while len(s) > k:\n temp = "" # Will contain the next round\'s string s\n i = 0 # counter to exit while loop below, when out of groups\n \n ... | 1 | 0 | ['String', 'Python'] | 1 |
calculate-digit-sum-of-a-string | Short Python while loop | short-python-while-loop-by-hszh-kkgu | \nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n while len(s) > k:\n curr = []\n for i in range(0, len | hsZh | NORMAL | 2022-04-17T15:01:20.731836+00:00 | 2022-04-17T15:04:06.190380+00:00 | 40 | false | ```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n while len(s) > k:\n curr = []\n for i in range(0, len(s), k):\n curr.append(str(sum(int(s[j]) for j in range(i, min(len(s),i+k)))))\n s = \'\'.join(curr)\n \n return ... | 1 | 0 | ['Python'] | 0 |
calculate-digit-sum-of-a-string | ✅ C++ | Easy to understand | c-easy-to-understand-by-rupam66-2pfu | \nclass Solution {\npublic:\n string digitSum(string s, int k) {\n if(s.size()<=1) return s;\n while(s.size()>k){\n string tmp;\n | rupam66 | NORMAL | 2022-04-17T14:09:55.296264+00:00 | 2022-04-17T14:09:55.296304+00:00 | 56 | false | ```\nclass Solution {\npublic:\n string digitSum(string s, int k) {\n if(s.size()<=1) return s;\n while(s.size()>k){\n string tmp;\n int num=0;\n for(int i=0;i<s.size();i++){\n if(i && i%k==0){\n tmp+=to_string(num);\n ... | 1 | 0 | ['String', 'C'] | 0 |
calculate-digit-sum-of-a-string | 5-Lines Python Solution || 100% Faster || Memory less than 85% | 5-lines-python-solution-100-faster-memor-bwr7 | \n\n\n### Solution: TC O((n/k)\n) / SC O(n)\n\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n while len(s)>k:\n tmp=\'\'\n | Taha-C | NORMAL | 2022-04-17T14:06:40.417026+00:00 | 2022-04-17T14:41:10.528848+00:00 | 88 | false | \n\n\n### ***Solution: TC O((n/k)\\*n) / SC O(n)***\n```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n while len(s)>k:\n tmp=\'\'\n for i in range(0,len(s),k): t... | 1 | 0 | ['Python', 'Python3'] | 0 |
calculate-digit-sum-of-a-string | Simple Python Solution | simple-python-solution-by-lcshiung-ix9k | \nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n final_s = s\n while len(final_s) > k:\n new_s = \'\' | lcshiung | NORMAL | 2022-04-17T13:37:39.820906+00:00 | 2022-04-17T17:19:15.599422+00:00 | 25 | false | ```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n \n final_s = s\n while len(final_s) > k:\n new_s = \'\' # newly formed string\n k_length_sub_s = \'\' # k-length substring\n for i in range(len(final_s)):\n ... | 1 | 0 | [] | 0 |
calculate-digit-sum-of-a-string | Java | java-by-ryann10-o1an | \nclass Solution {\n public String digitSum(String s, int k) {\n while (s.length() > k) {\n int groups = s.length() / k + (s.length() % k = | ryann10 | NORMAL | 2022-04-17T12:59:34.302437+00:00 | 2022-04-17T12:59:34.302465+00:00 | 70 | false | ```\nclass Solution {\n public String digitSum(String s, int k) {\n while (s.length() > k) {\n int groups = s.length() / k + (s.length() % k == 0 ? 0 : 1);\n \n StringBuilder sb = new StringBuilder();\n \n for (int i = 0; i < groups; i++) {\n ... | 1 | 0 | ['String', 'Java'] | 0 |
calculate-digit-sum-of-a-string | Python | Recursive Solution | Easy to understand | python-recursive-solution-easy-to-unders-vozq | ```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n def util(s):\n\t\t\t# if length of s is less than or equal to k, then we can not m | parmardiwakar150 | NORMAL | 2022-04-17T12:23:45.641159+00:00 | 2022-04-17T12:25:57.604575+00:00 | 36 | false | ```\nclass Solution:\n def digitSum(self, s: str, k: int) -> str:\n def util(s):\n\t\t\t# if length of s is less than or equal to k, then we can not make s any shorter\n if len(s) <= k:\n return s\n ans = \'\'\n for i in range(0, len(s), k):\n ans... | 1 | 0 | ['Recursion'] | 0 |
calculate-digit-sum-of-a-string | Simple Recursive Java Solution with comments | simple-recursive-java-solution-with-comm-4z8l | \nclass Solution {\n public String digitSum(String s, int k) {\n if (s.length() <= k) return s;\n \n StringBuilder sb = new StringBuilde | pure_aloha | NORMAL | 2022-04-17T10:51:46.913716+00:00 | 2022-04-17T10:51:46.913756+00:00 | 45 | false | ```\nclass Solution {\n public String digitSum(String s, int k) {\n if (s.length() <= k) return s;\n \n StringBuilder sb = new StringBuilder();\n for (int i=0; i < s.length(); i += k) {\n int start = i;\n int end = Math.min(i + k, s.length()); // To avoid indexOutOfB... | 1 | 0 | ['Recursion', 'Java'] | 1 |
find-maximum-number-of-string-pairs | Explained using map || Very simple & easy to understand solution | explained-using-map-very-simple-easy-to-j0ekd | Upvote if you like the solution\n# Approach\n1. Take a map with seen string as the index and its count as its value.\n2. Iterate the words vector and check if i | kreakEmp | NORMAL | 2023-06-24T16:07:27.638358+00:00 | 2023-06-24T19:14:19.951377+00:00 | 4,159 | false | #### Upvote if you like the solution\n# Approach\n1. Take a map with seen string as the index and its count as its value.\n2. Iterate the words vector and check if its reverse exist in the map or not. \n3. If exist in the map then update the ans and decrement the map value as the we have considered this value for pairi... | 25 | 1 | ['C++'] | 4 |
find-maximum-number-of-string-pairs | 676 | 676-by-votrubac-8zsx | There are 676 possible 2-character strings.\n\nSo, we use a boolean array to track if we\'ve seen a string before.\n\n> Note that we track reversed strings in t | votrubac | NORMAL | 2023-06-24T16:20:11.653861+00:00 | 2024-06-18T21:41:57.687449+00:00 | 1,483 | false | There are 676 possible 2-character strings.\n\nSo, we use a boolean array to track if we\'ve seen a string before.\n\n> Note that we track reversed strings in the boolean array. \n\n**C++**\n```cpp\nint maximumNumberOfStringPairs(vector<string>& words) {\n int vis[676] = {}, res = 0;\n for (const auto &w : words... | 23 | 1 | ['C'] | 7 |
find-maximum-number-of-string-pairs | ✅[Python] Simple and Clean, beats 88%✅ | python-simple-and-clean-beats-88-by-_tan-14nq | Please upvote if you find this helpful. \u270C\n\n\n\n# Intuition\nThe problem asks us to find the maximum number of pairs that can be formed from a given list | _Tanmay | NORMAL | 2023-07-05T09:54:04.927628+00:00 | 2023-07-05T09:54:04.927659+00:00 | 1,175 | false | ### Please upvote if you find this helpful. \u270C\n<img src="https://assets.leetcode.com/users/images/b8e25620-d320-420a-ae09-94c7453bd033_1678818986.7001078.jpeg" alt="Cute Robot - Stable diffusion" width="200"/>\n\n\n# Intuition\nThe problem asks us to find the maximum number of pairs that can be formed from a given... | 15 | 0 | ['Array', 'Hash Table', 'String', 'Ordered Set', 'Python3'] | 5 |
find-maximum-number-of-string-pairs | [C++/JAVA/PYTHON] O(1) Space || Easy to Understand || With Explanation | cjavapython-o1-space-easy-to-understand-khcrl | Time complexity : O(n^2)\n- Space complexity : O(1)\n# C++ Code\n\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& v) {\n i | Aditya00 | NORMAL | 2023-06-24T16:15:24.571875+00:00 | 2023-06-24T18:22:04.297415+00:00 | 2,217 | false | - Time complexity : O(n^2)\n- Space complexity : O(1)\n# C++ Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& v) {\n int ans = 0; // Initialize the variable to store the answer\n int n = v.size(); \n\n // Iterate over all pairs of strings\n for (int i ... | 15 | 1 | ['C++'] | 2 |
find-maximum-number-of-string-pairs | Python 3 || 3 lines, w/ explanation || T/S: 99% / 96% . | python-3-3-lines-w-explanation-ts-99-96-huga6 | Here\'s how the code works:\n\nThe code initializesdto keep track of the count of each word or its reverse.\n\nWe iterate over each word inwords, and we use the | Spaulding_ | NORMAL | 2023-07-29T16:57:46.041055+00:00 | 2024-06-23T15:34:20.724193+00:00 | 1,035 | false | Here\'s how the code works:\n\nThe code initializes`d`to keep track of the count of each word or its reverse.\n\nWe iterate over each word in`words`, and we use the lexicographic minimum of the word and its reverse (`word[::-1]`)as the key for loading `words` into `d`, which ensures that the same pair is counted only o... | 9 | 0 | ['Python3'] | 0 |
find-maximum-number-of-string-pairs | Easy c++ solution using unordered set | easy-c-solution-using-unordered-set-by-r-z0v4 | Intuition\n Describe your first thoughts on how to solve this problem. \nreverse and check the string is present in the set\n\n# Approach\n Describe your appr | rajgaurav95 | NORMAL | 2023-06-24T16:08:01.238133+00:00 | 2023-06-24T16:15:52.073323+00:00 | 1,862 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nreverse and check the string is present in the set\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. $$O(n)$$ -->\nO(n)\n\n- Space comple... | 9 | 0 | ['C++'] | 2 |
find-maximum-number-of-string-pairs | 🧩 [VIDEO] Maximum Number of String Pairs 🔀 - Reversed String Pairing | video-maximum-number-of-string-pairs-rev-qwzf | Intuition\nUpon reading the problem, it\'s apparent that we need to pair strings that are the reverse of each other. The first instinct is to use a data structu | vanAmsen | NORMAL | 2023-07-25T21:50:53.116825+00:00 | 2023-07-25T22:06:57.504868+00:00 | 844 | false | # Intuition\nUpon reading the problem, it\'s apparent that we need to pair strings that are the reverse of each other. The first instinct is to use a data structure that allows quick lookups to check for the existence of a reversed word. A Python dictionary fits this need perfectly.\n\nhttps://www.youtube.com/watch?v=z... | 8 | 0 | ['C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 1 |
find-maximum-number-of-string-pairs | How to be in top by time complexity | how-to-be-in-top-by-time-complexity-by-n-hej5 | Code\n\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n total = 0\n c = set()\n for i in range(len(wo | NurzhanSultanov | NORMAL | 2024-02-12T12:04:54.864126+00:00 | 2024-02-12T12:04:54.864168+00:00 | 415 | false | # Code\n```\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n total = 0\n c = set()\n for i in range(len(words)):\n for j in range(i + 1, len(words)):\n word1 = words[i]\n word2 = words[j]\n\n if word1 == w... | 6 | 0 | ['Python3'] | 1 |
find-maximum-number-of-string-pairs | ✅Runtime 12 ms Beats 🚀100% and 💡Memory 13 MB Beats 📉99.67% | runtime-12-ms-beats-100-and-memory-13-mb-vtjt | 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 | deleted_user | NORMAL | 2023-10-19T16:33:27.068540+00:00 | 2023-10-19T16:33:27.068570+00:00 | 263 | 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. $$O(n)$$ --... | 6 | 0 | ['Python'] | 0 |
find-maximum-number-of-string-pairs | Java Solution || Using StringBuilder | java-solution-using-stringbuilder-by-ank-vjyz | Intuition\n Describe your first thoughts on how to solve this problem. \nWe have to find whether the reverse of a string at words[i] is present at index j such | Ankur_Chhillar | NORMAL | 2023-07-08T15:15:16.909704+00:00 | 2023-07-08T15:15:16.909738+00:00 | 478 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWe have to find whether the reverse of a string at words[i] is present at index j such that 0<=i<j<words.length and return the count of such pairs found.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nI have used ... | 6 | 0 | ['Java'] | 1 |
find-maximum-number-of-string-pairs | Simple Java with Iterator | simple-java-with-iterator-by-dafteeer-hj7l | Approach\nI used Iterator and StringBuffer.reverse() for easy and simple solution.\n\n# Code\n\nclass Solution {\n public int maximumNumberOfStringPairs(Stri | dafteeer | NORMAL | 2023-07-08T21:04:44.059144+00:00 | 2023-07-08T21:04:44.059166+00:00 | 662 | false | # Approach\nI used Iterator and StringBuffer.reverse() for easy and simple solution.\n\n# Code\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n List<String> wordsList = new ArrayList(Arrays.asList(words));\n int count = 0;\n for (Iterator<String> wordsIterator =... | 5 | 0 | ['String', 'Iterator', 'Java'] | 1 |
find-maximum-number-of-string-pairs | Easy Python Solution | easy-python-solution-by-vistrit-qc01 | Code\n\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n c=0\n for i in range(len(words)):\n for j | vistrit | NORMAL | 2023-06-27T18:24:46.564103+00:00 | 2023-06-27T18:24:46.564132+00:00 | 976 | false | # Code\n```\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n c=0\n for i in range(len(words)):\n for j in range(i+1,len(words)):\n if words[i]==words[j][::-1]:\n c+=1\n return c\n``` | 5 | 0 | ['Python3'] | 2 |
find-maximum-number-of-string-pairs | [C++] Set Vs. Hashing String to Int Vs. Bitmask Hashing, 4ms, 20.5MB | c-set-vs-hashing-string-to-int-vs-bitmas-eqck | Let\'s start with the simplest approach, which means we will use a hashset (seen) to keep track of already encountered strings.\n\nMore specifically, we will st | Ajna2 | NORMAL | 2023-06-24T20:39:20.946592+00:00 | 2023-06-24T20:39:20.946616+00:00 | 308 | false | Let\'s start with the simplest approach, which means we will use a hashset (`seen`) to keep track of already encountered strings.\n\nMore specifically, we will start declaring:\n* `res` as our usual counter variable, initially set to be `0`;\n* `seen`, our set where we will store what we already saw (or its reverse, in... | 5 | 0 | ['String', 'Bit Manipulation', 'Hash Function', 'Bitmask', 'C++'] | 1 |
find-maximum-number-of-string-pairs | Simple || Short || Clean || Java Solution | simple-short-clean-java-solution-by-hima-qrcj | \njava []\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n Map<String, Integer> map = new HashMap<>();\n for (Stri | HimanshuBhoir | NORMAL | 2023-06-24T16:04:08.759853+00:00 | 2023-06-24T16:04:35.217089+00:00 | 2,007 | false | \n```java []\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n Map<String, Integer> map = new HashMap<>();\n for (String s : words) {\n String rev = new StringBuilder(s).reverse().toString();\n if(map.containsKey(rev)) map.put(rev, map.get(rev)+1);\n... | 5 | 1 | ['Java'] | 2 |
find-maximum-number-of-string-pairs | Best and simple java approach | best-and-simple-java-approach-by-venkate-9rvz | Code\njava []\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int ans=0;\n for(int i=0;i<words.length;i++) {\n | VenkateshThantapureddy | NORMAL | 2024-09-07T09:15:56.647899+00:00 | 2024-09-07T09:15:56.647937+00:00 | 155 | false | # Code\n```java []\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int ans=0;\n for(int i=0;i<words.length;i++) {\n \tfor(int j=i+1;j<words.length;j++) {\n \t\tif(words[i].charAt(0)==words[j].charAt(1) && words[i].charAt(1)==words[j].charAt(0)) \n \t\t... | 4 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | Good To Go ✅🚀 | good-to-go-by-dixon_n-v7o0 | \n# Code\njava []\n\n// This approcah is used to solve 2131. Longest Palindrome by Concatenating Two Letter Words\n\n\nclass Solution {\n public int maximumN | Dixon_N | NORMAL | 2024-08-03T00:35:32.923307+00:00 | 2024-08-03T00:35:32.923339+00:00 | 473 | false | \n# Code\n```java []\n\n// This approcah is used to solve 2131. Longest Palindrome by Concatenating Two Letter Words\n\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int ans = 0;\n HashMap<String, Integer> map = new HashMap<>();\n \n for (String w : words)... | 4 | 0 | ['Hash Table', 'Java'] | 4 |
find-maximum-number-of-string-pairs | One line solution with Runtime 98% | one-line-solution-with-runtime-98-by-smd-3rdm | \n\n/**\n * @param {string[]} words\n * @return {number}\n */\nvar maximumNumberOfStringPairs = function(words) {\n return words.length - new Set(words.map((a) | smd_94 | NORMAL | 2023-11-13T20:44:04.765751+00:00 | 2023-11-13T20:44:04.765773+00:00 | 678 | false | \n```\n/**\n * @param {string[]} words\n * @return {number}\n */\nvar maximumNumberOfStringPairs = function(words) {\n return words.length - new Set(words.map((a) => a[0]>a[1]? a[0]+a[1]: a[1]+a[0] )).size;\n};\n``` | 4 | 0 | ['JavaScript'] | 1 |
find-maximum-number-of-string-pairs | Easy Peasy | easy-peasy-by-ridamexe-3ofq | Code\n\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n int n=words.size();\n\n for | ridamexe | NORMAL | 2023-06-25T13:41:52.677295+00:00 | 2023-06-25T13:41:52.677315+00:00 | 393 | false | # Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n int n=words.size();\n\n for(int i=0; i<n-1; i++){\n string s1=words[i];\n\n for(int j=i+1; j<n; j++){\n string s2=words[j];\n\n if... | 4 | 0 | ['C++'] | 1 |
find-maximum-number-of-string-pairs | 1 line using Set | 1-line-using-set-by-bariscan97-r9s5 | 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 | Bariscan97 | NORMAL | 2023-06-24T23:54:56.667411+00:00 | 2023-06-24T23:54:56.667440+00:00 | 787 | 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. $$O(n)$$ --... | 4 | 0 | ['Hash Table', 'Python', 'Python3'] | 2 |
find-maximum-number-of-string-pairs | Java || O(N) runtime, O(N) space || Scalable solution using StringBuilder.reverse() | java-on-runtime-on-space-scalable-soluti-fpe0 | O(N) runtime due to iteration through words.\nO(N) space due to HashSet wordSet that may go up to N, the input words length.\n\nPairs are unique so removing fro | ZionWilliamson1 | NORMAL | 2023-06-24T16:23:51.857262+00:00 | 2023-06-24T16:23:51.857289+00:00 | 1,085 | false | O(N) runtime due to iteration through `words`.\nO(N) space due to HashSet `wordSet` that may go up to N, the input `words` length.\n\nPairs are unique so removing from HashSet is never needed.\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int maxPairs = 0;\n \n ... | 4 | 0 | ['Java'] | 5 |
find-maximum-number-of-string-pairs | 🔥MAP🔥||🔥C++🔥||🔥MOST SIMPLE🔥||🔥EASY TO UNDERSTAND🔥 | mapcmost-simpleeasy-to-understand-by-gan-90r2 | if this code helps you, please upvote\n# Code\n\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int ans = 0;\n | ganeshkumawat8740 | NORMAL | 2023-06-24T16:04:39.465737+00:00 | 2023-06-25T01:38:02.078879+00:00 | 803 | false | # if this code helps you, please upvote\n# Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int ans = 0;\n unordered_map<string,int> mp;\n string s;\n for(auto &i: words){\n s = i;\n reverse(s.begin(),s.end());\n ... | 4 | 0 | ['Hash Table', 'String', 'String Matching', 'C++'] | 0 |
find-maximum-number-of-string-pairs | 1 ms Beats 100.00% of users with Java | 1-ms-beats-10000-of-users-with-java-by-h-a2gj | 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 | hojiakbarmadaminov45 | NORMAL | 2024-02-23T14:49:17.197847+00:00 | 2024-02-23T14:49:17.197875+00:00 | 288 | 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: 1ms \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: 41.40 mb\n<!-- Add your space complexity here, e... | 3 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | String Pairs - JavaScript - Beats 99.46 % ( 56 ms ) 🔥 | string-pairs-javascript-beats-9946-56-ms-37is | \n\n1. Solution with two for loops\n\n/**\n * @param {string[]} words\n * @return {number}\n */\nvar maximumNumberOfStringPairs = function(words) {\n count = | zemamba | NORMAL | 2023-06-26T20:03:52.920086+00:00 | 2023-06-26T23:49:41.039596+00:00 | 397 | false | \n\n1. Solution with two for loops\n```\n/**\n * @param {string[]} words\n * @return {number}\n */\nvar maximumNumberOfStringPairs = function(words) {\n count = 0 \n\n for (let i = 0; i < words.length... | 3 | 0 | ['Hash Table', 'JavaScript'] | 1 |
find-maximum-number-of-string-pairs | C++ -> Brute Force -> Set Solution [2 Solution] | c-brute-force-set-solution-2-solution-by-56w1 | Intuition\nThink of reverse the each words of the given string and check if the reverse words available or not.\n\n---\n\n\n# Complexity\n- Time complexity:\n1. | amit24x | NORMAL | 2023-06-24T17:08:30.976141+00:00 | 2023-06-24T17:08:30.976173+00:00 | 702 | false | # Intuition\nThink of reverse the each words of the given string and check if the reverse words available or not.\n\n---\n\n\n# Complexity\n- Time complexity:\n1. *Brute Force:* **O(n^2)**\n2. *Set:* **O(n)**\n\n- Space complexity:**O(n)**\n\n---\n\n\n**Solution 1 - Brute Force**\n\n```\nclass Solution {\npublic:\n ... | 3 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | JavaScript simple solution with explanation | javascript-simple-solution-with-explanat-gqh1 | Approach\n\n1. Initialize maxPairs to keep track of the maximum number of pairs\n2. Initialize wordSet to track words that can potentially form pairs\n3. Iterat | js_pro | NORMAL | 2023-06-24T16:03:51.384912+00:00 | 2023-08-29T10:10:36.201827+00:00 | 408 | false | # Approach\n\n1. Initialize `maxPairs` to keep track of the maximum number of pairs\n2. Initialize `wordSet` to track words that can potentially form pairs\n3. Iterate through the words:\n - sort the word lexicographically\n - checks if it is in `wordSet`\n - if found, a pair is formed, increment `maxPairs` an... | 3 | 0 | ['JavaScript'] | 1 |
find-maximum-number-of-string-pairs | 100% beat in java very easy code for beginners 😊. | 100-beat-in-java-very-easy-code-for-begi-kmgn | Code | Galani_jenis | NORMAL | 2025-01-07T12:43:08.929376+00:00 | 2025-01-07T12:43:08.929376+00:00 | 269 | false | 
# Code
```java []
class Solution {
public int maximumNumberOfStringPairs(String[] words) {
int ans = 0;
for (int i = 0; i < words.le... | 2 | 0 | ['Java'] | 1 |
find-maximum-number-of-string-pairs | EASY C++ SOLUTION || 100% BEATS || 0 ms | easy-c-solution-100-beats-0-ms-by-yash_a-jxbu | 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 | Yash_Akabari | NORMAL | 2024-08-20T15:30:57.386323+00:00 | 2024-08-20T15:30:57.386369+00:00 | 97 | 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- O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n- O(N)\n<!-- Add your space complexity here, ... | 2 | 0 | ['Array', 'Hash Table', 'C++'] | 0 |
find-maximum-number-of-string-pairs | ✅ One Line Solution | one-line-solution-by-mikposp-2avr | Code #1\nTime complexity: O(n). Space complexity: O(n).\n\nclass Solution:\n def maximumNumberOfStringPairs(self, w: List[str]) -> int:\n return len(w | MikPosp | NORMAL | 2024-05-12T11:52:07.303409+00:00 | 2024-05-12T11:52:07.303443+00:00 | 140 | false | # Code #1\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def maximumNumberOfStringPairs(self, w: List[str]) -> int:\n return len(w)-len({*map(frozenset,w)})\n```\n\n# Code #2\nTime complexity: $$O(n)$$. Space complexity: $$O(n)$$.\n```\nclass Solution:\n def maximumNumberOf... | 2 | 0 | ['Array', 'Hash Table', 'String', 'Simulation', 'Python', 'Python3'] | 0 |
find-maximum-number-of-string-pairs | [Java] Easy 100% solution | java-easy-100-solution-by-ytchouar-9e0p | java\nclass Solution {\n public int maximumNumberOfStringPairs(final String[] words) {\n int count = 0;\n\n for (int i = 0; i< words.length; i+ | YTchouar | NORMAL | 2024-05-05T04:08:50.605584+00:00 | 2024-05-05T04:08:50.605601+00:00 | 231 | false | ```java\nclass Solution {\n public int maximumNumberOfStringPairs(final String[] words) {\n int count = 0;\n\n for (int i = 0; i< words.length; i++)\n for(int j=i+1; j<words.length; j++)\n if(words[i].charAt(0) == words[j].charAt(1) && words[i].charAt(1) == words[j].charAt(0))... | 2 | 0 | ['Java'] | 1 |
find-maximum-number-of-string-pairs | 🔒 Use constraints to solve with vector of size 677 🔢 | use-constraints-to-solve-with-vector-of-j22go | Intuition\nThe problem seems to involve counting pairs of words in a given list of strings. Each word in the list is potentially paired with its reverse. The ob | sambhav22435 | NORMAL | 2024-04-06T06:30:23.274279+00:00 | 2024-04-06T06:30:23.274307+00:00 | 97 | false | # Intuition\nThe problem seems to involve counting pairs of words in a given list of strings. Each word in the list is potentially paired with its reverse. The objective is to find the maximum number of such pairs that can be formed.\n\n# Approach\n1. We\'ll iterate through the list of words, counting occurrences of ea... | 2 | 0 | ['Array', 'Hash Table', 'Hash Function', 'C++'] | 0 |
find-maximum-number-of-string-pairs | Java || Simple string solution ♨️♨️ | java-simple-string-solution-by-ritabrata-tara | \n\n# Code\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n List<String> list = new ArrayList<>();\n StringBuild | Ritabrata_1080 | NORMAL | 2023-11-03T18:37:13.768094+00:00 | 2023-11-03T18:37:13.768123+00:00 | 119 | false | \n\n# Code\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n List<String> list = new ArrayList<>();\n StringBuilder sb = new StringBuilder();\n int count = 0;\n for(String p : words){\n char c[] = p.toCharArray();\n if(c[0] != c[1]){\... | 2 | 0 | ['String', 'Java'] | 0 |
find-maximum-number-of-string-pairs | 100% and 78% beats | Java Solution | 100-and-78-beats-java-solution-by-s_u_04-hofa | \nclass Solution \n{\n public int maximumNumberOfStringPairs(String[] words) \n {\n // creating a dic for each unique pair(26 * 26)\n boolea | S_U_04 | NORMAL | 2023-10-02T20:11:36.208835+00:00 | 2023-10-12T19:02:53.370126+00:00 | 496 | false | ```\nclass Solution \n{\n public int maximumNumberOfStringPairs(String[] words) \n {\n // creating a dic for each unique pair(26 * 26)\n boolean[] dictionary = new boolean[676];\n int count = 0;\n\n for (String w : words) \n {\n // current word index\n int ... | 2 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | Find Maximum Number Of String Pairs Solution in C++ | find-maximum-number-of-string-pairs-solu-2n3q | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nUsing Unordered map\n# | The_Kunal_Singh | NORMAL | 2023-09-15T18:08:13.615655+00:00 | 2023-09-15T18:08:13.615691+00:00 | 42 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nUsing Unordered map\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(n^2)\n- Space complexity:\n<!-- Add your space complexi... | 2 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Easy Java solution - Beats 100% | easy-java-solution-beats-100-by-karansid-tepv | We can use a nested for-loop to iterate through our \'words\' array and compare any two words in our array without worrying about it being counted as a duplicat | karansidz | NORMAL | 2023-08-24T22:40:52.024645+00:00 | 2023-08-24T22:40:52.024665+00:00 | 148 | false | We can use a nested for-loop to iterate through our \'words\' array and compare any two words in our array without worrying about it being counted as a duplicate (because the second for loop starts at \'i+1\' to ensure it can not be compared again). \n We can make use of String\'s charAt function to compare the char... | 2 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | Finding Maximum Number of String Pairs | finding-maximum-number-of-string-pairs-b-lj67 | Intuition\n Describe your first thoughts on how to solve this problem. \nWHEN I SEE THIS PROBLEM: Find Maximum Number of (String Pairs)\n\n\n\n# Code\n\n\nclass | aadhiganapathy8 | NORMAL | 2023-08-16T11:59:54.131224+00:00 | 2023-08-16T11:59:54.131251+00:00 | 626 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nWHEN I SEE THIS PROBLEM: Find Maximum Number of (String Pairs)\n\n\n\n# Code\n```\n\nclass Solution {\n public int maximu... | 2 | 0 | ['Array', 'Java'] | 1 |
find-maximum-number-of-string-pairs | Simple Solution using Hash map . Easy to understand for Beginners . Beats about 99 % . | simple-solution-using-hash-map-easy-to-u-xvjr | 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 | BinaryBrawler | NORMAL | 2023-08-07T02:02:06.160134+00:00 | 2023-08-07T02:02:06.160162+00:00 | 622 | 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 mem:O(1) || Boolean flags, No HashMap | java-1ms-runon-memo1-boolean-flags-no-ha-9paz | \nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n boolean[] found = new boolean[26 * 26];\n int pairCount = 0;\n | dudeandcat | NORMAL | 2023-08-02T07:10:16.223568+00:00 | 2023-08-02T07:10:16.223607+00:00 | 261 | false | ```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n boolean[] found = new boolean[26 * 26];\n int pairCount = 0;\n for (String s : words) {\n if (found[(s.charAt(1) - \'a\') * 26 + s.charAt(0) - \'a\']) pairCount++;\n found[(s.charAt(0) - \'a\... | 2 | 0 | [] | 1 |
find-maximum-number-of-string-pairs | Easy Approach | easy-approach-by-noor88-2qwq | 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 | Noor88 | NORMAL | 2023-07-24T18:48:05.580157+00:00 | 2023-07-24T18:48:05.580178+00:00 | 163 | 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. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | 2 C++ solutions || Using hash-map and hash-set approach || Beats 100% !! | 2-c-solutions-using-hash-map-and-hash-se-w7wc | Code\n\n// Soution 1 (HashMap)\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n unordered_map<string, int> mp;\n | prathams29 | NORMAL | 2023-07-19T09:19:26.712972+00:00 | 2023-07-19T09:19:26.713002+00:00 | 133 | false | # Code\n```\n// Soution 1 (HashMap)\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n unordered_map<string, int> mp;\n int ans = 0;\n for(auto i : words){\n string rev = i;\n reverse(rev.begin(), rev.end());\n if(mp[rev] > 0)\... | 2 | 0 | ['Hash Table', 'C++'] | 0 |
find-maximum-number-of-string-pairs | Easy c++ solution using linear search. | easy-c-solution-using-linear-search-by-j-y1pi | Intuition\n\n# Approach\nBrute Force Solution:\nUsing two loops firstly reverse the string of words and compare string of entire words.\n\n# Complexity\n- Time | jhasudarshan_07 | NORMAL | 2023-07-15T14:18:23.411509+00:00 | 2023-07-15T14:18:23.411537+00:00 | 288 | false | # Intuition\n\n# Approach\nBrute Force Solution:\nUsing two loops firstly reverse the string of words and compare string of entire words.\n\n# Complexity\n- Time complexity:O(n^2)\n\n- Space complexity:O(n)\n\n# Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n s... | 2 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Q2744 Accepted C++ ✅ HashMap O(n) | Easiest Method | q2744-accepted-c-hashmap-on-easiest-meth-m3au | \nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int n = words.size();\n int count = 0;\n unorder | adityasrathore | NORMAL | 2023-07-09T10:07:33.492173+00:00 | 2023-07-09T10:07:33.492193+00:00 | 176 | false | ```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int n = words.size();\n int count = 0;\n unordered_map <string,int> mp;\n for(int i=0;i<n;i++){\n if(words[i][0] > words[i][1]){\n swap(words[i][0],words[i][1]);\n ... | 2 | 0 | ['C'] | 2 |
find-maximum-number-of-string-pairs | ||EASY||SIMPLE||JAVA|| | easysimplejava-by-himakshikohli-2b46 | \n\n# Code\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int count =0;\n HashMap<String,Integer> arr = new Ha | himakshikohli | NORMAL | 2023-07-08T13:15:46.546040+00:00 | 2023-07-08T13:15:46.546067+00:00 | 627 | false | \n\n# Code\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int count =0;\n HashMap<String,Integer> arr = new HashMap<>();\n\n for(int i=0 ;i<words.length;i++){\n String r = new StringBuilder(words[i]).reverse().toString();\n if(arr.contai... | 2 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | Most easy c++ ode | most-easy-c-ode-by-ayush07khurana-m0sh | Code\n\nclass Solution \n{\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n for(int i=0;i<words.size();i++)\ | ayush07khurana | NORMAL | 2023-06-27T11:20:36.891793+00:00 | 2023-06-27T11:20:36.891826+00:00 | 45 | false | # Code\n```\nclass Solution \n{\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n for(int i=0;i<words.size();i++)\n {\n for(int j=i+1;j<words.size();j++)\n {\n string a=words[i];\n string b=words[j];\n ... | 2 | 0 | ['Array', 'String', 'C++'] | 0 |
find-maximum-number-of-string-pairs | Python3 Solution | python3-solution-by-motaharozzaman1996-6cud | \n\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n ans = 0\n for i in range(len(words)):\n for j | Motaharozzaman1996 | NORMAL | 2023-06-25T01:12:29.831282+00:00 | 2023-06-25T01:12:29.831311+00:00 | 1,107 | false | \n```\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n ans = 0\n for i in range(len(words)):\n for j in range(i+1, len(words)):\n if words[i][0] == words[j][1] and words[i][1] == words[j][0]:\n ans += 1\n return ans\... | 2 | 0 | ['Python', 'Python3'] | 0 |
find-maximum-number-of-string-pairs | Easy Bruteforce | easy-bruteforce-by-deveshpatel11-v3oh | \nclass Solution {\n public static int maximumNumberOfStringPairs(String[] words) {\n int ans=0;\n for(int i=0;i<words.length;i++){\n | kanishkpatel1 | NORMAL | 2023-06-24T16:37:49.077426+00:00 | 2023-06-24T16:37:49.077445+00:00 | 1,114 | false | ```\nclass Solution {\n public static int maximumNumberOfStringPairs(String[] words) {\n int ans=0;\n for(int i=0;i<words.length;i++){\n for(int j=i+1;j<words.length;j++){\n if(words[i].equals(reverse(words[j]))){\n // System.out.println(words[i]);\n ... | 2 | 0 | ['Array', 'Java'] | 0 |
find-maximum-number-of-string-pairs | [Python] easy using of set() | python-easy-using-of-set-by-pbelskiy-trqq | During this beweekly contest leetcode server is down again \uD83D\uDE2D\n\npython\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) - | pbelskiy | NORMAL | 2023-06-24T16:00:42.079119+00:00 | 2023-06-25T08:51:32.221297+00:00 | 87 | false | During this beweekly contest leetcode server is down again \uD83D\uDE2D\n\n```python\nclass Solution:\n def maximumNumberOfStringPairs(self, words: List[str]) -> int:\n v = set()\n \n for i in range(len(words)):\n for j in range(i + 1, len(words)):\n if words[i] == word... | 2 | 1 | [] | 0 |
find-maximum-number-of-string-pairs | Simple Python Solution | simple-python-solution-by-ascending-3ccb | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | ascending | NORMAL | 2025-04-06T12:49:07.698723+00:00 | 2025-04-06T12:49:07.698723+00:00 | 10 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Python3'] | 0 |
find-maximum-number-of-string-pairs | Beats 100% C++ Solution, Using Set. T.C = O(n) & S.C = O(n). | beats-100-c-solution-using-set-tc-on-sc-u1aqk | IntuitionUse of Set (Unordered Set)ApproachIterate through the vector from start to finish, checking each element to see if its reverse exists in the set. If th | yugshah7777 | NORMAL | 2025-04-01T10:57:01.433667+00:00 | 2025-04-01T10:57:01.433667+00:00 | 24 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Use of Set (Unordered Set)
# Approach
<!-- Describe your approach to solving the problem. -->
Iterate through the vector from start to finish, checking each element to see if its reverse exists in the set. If the reversed string is found, i... | 1 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | EASY TO UNDERSTAND | BEGINNER FRIENDLY | 3ms Runtime | Beats 100% 🥷🏻🔥 | easy-to-understand-beginner-friendly-3ms-36nu | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | tyagideepti9 | NORMAL | 2025-03-25T17:01:54.937130+00:00 | 2025-03-25T17:01:54.937130+00:00 | 54 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Array', 'Hash Table', 'String', 'Simulation', 'Java', 'C#', 'MS SQL Server'] | 0 |
find-maximum-number-of-string-pairs | Simple Approach for beginners | simple-approach-for-beginners-by-srinath-rvga | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Srinath_Y | NORMAL | 2025-02-14T15:08:39.724528+00:00 | 2025-02-14T15:08:39.724528+00:00 | 74 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['String', 'Python'] | 0 |
find-maximum-number-of-string-pairs | Beginner Friendly | Solution with approach🐐 | 0ms Beats 100% | beginner-friendly-solution-with-approach-7jsy | IntuitionHere we will take use of unordered_set as by doing this we can do this problem in one pass with help of it, else it will take nested loops that's not t | ujjwalBuilds | NORMAL | 2025-02-13T19:17:58.093721+00:00 | 2025-02-13T19:17:58.093721+00:00 | 79 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
Here we will take use of unordered_set as by doing this we can do this problem in one pass with help of it, else it will take nested loops that's not the perfect approach
# Approach
<!-- Describe your approach to solving the problem. -->
1... | 1 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | C++ solution using unordered set | c-solution-using-unordered-set-by-k_dadd-yfrp | Intuitionsets consists of unique elementsApproachtraverse on the vector, store the reverse of the element
if reversed element is present in set => count++
else | k_daddyyy | NORMAL | 2025-02-13T18:46:39.305643+00:00 | 2025-02-13T18:46:39.305643+00:00 | 52 | false | # Intuition
sets consists of unique elements
# Approach
traverse on the vector, store the reverse of the element
if reversed element is present in set => count++
else insert that element into the set
# Complexity
- Time complexity:
$$O(n)$$
- Space complexity:
$$O(n)$$
# Code
```cpp []
class Solution {
public:
... | 1 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | best and fast code | best-and-fast-code-by-diptanshu_888-zbaw | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | diptanshu_888 | NORMAL | 2025-02-12T19:32:30.705173+00:00 | 2025-02-12T19:32:30.705173+00:00 | 98 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Array', 'Hash Table', 'String', 'Simulation', 'Java'] | 0 |
find-maximum-number-of-string-pairs | Runtime 2 ms Beats 70.40% Memory 17.55 MB Beats 78.50% | runtime-2-ms-beats-7040-memory-1755-mb-b-0w65 | :):Code | ozodbek_bosimov | NORMAL | 2025-02-07T12:38:59.824549+00:00 | 2025-02-07T12:38:59.824549+00:00 | 73 | false | :):
# Code
```python3 []
class Solution:
def maximumNumberOfStringPairs(self, words: List[str]) -> int:
ans = 0
n = len(words)
for i in range(n):
for j in range(i+1,n):
if words[i][::-1] == words[j]:
ans +=1
continue
... | 1 | 0 | ['Python3'] | 0 |
find-maximum-number-of-string-pairs | Easy Solution with 100% beats in Java | easy-solution-with-100-beats-in-java-by-i1ilt | The problem is asking us to find the number of pairs of strings in the array words such that one string is the reverse of another. The brute-force approach woul | SANTHOSH_S_AIDS | NORMAL | 2025-01-28T08:02:03.559198+00:00 | 2025-01-28T08:02:03.559198+00:00 | 166 | false | The problem is asking us to find the number of pairs of strings in the array words such that one string is the reverse of another. The brute-force approach would involve checking every string and comparing it with the reversed form of every other string in the list. However, we can optimize this solution.
Approach
We ... | 1 | 0 | ['Array', 'Java'] | 1 |
find-maximum-number-of-string-pairs | simple c++ code 👻 | simple-c-code-by-varuntyagig-m52u | Complexity
Time complexity:
O(n2)
Space complexity:
O(1)
Code | varuntyagig | NORMAL | 2025-01-27T09:26:41.174990+00:00 | 2025-01-27T09:26:41.174990+00:00 | 63 | false | 
# Complexity
- Time complexity:
$$O(n^2)$$
- Space complexity:
$$O(1)$$
# Code
```cpp []
class Solution {
public:
int maximumNumberOfStringPairs(vector<string>& words) ... | 1 | 0 | ['Array', 'String', 'Counting', 'C++'] | 0 |
find-maximum-number-of-string-pairs | Simple C Solution | simple-c-solution-by-mukesh1855-afpz | IntuitionApproach
Write a function that takes two strings as paramter to check whether the two strings are palindrome or not.
Then pass the two words to the fun | mukesh1855 | NORMAL | 2025-01-20T16:50:01.555165+00:00 | 2025-01-20T16:50:01.555165+00:00 | 22 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
- Write a function that takes two strings as paramter to check whether the two strings are palindrome or not.
- Then pass the two words to the function to find the given string is palindrome, if it is palindrome then increase th... | 1 | 0 | ['C'] | 0 |
find-maximum-number-of-string-pairs | Count Maximum Reversed String Pairs in an Array | count-maximum-reversed-string-pairs-in-a-5bff | IntuitionTo solve this problem, the main idea is to identify pairs of strings where one is the reverse of the other. Since the words array consists of distinct | aljo00 | NORMAL | 2025-01-17T04:13:41.534627+00:00 | 2025-01-17T04:13:41.534627+00:00 | 52 | false | # Intuition
To solve this problem, the main idea is to identify pairs of strings where one is the reverse of the other. Since the `words` array consists of distinct strings, we can safely check all possible pairs without worrying about duplicates.
The most intuitive approach is to use two nested loops:
1. Fix one st... | 1 | 0 | ['JavaScript'] | 1 |
find-maximum-number-of-string-pairs | Simple answer using HashSet | simple-answer-using-hashset-by-saksham_g-4q99 | Complexity
Time complexity: O(n.m)
(n for loop and m for reverse)
Space complexity: O(n+m)
(n for Set and m for StringBuilder)
Code | Saksham_Gupta_ | NORMAL | 2025-01-03T11:11:53.877446+00:00 | 2025-01-03T11:11:53.877446+00:00 | 83 | false |
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: ***O(n.m)***
(n for loop and m for reverse)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: ***O(n+m)***
(n for Set and m for StringBuilder)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Cod... | 1 | 0 | ['Array', 'String', 'Java'] | 0 |
find-maximum-number-of-string-pairs | Find maximum number of string pairs -Easy solution for beginner --> | find-maximum-number-of-string-pairs-easy-vpu0 | Intuition\n\n\n# Approach\nreversing the string and placing it in the words array and checking it with variable a.\n\n# Complexity\n- Time complexity:\n\n\n- Sp | Codeia | NORMAL | 2024-08-27T00:55:55.563208+00:00 | 2024-08-27T00:55:55.563254+00:00 | 16 | false | # Intuition\n\n\n# Approach\nreversing the string and placing it in the words array and checking it with variable a.\n\n# Complexity\n- Time complexity:\n\n\n- Space complexity:\n\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int c = 0;\n fo... | 1 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Easy code | easy-code-by-shubhadip_009-g28w | 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 | Shubhadip_009 | NORMAL | 2024-08-22T18:22:24.187609+00:00 | 2024-08-22T18:22:24.187680+00:00 | 133 | 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 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | two_loops_check _ki_iska reverse == other elemnt and yes count++ | two_loops_check-_ki_iska-reverse-other-e-1npq | 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 | yesyesem | NORMAL | 2024-08-22T17:19:20.272311+00:00 | 2024-08-22T17:19:20.272336+00:00 | 40 | 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 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Beats 100% | beats-100-by-manyaagargg-jd0k | 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 | 2024-07-25T16:34:54.472637+00:00 | 2024-07-25T16:34:54.472668+00:00 | 10 | 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 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Easy to Understand || Solution by Navyendhu Menon | easy-to-understand-solution-by-navyendhu-h0wd | 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 | navyendhummenon | NORMAL | 2024-07-08T05:06:16.062519+00:00 | 2024-07-08T05:06:16.062550+00:00 | 7 | 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 | ['JavaScript'] | 0 |
find-maximum-number-of-string-pairs | js solution🔥 | js-solution-by-mhdsabeeh-zaby | 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 | MHDSabeeh | NORMAL | 2024-05-04T07:27:49.688968+00:00 | 2024-05-04T07:27:49.689015+00:00 | 182 | 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 | ['JavaScript'] | 0 |
find-maximum-number-of-string-pairs | Brute force + O(n) Hashmap solution | brute-force-on-hashmap-solution-by-jaime-ec9h | Brute Force\nSimple brute force method by computing the reverse of every single word and looking ahead the rest of the words to see if any of those match. This | jaimeloeuf | NORMAL | 2024-04-11T13:13:19.810671+00:00 | 2024-04-11T13:13:19.810702+00:00 | 19 | false | ### Brute Force\nSimple brute force method by computing the reverse of every single word and looking ahead the rest of the words to see if any of those match. This has a TC of O(n^2) where n is the number of elements in `words`, and a SC of O(1).\n\n```typescript\nfunction maximumNumberOfStringPairs(words: string[]): n... | 1 | 0 | ['Array', 'Hash Table', 'Simulation', 'TypeScript'] | 0 |
find-maximum-number-of-string-pairs | Simple java code 1 ms beats 100 % && 41 mb beats 89.25 % | simple-java-code-1-ms-beats-100-41-mb-be-pile | \n# Complexity\n- \n\n# Code\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int n = words.length;\n int count | Arobh | NORMAL | 2024-01-23T04:27:44.701168+00:00 | 2024-01-23T04:27:44.701187+00:00 | 4 | false | \n# Complexity\n- \n\n# Code\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int n = words.length;\n int count = 0;\n for (int i = 0; i < n; i++) {... | 1 | 0 | ['Array', 'String', 'Simulation', 'Java'] | 0 |
find-maximum-number-of-string-pairs | Simple Solution | simple-solution-by-adhilalikappil-1prs | 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 | adhilalikappil | NORMAL | 2024-01-20T04:53:27.573234+00:00 | 2024-01-20T04:53:27.573281+00:00 | 358 | 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 | ['JavaScript'] | 0 |
find-maximum-number-of-string-pairs | C++ || HashSet | c-hashset-by-luisfrodriguezr-1ing | \nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n unordered_set<string> hash_set;\n \n int ans = 0;\n for (au | ergodico | NORMAL | 2024-01-19T19:44:00.165943+00:00 | 2024-01-19T19:44:00.165970+00:00 | 0 | false | ```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n unordered_set<string> hash_set;\n \n int ans = 0;\n for (auto& word: words) {\n sort(begin(word), end(word));\n if (hash_set.find(word) != end(hash_set) ) {\n hash_set.erase(word);\n ans++;\n ... | 1 | 0 | [] | 0 |
find-maximum-number-of-string-pairs | Easy solution || C++ | easy-solution-c-by-garimapachori827-iv3y | Code\n\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n for(int i=0;i<words.size();i++){\n | garimapachori827 | NORMAL | 2023-12-13T16:16:55.983627+00:00 | 2023-12-13T16:16:55.983662+00:00 | 92 | false | # Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n int count=0;\n for(int i=0;i<words.size();i++){\n for(int j=i+1;j<words.size();j++){\n if(words[i]==palindrome(words[j])){\n count++;\n }\n ... | 1 | 0 | ['Array', 'String', 'C++'] | 0 |
find-maximum-number-of-string-pairs | EASY C++ SOLUTION WITH MINIMAL CONCEPTS | easy-c-solution-with-minimal-concepts-by-405g | EASY C++ SOLUTION WITH MINIMAL CONCEPTS\n\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to | codermal_ | NORMAL | 2023-11-08T13:13:52.462054+00:00 | 2023-11-08T13:13:52.462085+00:00 | 4 | false | **EASY C++ SOLUTION WITH MINIMAL CONCEPTS**\n\n\n# 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<!--... | 1 | 0 | ['C++'] | 0 |
find-maximum-number-of-string-pairs | Fcuk strrev | fcuk-strrev-by-parthapratimjana99-o9t7 | Intuition\n Describe your first thoughts on how to solve this problem. \nstrlen(string_name)\treturns the length of string name.\n2)\tstrcpy(destination, source | parthapratimjana99 | NORMAL | 2023-10-08T07:52:45.214225+00:00 | 2023-10-08T07:52:45.214247+00:00 | 299 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nstrlen(string_name)\treturns the length of string name.\n2)\tstrcpy(destination, source)\tcopies the contents of source string to destination string.\n3)\tstrcat(first_string, second_string)\tconcats or joins first string with second stri... | 1 | 0 | ['C'] | 1 |
find-maximum-number-of-string-pairs | JAVA SOLUTION || 100% FASTER || 1MS SOLUTION | java-solution-100-faster-1ms-solution-by-8wk9 | 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 | viper_01 | NORMAL | 2023-09-06T11:40:05.971361+00:00 | 2023-09-06T11:40:05.971382+00:00 | 5 | 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 |
find-maximum-number-of-string-pairs | Beats 95.59% / 64.74%. The most simple js solution | beats-9559-6474-the-most-simple-js-solut-z9v0 | \nvar maximumNumberOfStringPairs = function(words) {\n const set = new Set()\n let cnt = 0\n \n for (let word of words) {\n if (set.has(word. | tishka | NORMAL | 2023-08-23T11:31:42.213235+00:00 | 2023-08-23T11:31:42.213262+00:00 | 80 | false | ```\nvar maximumNumberOfStringPairs = function(words) {\n const set = new Set()\n let cnt = 0\n \n for (let word of words) {\n if (set.has(word.split(\'\').reverse().join(\'\'))) cnt++\n \n set.add(word)\n }\n \n return cnt\n};\n``` | 1 | 0 | ['JavaScript'] | 1 |
find-maximum-number-of-string-pairs | Rust/Python Linear time/space with explanation | rustpython-linear-timespace-with-explana-iz8u | Intuition\nStore the counts of all strings in the hashmap. Now iterate over all our strings and find their reversed strings. Check if it exists in a hashmap. \n | salvadordali | NORMAL | 2023-07-25T23:40:58.759144+00:00 | 2023-07-25T23:40:58.759192+00:00 | 45 | false | # Intuition\nStore the counts of all strings in the hashmap. Now iterate over all our strings and find their reversed strings. Check if it exists in a hashmap. \n\nNow we have two possible options:\n - reverse string is equal to our original string. To be able to find a reverse we need to have at least two values in a ... | 1 | 0 | ['Python3', 'Rust'] | 0 |
find-maximum-number-of-string-pairs | Java Easy to Understand Solution Using HashSet | java-easy-to-understand-solution-using-h-3iph | Code\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int count = 0;\n\t\tSet<String> ans = new HashSet<>();\n\t\tfor ( | brothercode | NORMAL | 2023-07-22T17:58:28.183301+00:00 | 2023-07-22T18:01:50.860433+00:00 | 5 | false | # Code\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n int count = 0;\n\t\tSet<String> ans = new HashSet<>();\n\t\tfor (String str : words) {\n\t\t\tchar ch[] = str.toCharArray();\n\t\t\tArrays.sort(ch); // Sort the Characters of each string.\n\t\t\tif (!ans.add(String.value... | 1 | 0 | ['Java'] | 0 |
find-maximum-number-of-string-pairs | [JAVA] easy solution without StringBuilder/Map | java-easy-solution-without-stringbuilder-y18x | Intuition\n Describe your first thoughts on how to solve this problem. \nIn the entire array words count the number of times a word and its reverse is present i | Jugantar2020 | NORMAL | 2023-07-18T18:03:41.907219+00:00 | 2023-07-18T18:03:41.907244+00:00 | 370 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIn the entire array words count the number of times a word and its reverse is present in the array\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nfor all the words check if the first letter of a word equals the se... | 1 | 0 | ['Array', 'String', 'Simulation', 'Java'] | 1 |
find-maximum-number-of-string-pairs | Kotlin 1 line | kotlin-1-line-by-georgcantor-iwp8 | \nfun maximumNumberOfStringPairs(a: Array<String>) = a.groupingBy { it.toSortedSet() }.eachCount().count { it.value > 1 }\n | GeorgCantor | NORMAL | 2023-07-16T10:58:51.919407+00:00 | 2023-07-16T12:19:55.416481+00:00 | 9 | false | ```\nfun maximumNumberOfStringPairs(a: Array<String>) = a.groupingBy { it.toSortedSet() }.eachCount().count { it.value > 1 }\n``` | 1 | 0 | ['Kotlin'] | 0 |
find-maximum-number-of-string-pairs | One line solution | one-line-solution-by-likils-gaul | Code\n\nclass Solution {\n func maximumNumberOfStringPairs(_ words: [String]) -> Int {\n return words.count - Set(words.map(Set.init)).count\n }\n} | likils | NORMAL | 2023-07-14T07:55:08.488844+00:00 | 2023-07-14T07:55:08.488870+00:00 | 38 | false | # Code\n```\nclass Solution {\n func maximumNumberOfStringPairs(_ words: [String]) -> Int {\n return words.count - Set(words.map(Set.init)).count\n }\n}\n``` | 1 | 0 | ['Swift'] | 1 |
find-maximum-number-of-string-pairs | Simple PHP solution | simple-php-solution-by-vlados117-ycah | Complexity\n- Time complexity: O(n)\n Add your time complexity here, e.g. O(n) \n\n\n# Code\n\nclass Solution {\n\n /**\n * @param String[] $words\n | Vlados117 | NORMAL | 2023-07-14T07:47:08.622600+00:00 | 2023-07-14T07:51:12.011558+00:00 | 15 | false | # Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n# Code\n```\nclass Solution {\n\n /**\n * @param String[] $words\n * @return Integer\n */\n function maximumNumberOfStringPairs($words) {\n //Here we\'ll keep words as array keys\n $hash... | 1 | 0 | ['PHP'] | 0 |
find-maximum-number-of-string-pairs | Java || Simple Solution ✅ || Rutime - 100% 💥 || Memory - 94% 💥 || With Explanation ✅ | java-simple-solution-rutime-100-memory-9-ljni | Approach\n\n1. Initialize two variables: n to store the length of the words array and cnt to keep track of the count of valid pairs.\n2. Use nested loops to ite | akobirswe | NORMAL | 2023-07-11T05:34:29.553852+00:00 | 2023-07-11T05:34:29.553869+00:00 | 379 | false | # Approach\n\n1. Initialize two variables: `n` to store the length of the `words` array and `cnt` to keep track of the count of valid pairs.\n2. Use nested loops to iterate over all possible pairs of strings in the `words` array. The outer loop variable `i` represents the index of the first word, and the inner loop var... | 1 | 0 | ['Array', 'String', 'Simulation', 'Java', 'C#'] | 0 |
find-maximum-number-of-string-pairs | C++ | Easy Solution Using Unordered Set | c-easy-solution-using-unordered-set-by-s-tx4j | Complexity\n- Time complexity: O(N)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(N)\n Add your space complexity here, e.g. O(n) \n\n# Co | shweta0098 | NORMAL | 2023-07-08T10:15:40.599402+00:00 | 2023-07-08T10:15:40.599428+00:00 | 17 | false | # Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(N)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int maximumNumberOfStringPairs(vector<string>& words) {\n unordered_set<string> s;\n ... | 1 | 0 | ['Hash Table', 'C++'] | 0 |
find-maximum-number-of-string-pairs | ✅👌JAVA | 2 DIFFERENT APPROACH | EASY | java-2-different-approach-easy-by-dipesh-d691 | 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 | dipesh_12 | NORMAL | 2023-07-07T14:46:33.593289+00:00 | 2023-07-07T14:46:33.593317+00:00 | 703 | 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 |
find-maximum-number-of-string-pairs | 📌Solution with shift() method | solution-with-shift-method-by-m3f-smq1 | Intuition\n Describe your first thoughts on how to solve this problem. \n1. Remove the first item from the array words and returns that removed item;\n2. Reverc | M3f | NORMAL | 2023-06-29T11:34:53.198083+00:00 | 2023-06-29T11:34:53.198112+00:00 | 235 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n1. Remove the first `item` from the array `words` and returns that removed `item`;\n2. Reverce letters in removed `item`;\n3. Determine whether the array `words` includes a reverce `item` among its entries;\n4. Repeat until the array `wor... | 1 | 0 | ['JavaScript'] | 0 |
find-maximum-number-of-string-pairs | [scala] - make pairs by indices, count | scala-make-pairs-by-indices-count-by-nik-2mwk | Code\n\nobject Solution {\n def maximumNumberOfStringPairs(words: Array[String]): Int = {\n val counts =\n words.indices.map { i =>\n val revers | nikiforo | NORMAL | 2023-06-29T09:18:06.001054+00:00 | 2023-06-29T09:18:06.001087+00:00 | 16 | false | # Code\n```\nobject Solution {\n def maximumNumberOfStringPairs(words: Array[String]): Int = {\n val counts =\n words.indices.map { i =>\n val reversed = words(i).reverse\n words.iterator.drop(i + 1).count(_ == reversed)\n }\n counts.sum\n }\n}\n``` | 1 | 0 | ['Scala'] | 0 |
find-maximum-number-of-string-pairs | JAVA | HashSet | java-hashset-by-sourin_bruh-k6iq | Solution:\n\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n Set<String> set = new HashSet<>();\n int cnt = 0;\n | sourin_bruh | NORMAL | 2023-06-28T13:30:00.172784+00:00 | 2023-06-28T13:30:00.172818+00:00 | 223 | false | # Solution:\n```\nclass Solution {\n public int maximumNumberOfStringPairs(String[] words) {\n Set<String> set = new HashSet<>();\n int cnt = 0;\n for (String s : words) {\n StringBuilder sb = new StringBuilder(s).reverse();\n if (set.contains(sb.toString())) {\n ... | 1 | 0 | ['Array', 'Hash Table', 'String', 'Java'] | 0 |
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