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maximum-candies-allocated-to-k-children | Binary Search, C++ With Explanation ✅✅ | binary-search-c-with-explanation-by-deep-3j1t | \n# Approach\n Describe your approach to solving the problem. \nHere i apply Binary Search Algorithm to find out Maximum Number of candies.\n\nMaximum Number of | Deepak_5910 | NORMAL | 2023-06-05T16:45:51.170565+00:00 | 2023-06-05T16:47:29.299886+00:00 | 129 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nHere i apply **Binary Search Algorithm** to find out Maximum Number of candies.\n\nMaximum Number of Candies Always Belong to range **[1 , Maximum of array]** , number of Maximum Candies can be any Element from this sorted array that satisfy the giv... | 2 | 0 | ['Binary Search', 'C++'] | 0 |
maximum-candies-allocated-to-k-children | using binary search on answer , explanation | using-binary-search-on-answer-explanatio-e3jt | instead of applying binary search directly on array we will apply binary search on the range of the answer.\n\nlow = 1 because there can be minimum 1 candy in t | AbhayDutt | NORMAL | 2023-04-01T09:55:27.451852+00:00 | 2023-04-01T09:55:27.451892+00:00 | 983 | false | instead of applying binary search directly on array we will apply binary search on the range of the answer.\n\nlow = 1 because there can be minimum 1 candy in the array\nhigh = maximum in the array because that is the maaximum amount of candies that can be present in the pile\n\n\nnow we want to validate each mid that ... | 2 | 0 | ['Binary Search', 'Java'] | 1 |
maximum-candies-allocated-to-k-children | C++ || Binary Search on Search Space || Easy Solution | c-binary-search-on-search-space-easy-sol-6npp | Time Complexity : O(n*logm) , where m is the maximum number of candies in the single pile.\n\nclass Solution {\nprivate:\n bool isValid(vector<int>& candies, | aniketk2k | NORMAL | 2023-03-08T10:45:20.341807+00:00 | 2023-03-08T11:32:47.681824+00:00 | 872 | false | **Time Complexity : O(n*logm)** , where m is the maximum number of candies in the single pile.\n```\nclass Solution {\nprivate:\n bool isValid(vector<int>& candies, long long k, int max_candies){\n long long child_cnt = 0;\n \n for(auto candies: candies)\n child_cnt += candies/max_can... | 2 | 0 | ['C', 'Binary Tree', 'C++'] | 1 |
maximum-candies-allocated-to-k-children | ✅✅Easy Java code|| Beats 99%||✅✅ | easy-java-code-beats-99-by-princebharti-zfmp | \n# Approach\n Describe your approach to solving the problem. \nFirst find the range of candies to be distributed then,\nUse Binary Search to find the maximum n | PRINCEBharti | NORMAL | 2023-03-07T14:55:04.602965+00:00 | 2023-03-07T14:55:04.603011+00:00 | 566 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nFirst find the range of candies to be distributed then,\nUse Binary Search to find the maximum number of candies distributed.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n$$O(nlogn)$$\n- Space complexit... | 2 | 0 | ['Java'] | 0 |
maximum-candies-allocated-to-k-children | Java Sol || Easy || Binary Search || Simple | java-sol-easy-binary-search-simple-by-hi-d2ru | Code\n\nclass Solution {\n public int maximumCandies(int[] candies, long k) {\n int low = 1;\n int high = Arrays.stream(candies).max().getAsInt | himanshu_gupta_ | NORMAL | 2023-02-07T20:45:06.704385+00:00 | 2023-02-07T20:45:06.704412+00:00 | 624 | false | # Code\n```\nclass Solution {\n public int maximumCandies(int[] candies, long k) {\n int low = 1;\n int high = Arrays.stream(candies).max().getAsInt();\n int ans = 0;\n while (low<=high) {\n int mid = low + (high-low)/2;\n long currAns = 0;\n for (int i : ... | 2 | 0 | ['Java'] | 0 |
maximum-candies-allocated-to-k-children | Binary Search method | binary-search-method-by-optimal_ns65-3v91 | class Solution {\npublic:\n bool allocation(vector& candies,long long k,long long mid){\n if(mid==0) return true;\n \n long long q=0;\n | Optimal_NS65 | NORMAL | 2022-04-03T18:29:07.665170+00:00 | 2022-04-03T18:29:07.665219+00:00 | 24 | false | class Solution {\npublic:\n bool allocation(vector<int>& candies,long long k,long long mid){\n if(mid==0) return true;\n \n long long q=0;\n long long count=0;\n for(int i=0;i<candies.size();i++){\n int q=candies[i]/mid;\n count+=q;\n if(count>=k){... | 2 | 0 | [] | 0 |
maximum-candies-allocated-to-k-children | C++ | Simple Binary Search Solution O(n*log(max(candies))) | c-simple-binary-search-solution-onlogmax-xcre | \nclass Solution {\npublic:\n bool check(vector<int>& arr,int mid, long long k){\n long long count = 0;\n for(int i : arr){\n count | LastStyleBender | NORMAL | 2022-04-03T16:13:07.879198+00:00 | 2022-04-03T16:13:07.879243+00:00 | 74 | false | ```\nclass Solution {\npublic:\n bool check(vector<int>& arr,int mid, long long k){\n long long count = 0;\n for(int i : arr){\n count += (i/mid);\n }\n \n return (count >= k);\n }\n \n int maximumCandies(vector<int>& arr, long long k) {\n int lo = 1, hi ... | 2 | 0 | ['C', 'Binary Tree'] | 1 |
maximum-candies-allocated-to-k-children | C++ || Binary Search | c-binary-search-by-kr_sk_01_in-u8z5 | \n bool is_possible(vector<int>& candies, int mid, long long k)\n {\n long long count = 0;\n \n for(int i = 0; i < candies.size(); i++) | KR_SK_01_In | NORMAL | 2022-04-03T07:27:30.530943+00:00 | 2022-04-03T07:27:30.530972+00:00 | 160 | false | ```\n bool is_possible(vector<int>& candies, int mid, long long k)\n {\n long long count = 0;\n \n for(int i = 0; i < candies.size(); i++)\n {\n count += candies[i] / mid;\n }\n \n if(count >= k)\n return true;\n \n return false;\n... | 2 | 0 | ['C', 'Binary Tree', 'C++'] | 0 |
maximum-candies-allocated-to-k-children | C++ Binary Search Solution | c-binary-search-solution-by-mayanksamadh-2bno | Please Upvote If It Helps\n\nclass Solution {\npublic:\n bool valid(int mid,vector<int>& candies,long long k)\n {\n // it will containg the count t | mayanksamadhiya12345 | NORMAL | 2022-04-03T06:41:35.741340+00:00 | 2022-04-03T06:42:00.450260+00:00 | 56 | false | **Please Upvote If It Helps**\n```\nclass Solution {\npublic:\n bool valid(int mid,vector<int>& candies,long long k)\n {\n // it will containg the count that will tell we can assign or not assign candies to children \n long sum = 0;\n for(auto it : candies)\n {\n sum += (it/... | 2 | 0 | [] | 0 |
maximum-candies-allocated-to-k-children | Java | Binary Search | Easy To Understand | java-binary-search-easy-to-understand-by-zrap | Here I am discussing my approach to this problem\nApproach:\n1. First calculate total number of candies. Let it be sum.\n2. If sum < k, return 0 we have more ch | SudhanshuMallick | NORMAL | 2022-04-03T04:17:21.909428+00:00 | 2022-04-03T04:17:21.909472+00:00 | 181 | false | **Here I am discussing my approach to this problem**\n**Approach:**\n1. First calculate **total number** of candies. Let it be `sum`.\n2. If `sum < k`, **return 0** we have **more children than candies**.\n3. If `sum = k`, **return 1** as we have** as many children as candies**.\n4. Else, perform **binary search** on *... | 2 | 0 | ['Binary Tree', 'Java'] | 0 |
maximum-candies-allocated-to-k-children | 🎁C++ Binary Search Easy Explaination Code | c-binary-search-easy-explaination-code-b-6hnt | \n\nclass Solution\n{\npublic:\n long long int maxPiles(vector<int> candies, long long int k)\n {\n long long int l = 1; // We can select an minimu | omhardaha1 | NORMAL | 2022-04-03T04:13:28.672021+00:00 | 2022-04-03T04:15:57.211359+00:00 | 66 | false | ```\n\nclass Solution\n{\npublic:\n long long int maxPiles(vector<int> candies, long long int k)\n {\n long long int l = 1; // We can select an minimum 1 element\n long long int h = 0;\n for (auto i : candies)\n {\n h += i;\n }\n if (k > h)\n {\n ... | 2 | 0 | ['C', 'Binary Tree', 'C++'] | 0 |
maximum-candies-allocated-to-k-children | [CPP] Binary Search || Easy to Understand | cpp-binary-search-easy-to-understand-by-74725 | \nclass Solution {\npublic:\n //checks if each children can get C candies\n bool isPossible(vector<int>& candies, long long C, long long k) {\n lon | arsator | NORMAL | 2022-04-03T04:02:04.812294+00:00 | 2022-04-03T04:02:04.812341+00:00 | 67 | false | ```\nclass Solution {\npublic:\n //checks if each children can get C candies\n bool isPossible(vector<int>& candies, long long C, long long k) {\n long long cnt = 0;\n for(auto c : candies) {\n cnt += (long long)(c/C);\n }\n \n return cnt >= k;\n }\n \n //Bin... | 2 | 0 | [] | 0 |
maximum-candies-allocated-to-k-children | 🧠 Greedy Candy Splitting: The Binary Search Trick Everyone Misses! | greedy-candy-splitting-the-binary-search-yqp6 | IntuitionWe want to split piles of candies into exactly k children — each getting the maximum equal amount possible.
Trying all possible values is too slow. But | loginov-kirill | NORMAL | 2025-03-30T07:01:14.353214+00:00 | 2025-04-01T18:55:04.384759+00:00 | 1,140 | false |
### Intuition
We want to split piles of candies into exactly `k` children — each getting the **maximum equal amount possible**.
Trying all possible values is too slow. But what if we reverse the problem and ask:
**“Can we give each child `x` candies?”**
This naturally leads us to **binary search**!
---
### Ap... | 1 | 0 | ['Array', 'Math', 'Binary Search', 'Heap (Priority Queue)', 'Python', 'JavaScript'] | 1 |
count-prefixes-of-a-given-string | [Java/C++/Python] Starts With | javacpython-starts-with-by-lee215-lw6y | Explanation\nfor each word w in words list,\ncheck if word w startsWith the string s\n\n\n# Complexity\nTime O(NS)\nSpace O(1)\n\n\nJava\njava\n public int c | lee215 | NORMAL | 2022-04-30T16:16:13.275791+00:00 | 2022-05-03T06:43:22.216403+00:00 | 6,395 | false | # **Explanation**\nfor each word `w` in `words` list,\ncheck if word `w` `startsWith` the string `s`\n<br>\n\n# **Complexity**\nTime `O(NS)`\nSpace `O(1)`\n<br>\n\n**Java**\n```java\n public int countPrefixes(String[] words, String s) {\n int res = 0;\n for (String w : words)\n if (s.startsW... | 46 | 0 | ['C', 'Python', 'Java'] | 12 |
count-prefixes-of-a-given-string | C++ || Easy O(NxS) solution || String | c-easy-onxs-solution-string-by-yash2arma-3lb0 | \nclass Solution {\npublic:\n //iterate over each words[i] and take the sub-string str from s of size equals to words[i].\n //and compare str with words[i | Yash2arma | NORMAL | 2022-04-30T16:13:17.949594+00:00 | 2022-05-01T07:29:15.665748+00:00 | 2,487 | false | ```\nclass Solution {\npublic:\n //iterate over each words[i] and take the sub-string str from s of size equals to words[i].\n //and compare str with words[i] if both are equal increase count by 1.\n int countPrefixes(vector<string>& words, string s) \n {\n int count=0;\n for(auto it: words)\n... | 28 | 0 | ['String', 'C', 'Iterator', 'C++'] | 3 |
count-prefixes-of-a-given-string | Java 1-Liner | java-1-liner-by-fllght-z8l9 | java\npublic int countPrefixes(String[] words, String s) {\n return (int) Arrays.stream(words).filter(s::startsWith).count();\n }\n\n\nMy repositories | FLlGHT | NORMAL | 2022-05-01T08:12:39.278135+00:00 | 2023-05-12T05:22:01.753813+00:00 | 1,724 | false | ```java\npublic int countPrefixes(String[] words, String s) {\n return (int) Arrays.stream(words).filter(s::startsWith).count();\n }\n```\n\nMy repositories with leetcode problems solving - [Java](https://github.com/FLlGHT/algorithms/tree/master/j-algorithms/src/main/java), [C++](https://github.com/FLlGHT/al... | 27 | 0 | ['Java'] | 1 |
count-prefixes-of-a-given-string | Easy python solution | easy-python-solution-by-tusharkhanna575-rxrk | \nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count=0\n for i in words:\n if (s[:len(i)]==i):\n | tusharkhanna575 | NORMAL | 2022-05-26T14:42:38.412749+00:00 | 2022-05-26T14:42:38.412777+00:00 | 1,521 | false | ```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count=0\n for i in words:\n if (s[:len(i)]==i):\n count+=1\n return count\n``` | 15 | 0 | ['Python', 'Python3'] | 1 |
count-prefixes-of-a-given-string | C++ || EASY TO UNDERSTAND || Simple Solution | c-easy-to-understand-simple-solution-by-drkqu | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) \n {\n int counter = 0;\n \n for(int i=0;i<words.siz | mayanksamadhiya12345 | NORMAL | 2022-04-30T16:10:41.260290+00:00 | 2022-05-01T16:15:53.055127+00:00 | 1,085 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) \n {\n int counter = 0;\n \n for(int i=0;i<words.size();i++)\n {\n if(words[i] == s.substr(0,words[i].size())) \n counter++;\n }\n\n return counter;\n }\n};\n... | 11 | 0 | [] | 2 |
count-prefixes-of-a-given-string | count_if | count_if-by-votrubac-cb8s | C++\ncpp\nint countPrefixes(vector<string>& words, string s) {\n return count_if(begin(words), end(words), [&](const auto &w){ return s.compare(0, w.size(), | votrubac | NORMAL | 2022-05-01T22:06:35.443879+00:00 | 2022-05-03T03:55:36.094024+00:00 | 798 | false | **C++**\n```cpp\nint countPrefixes(vector<string>& words, string s) {\n return count_if(begin(words), end(words), [&](const auto &w){ return s.compare(0, w.size(), w) == 0; });\n}\n``` | 8 | 0 | ['C'] | 3 |
count-prefixes-of-a-given-string | Python simple solution | python-simple-solution-by-tovam-xyjb | Approach\nCheck for each word if is prefix of s.\n\n# Code\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n res = 0\n\ | TovAm | NORMAL | 2023-01-05T15:28:38.090957+00:00 | 2023-01-05T15:28:38.090995+00:00 | 1,139 | false | # Approach\nCheck for each word if is prefix of s.\n\n# Code\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n res = 0\n\n for word in words:\n if s.startswith(word):\n res += 1\n\n return res\n```\n\nLike it? Please upvote! | 7 | 0 | ['Python3'] | 1 |
count-prefixes-of-a-given-string | Java Easy to Understand Solution | java-easy-to-understand-solution-by-janh-wflx | \nclass Solution {\n public int countPrefixes(String[] words, String s) {\n \n int result = 0;\n for (int i=0; i<=s.length(); i++) { | Janhvi__28 | NORMAL | 2022-10-20T18:06:56.439131+00:00 | 2022-10-20T18:06:56.439175+00:00 | 814 | false | ```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n \n int result = 0;\n for (int i=0; i<=s.length(); i++) {\n String x = s.substring(0, i);\n for (int j = 0; j<words.length; j++) {\n if (x.equals(words[j]))\n ... | 7 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | ✅✅✅Straightforward Javascript Solution || Faster than 96.67% | straightforward-javascript-solution-fast-95qp | \nvar countPrefixes = function(words, s) {\n let count = 0;\n for (let word of words) {\n if (s.startsWith(word)) count++\n } return count;\n};\ | a-fr0stbite | NORMAL | 2022-05-28T17:24:05.888742+00:00 | 2022-08-02T00:42:34.923627+00:00 | 459 | false | ```\nvar countPrefixes = function(words, s) {\n let count = 0;\n for (let word of words) {\n if (s.startsWith(word)) count++\n } return count;\n};\n```\n\n | 7 | 0 | ['JavaScript'] | 1 |
count-prefixes-of-a-given-string | ✅ 2 Lines || C++ / java | 2-lines-c-java-by-xxvvpp-g6gk | Apply find() in case of c++ and indexof() in case of java on every word and check if return position is 0.\nReturn the count of such strings.\n# C++\n int co | xxvvpp | NORMAL | 2022-04-30T16:15:05.509693+00:00 | 2022-04-30T16:21:12.974176+00:00 | 975 | false | Apply **find()** in case of c++ and **indexof()** in case of java on every word and check if return position is 0.\nReturn the count of such strings.\n# C++\n int countPrefixes(vector<string>& words, string s, int cnt=0){\n for(auto i:words) if(s.find(i)==0) cnt++;\n return cnt;\n }\n\t\n# Java\n ... | 7 | 0 | ['C', 'Java'] | 3 |
count-prefixes-of-a-given-string | C++ || Basic string understanding | c-basic-string-understanding-by-shm_47-4r2k | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int cnt = 0;\n int i, j;\n for(i=0; i<words.size(); | shm_47 | NORMAL | 2022-04-30T16:11:13.336732+00:00 | 2022-04-30T16:11:13.336771+00:00 | 555 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int cnt = 0;\n int i, j;\n for(i=0; i<words.size(); i++){\n j = 0;\n while(j< words[i].size()){\n if(words[i][j] != s[j])\n break;\n j++;... | 7 | 0 | ['C'] | 0 |
count-prefixes-of-a-given-string | C++ Easy Solution using Only One For loop and One If Condition | c-easy-solution-using-only-one-for-loop-djv3c | Intuition\nThe intuition for the problem is to iterate through each word in the words array, extract a substring from s that matches the length of the word, and | _sxrthakk | NORMAL | 2024-06-29T12:16:30.243764+00:00 | 2024-06-29T12:16:30.243794+00:00 | 280 | false | # Intuition\nThe intuition for the problem is to iterate through each word in the words array, extract a substring from s that matches the length of the word, and compare it with the word itself. Increment a counter each time a match is found, returning the final count of matching prefixes.\n\n# Approach\n**Initialize ... | 6 | 0 | ['String', 'C++'] | 1 |
count-prefixes-of-a-given-string | 3 different fastest approaches with javascript - including one-liner | 3-different-fastest-approaches-with-java-6yk0 | // approach 1 \n\n\nvar countPrefixes = function(words, s) {\n let counter = 0\n for (let i = 0; i < words.length; i++) {\n if (words[i]) === s. | Apoorv-123 | NORMAL | 2022-08-22T00:15:17.242709+00:00 | 2022-08-22T00:22:25.549321+00:00 | 679 | false | // approach 1 \n\n```\nvar countPrefixes = function(words, s) {\n let counter = 0\n for (let i = 0; i < words.length; i++) {\n if (words[i]) === s.slice(0,words[i].length)) {\n counter++\n }\n }\n return counter\n};\n```\n\n// one liner approach - 2\n\n```\nvar countPrefixes = f... | 6 | 0 | ['JavaScript'] | 1 |
count-prefixes-of-a-given-string | JS easiest way | js-easiest-way-by-andrezzz-acee | \nvar countPrefixes = function(words, s) {\n \n let cont = 0;\n \n for(i = 0; i < words.length; i++){\n for(j = 1; j <= s.length; j++){\n | andrezzz | NORMAL | 2022-05-04T18:10:02.585610+00:00 | 2022-05-04T18:10:02.585649+00:00 | 547 | false | ```\nvar countPrefixes = function(words, s) {\n \n let cont = 0;\n \n for(i = 0; i < words.length; i++){\n for(j = 1; j <= s.length; j++){\n if(words[i] == s.slice(0, j)){\n cont++;\n }\n } \n }\n return cont;\n \n};\n``` | 6 | 0 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | Python straightforward with startswith. | python-straightforward-with-startswith-b-uw5f | \ndef countPrefixes(self, words: List[str], s: str) -> int:\n\n c = 0\n for w in words:\n if s.startswith(w):\n c += 1\n\n return c\n | vineeth_moturu | NORMAL | 2022-04-30T16:08:57.925966+00:00 | 2022-04-30T16:08:57.926001+00:00 | 588 | false | ```\ndef countPrefixes(self, words: List[str], s: str) -> int:\n\n c = 0\n for w in words:\n if s.startswith(w):\n c += 1\n\n return c\n``` | 6 | 0 | ['Python'] | 3 |
count-prefixes-of-a-given-string | Beats 100% of users with C++ || Step By Step Explain || Easy to Understand || | beats-100-of-users-with-c-step-by-step-e-bcxc | Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it....\n\n---\n\n# Intuition\n Describe your first thoughts on how to solve this problem | abhirajpratapsingh | NORMAL | 2024-02-05T15:29:05.835690+00:00 | 2024-02-05T15:29:05.835725+00:00 | 229 | false | # Abhiraj Pratap Singh\n\n---\n\n# if you like the solution please UPVOTE it....\n\n---\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- The problem involves counting the number of words in a vector that have a specific prefix equal to a given string s.\n\n---\n\n![Screenshot (47).... | 5 | 0 | ['Array', 'String', 'String Matching', 'C++'] | 1 |
count-prefixes-of-a-given-string | Java || 100% beats || 2 solutions || ❤️❤️ | java-100-beats-2-solutions-by-saurabh_ku-kjc1 | 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 | saurabh_kumar1 | NORMAL | 2023-08-24T18:56:09.555301+00:00 | 2023-08-24T18:56:09.555328+00:00 | 479 | 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:0(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:0(1)\n<!-- Add your space complexity here, e.g. $$O... | 5 | 0 | ['String', 'String Matching', 'Java'] | 0 |
count-prefixes-of-a-given-string | C++ | 3-Line solution | c-3-line-solution-by-vartika2207-wlj1 | \tclass Solution {\n\tpublic:\n\t\t//time:O(n), space: O(1)\n\t\tint countPrefixes(vector& words, string s) {\n\t\t\tint count = 0;\n\t\t\tfor(auto wd : words)\ | vartika2207 | NORMAL | 2022-11-30T22:07:23.519381+00:00 | 2022-11-30T22:07:23.519415+00:00 | 697 | false | \tclass Solution {\n\tpublic:\n\t\t//time:O(n), space: O(1)\n\t\tint countPrefixes(vector<string>& words, string s) {\n\t\t\tint count = 0;\n\t\t\tfor(auto wd : words)\n\t\t\t\tif(s.find(wd) < 1) count++; //if first occurrence of sub-string is in the specified string\n\t\t\treturn count;\n\t\t}\n\t}; | 5 | 0 | ['C', 'C++'] | 2 |
count-prefixes-of-a-given-string | C++ Simple Solution using substr() | c-simple-solution-using-substr-by-prasad-uxjd | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int n = words.size();\n int cnt = 0;\n for(int i=0; | Prasad264 | NORMAL | 2022-04-30T16:35:11.347911+00:00 | 2022-04-30T16:35:11.347949+00:00 | 426 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int n = words.size();\n int cnt = 0;\n for(int i=0; i<n; i++) {\n string str = words[i];\n if(str == s.substr(0, str.size())) cnt++;\n }\n return cnt;\n }\n};\n``` | 5 | 0 | ['String', 'C'] | 3 |
count-prefixes-of-a-given-string | Java || Easy Appraoch | java-easy-appraoch-by-ashwinj_984-orxi | \nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int i = 0;\n int j = 0;\n int count = 0;\n for(int k | ashwinj_984 | NORMAL | 2022-04-30T16:08:54.194604+00:00 | 2022-04-30T16:08:54.194656+00:00 | 768 | false | ```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int i = 0;\n int j = 0;\n int count = 0;\n for(int k = 0; k < words.length; k++){\n if(words[k].length() > s.length()){\n continue;\n }\n \n while(i < ... | 5 | 0 | ['String', 'Java'] | 0 |
count-prefixes-of-a-given-string | Best Solution Using set || Easy to Understand | best-solution-using-set-easy-to-understa-d8fu | Approach\nStep 1 : Define a map and empty stirng\nStep2 : insert prefixes in map\nStep3 : check how many array of words element is present in map\n\n\n\nclass S | ParthBabbar | NORMAL | 2022-04-30T16:06:30.606400+00:00 | 2022-04-30T16:15:36.091331+00:00 | 283 | false | Approach\nStep 1 : Define a map and empty stirng\nStep2 : insert prefixes in map\nStep3 : check how many array of words element is present in map\n\n\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n unordered_set<string> mp;\n string st = "";\n for(int i=0... | 5 | 0 | ['String', 'C'] | 0 |
count-prefixes-of-a-given-string | python3 code | python3-code-by-blackhole_tiub-c62q | Code\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return len([word for word in words if s.startswith(word)])\n\n\n | Little_Tiub | NORMAL | 2022-12-28T23:53:42.204520+00:00 | 2023-05-02T05:51:06.502800+00:00 | 271 | false | # Code\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return len([word for word in words if s.startswith(word)])\n```\n\n\n\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(1)\n Add your space comple | diyordev | NORMAL | 2023-12-07T16:10:53.721069+00:00 | 2023-12-07T16:10:53.721101+00:00 | 191 | false | # Please UPVOTE if helps.\n# Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(1)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int countPrefixes(String[] words, String s) {;\n int c... | 3 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | Java Solution|| Beats 100%💯 || 3 line code🔥 | java-solution-beats-100-3-line-code-by-v-cqx1 | \n# Approach\n Describe your approach to solving the problem. \n- The startsWith method returns true if the string it is called on starts with the provided pref | Vishu6403 | NORMAL | 2023-10-09T09:11:03.521445+00:00 | 2023-10-09T09:11:03.521471+00:00 | 151 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n- The startsWith method returns true if the string it is called on starts with the provided prefix (string s), otherwise it returns false.\n\n# Complexity\n- Time complexity:$$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space ... | 3 | 0 | ['Array', 'String', 'Java'] | 1 |
count-prefixes-of-a-given-string | one line solution | one-line-solution-by-yuliia_anatoliivna-nvsv | \n\n# Code\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return sum(1 for x in words if s.startswith(x))\n | Yuliia_Anatoliivna | NORMAL | 2023-05-31T17:17:11.986423+00:00 | 2023-05-31T17:17:11.986471+00:00 | 298 | false | \n\n# Code\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return sum(1 for x in words if s.startswith(x))\n``` | 3 | 0 | ['Python3'] | 1 |
count-prefixes-of-a-given-string | Simple | Easy | (100%) JAVA Solution | simple-easy-100-java-solution-by-pranavi-cwy0 | \n\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count=0;\n for(String word: words){\n if(s.startsW | pranavibhute | NORMAL | 2023-05-17T12:05:54.210279+00:00 | 2023-05-17T13:13:31.425017+00:00 | 558 | false | \n```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count=0;\n for(String word: words){\n if(s.startsWith(word)) count++;\n }\n return count;\n }\n}\n``` | 3 | 0 | ['String', 'Counting', 'Java'] | 0 |
count-prefixes-of-a-given-string | Simple Using Substr | C++ | simple-using-substr-c-by-mankaran_07-g3ym | Please Upvote If you Like the Solution !!!\n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int ans = 0;\ | mankaran_07 | NORMAL | 2023-04-22T10:16:10.736757+00:00 | 2023-04-22T10:16:10.736804+00:00 | 400 | false | `Please Upvote If you Like the Solution !!!`\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int ans = 0;\n for(auto &w : words) {\n int n = w.size();\n if(s.substr(0,n) == w) ans++;\n }\n return ans;\n }\n};\n``` | 3 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | ✅ Easy and Fast C++ Solution | easy-and-fast-c-solution-by-adarsh_shukl-1kdx | Approach\nUsing s.find(words[i]) function which returns the index of words[i] if it is present in the string s , and to check whether words[i] occurs as a prefi | Adarsh_Shukla | NORMAL | 2022-12-22T12:55:35.040102+00:00 | 2022-12-22T12:56:40.386175+00:00 | 605 | false | # Approach\nUsing **s.find(words[i])** function which returns the index of words[i] if it is present in the string s , and to check whether words[i] occurs as a prefix use **s.find(words[i])==0** ( since prefix always occurs at index 0).\n\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& ... | 3 | 0 | ['C++'] | 1 |
count-prefixes-of-a-given-string | Easy O(N) | Maintain a Set | easy-on-maintain-a-set-by-vkadu68-klwj | We maintian a set called seen and add all the prefixes in the set and then simply check if the word in word list is present in the set or not\n\nLeave an upvote | vkadu68 | NORMAL | 2022-11-01T21:29:24.471481+00:00 | 2022-11-01T21:29:24.471515+00:00 | 388 | false | We maintian a set called seen and add all the prefixes in the set and then simply check if the word in word list is present in the set or not\n\n***Leave an upvote if this helps!!!***\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n seen=set()\n count=0\n for i... | 3 | 0 | ['Ordered Set', 'Python'] | 0 |
count-prefixes-of-a-given-string | Easy Python solution | easy-python-solution-by-vistrit-igdd | \ndef countPrefixes(self, words: List[str], s: str) -> int:\n c=0\n for i in words:\n l=len(i)\n if i==s[:l]:\n | vistrit | NORMAL | 2022-05-17T18:53:59.607539+00:00 | 2022-05-17T18:53:59.607581+00:00 | 389 | false | ```\ndef countPrefixes(self, words: List[str], s: str) -> int:\n c=0\n for i in words:\n l=len(i)\n if i==s[:l]:\n c+=1\n return c\n``` | 3 | 0 | ['Python', 'Python3'] | 0 |
count-prefixes-of-a-given-string | C++ online with std::count_if and string::rfind | c-online-with-stdcount_if-and-stringrfin-jv4m | This stackover flow post descrives how to you string::rfind to implement startsWith, which leads to following over all code to solve to problem.\n\n\nint countP | heder | NORMAL | 2022-04-30T18:26:54.995781+00:00 | 2022-08-10T18:24:35.411643+00:00 | 93 | false | This [stackover flow post](https://stackoverflow.com/questions/1878001/how-do-i-check-if-a-c-stdstring-starts-with-a-certain-string-and-convert-a) descrives how to you ```string::rfind``` to implement ```startsWith```, which leads to following over all code to solve to problem.\n\n```\nint countPrefixes(const vector<st... | 3 | 0 | ['C'] | 0 |
count-prefixes-of-a-given-string | C++ || 5 Lines Code || Very Easy O(N) Solution | c-5-lines-code-very-easy-on-solution-by-cwkyx | \nint countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(string str:words){\n int len=str.length();\n string sub | Gagan-Chaudhary | NORMAL | 2022-04-30T17:53:41.019322+00:00 | 2022-04-30T17:53:41.019365+00:00 | 130 | false | ```\nint countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(string str:words){\n int len=str.length();\n string subs=s.substr(0,len);\n if(str==subs){\n c++;\n }\n } \n return c;\n }\n``` | 3 | 0 | ['String', 'C', 'Iterator', 'C++'] | 1 |
count-prefixes-of-a-given-string | java | java-by-sakshi2602-c4cs | int ans=0;\n HashMap l=new HashMap<>();\n for(String str:words)\n l.put(str,l.getOrDefault(str,0)+1);\n for(int i=0;i<=s.length( | sakshi2602 | NORMAL | 2022-04-30T16:13:44.527942+00:00 | 2022-04-30T16:15:48.564339+00:00 | 288 | false | int ans=0;\n HashMap<String,Integer> l=new HashMap<>();\n for(String str:words)\n l.put(str,l.getOrDefault(str,0)+1);\n for(int i=0;i<=s.length();i++)\n {\n String st=s.substring(0,i);\n if(l.containsKey(st))\n ans=ans+l.get(st);\n }\n ... | 3 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | C++ || EASY TO UNDERSTAND || Simple Solution | c-easy-to-understand-simple-solution-by-xz3qm | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(auto word:words)\n {\n | aarindey | NORMAL | 2022-04-30T16:02:53.460367+00:00 | 2022-04-30T16:02:53.460401+00:00 | 135 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(auto word:words)\n {\n string str=word;\n int l=word.size();\n string s2=s.substr(0,l);\n if(s2==word)\n c++;\n ... | 3 | 1 | [] | 0 |
count-prefixes-of-a-given-string | 7 Liner javaScript Easy Solution | 7-liner-javascript-easy-solution-by-ali-6tsvl | Code\njavascript []\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function (words, s) {\n let count = 0 | ALI-786 | NORMAL | 2024-08-27T12:29:58.729179+00:00 | 2024-08-27T12:29:58.729206+00:00 | 49 | false | # Code\n```javascript []\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function (words, s) {\n let count = 0;\n for (let i = 0; i < words.length; i++) {\n if (s.startsWith(words[i])) {\n count++;\n }\n }\n return count;\n};\n``` | 2 | 0 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | simplest one PYTHON | simplest-one-python-by-justingeorge-2hv1 | 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 | justingeorge | NORMAL | 2024-03-05T08:18:29.839533+00:00 | 2024-03-05T08:18:29.839560+00:00 | 216 | 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 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | Very easy logic , for beginners !! | very-easy-logic-for-beginners-by-user120-2rdh | Intuition\nInitialize a variable count to store the count of prefixes found in the string s.\nIterate through each word in the words array.\nFor each word encou | user1203oJ | NORMAL | 2024-02-24T06:22:53.272079+00:00 | 2024-02-24T06:22:53.272139+00:00 | 520 | false | # Intuition\nInitialize a variable count to store the count of prefixes found in the string s.\nIterate through each word in the words array.\nFor each word encountered, check if the string s starts with that word using the startsWith method.\nIf the word is found to be a prefix of s, increment the count.\nAfter iterat... | 2 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | Simple Solution | simple-solution-by-anshika_73-tgte | \n\n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count =0;\n for(auto str : words){\n | Anshika_73_ | NORMAL | 2023-12-04T13:21:28.114548+00:00 | 2023-12-04T13:21:28.114588+00:00 | 92 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count =0;\n for(auto str : words){\n int len = str.length();\n auto subs = s.substr(0,len);\n if(str == subs){\n count++;\n }\n }\n ... | 2 | 0 | ['C++'] | 2 |
count-prefixes-of-a-given-string | Easy one-liner C# | easy-one-liner-c-by-rxcore-v57d | \n# Approach\nUsing Count and StartsWith\n\n# Code\n\npublic class Solution {\n public int CountPrefixes(string[] words, string s) {\n return words.Co | rxcore | NORMAL | 2023-09-26T17:11:18.965491+00:00 | 2023-09-26T17:11:18.965518+00:00 | 38 | false | \n# Approach\nUsing Count and StartsWith\n\n# Code\n```\npublic class Solution {\n public int CountPrefixes(string[] words, string s) {\n return words.Count(i => s.StartsWith(i));\n }\n}\n``` | 2 | 0 | ['C#'] | 0 |
count-prefixes-of-a-given-string | Javascript easy one line solution | javascript-easy-one-line-solution-by-bri-5ylw | Complexity\n- Time complexity: O(n*m)\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# | brijesh178 | NORMAL | 2023-08-28T16:53:55.875454+00:00 | 2023-08-28T16:53:55.875475+00:00 | 209 | false | # Complexity\n- Time complexity: O(n*m)\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)$$ -->\n\n# Code\n```\n/**\n * @param {string[]} words\n * @param {string} s\n * @return {number}\n */\nvar countPrefixes = function (words, s) {\n ... | 2 | 0 | ['JavaScript'] | 0 |
count-prefixes-of-a-given-string | Go easy | go-easy-by-ganesh_raveendran-kw1n | \n\n# Code\n\nfunc countPrefixes(words []string, s string) int {\n\tvar count int\n\n\tfor _, v := range words {\n\t\tif strings.HasPrefix(s, v) {\n\t\t\tcount+ | ganesh_raveendran | NORMAL | 2023-08-16T11:56:07.177539+00:00 | 2023-08-16T11:56:07.177556+00:00 | 16 | false | \n\n# Code\n```\nfunc countPrefixes(words []string, s string) int {\n\tvar count int\n\n\tfor _, v := range words {\n\t\tif strings.HasPrefix(s, v) {\n\t\t\tcount++\n\t\t}\n\t}\n\n\treturn count\n}\n``` | 2 | 0 | ['Go'] | 0 |
count-prefixes-of-a-given-string | Python one line solution | python-one-line-solution-by-azazellooo-blf6 | \n\n# Code\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return len([i for i in words if s.startswith(i)])\n | azazellooo | NORMAL | 2023-07-28T07:01:57.409952+00:00 | 2023-07-28T07:01:57.409975+00:00 | 205 | false | \n\n# Code\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n return len([i for i in words if s.startswith(i)])\n``` | 2 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | Easy to understand | easy-to-understand-by-bhaskarkumar07-wius | \n\n# Approach\n Describe your approach to solving the problem. \n if it starts with element present in array count++;\n# Complexity\n- Time complexity:O | bhaskarkumar07 | NORMAL | 2023-04-24T15:26:33.969584+00:00 | 2023-04-24T15:26:33.969610+00:00 | 690 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n if it starts with element present in array count++;\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n... | 2 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | simple java 3 line code | simple-java-3-line-code-by-bhupendra0820-j0pm | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | bhupendra082002 | NORMAL | 2023-03-31T16:28:23.926044+00:00 | 2023-03-31T16:28:23.926078+00:00 | 311 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | Esay Java Solution | esay-java-solution-by-md_ashik786-epfi | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | MD_Ashik786 | NORMAL | 2023-03-25T01:52:40.291810+00:00 | 2023-03-25T01:52:40.291850+00:00 | 593 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | C++ || Easy || Simple | c-easy-simple-by-naoman-l9zb | \n\n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count = 0;\n for (string word : words)\n | Naoman | NORMAL | 2023-03-10T05:38:08.079838+00:00 | 2023-03-10T05:38:08.079884+00:00 | 725 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int count = 0;\n for (string word : words)\n {\n if(s.find(word) == 0) \n { \n count++;\n }\n }\n return count;\n }\n};\n\n``` | 2 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | Pyhton3 oneliner using startswith() | pyhton3-oneliner-using-startswith-by-ami-hnjc | \n\n# Code\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count = 0\n for i in words:\n if s.start | amitpandit03 | NORMAL | 2023-03-03T01:46:14.379198+00:00 | 2023-03-03T01:46:14.379236+00:00 | 59 | false | \n\n# Code\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count = 0\n for i in words:\n if s.startswith(i):\n count += 1\n return count\n\n #One Liner\n\n return sum(1 for i in words if s.startswith(i))\n``` | 2 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | ✔Simple C++ code || Easy || beats 100% ✌ | simple-c-code-easy-beats-100-by-kraken_0-8ae4 | \n\nRuntime: Beats 100%\n\n# Code\n\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(auto j:w | Kraken_02 | NORMAL | 2023-02-12T16:02:40.108526+00:00 | 2023-02-12T16:02:40.108564+00:00 | 545 | false | \n\nRuntime: Beats 100%\n\n# Code\n```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n int c=0;\n for(auto j:words){\n int b=1;\n for(int i=0;i<j.size();i++){\n if(j[i]!=s[i]){\n b=0;\n break;... | 2 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | Pooping solution :) | pooping-solution-by-kaushikp0908-tlra | Approach\n Describe your approach to solving the problem. \nPoop the loop\n\n# Complexity\n- Time complexity: O(N)\n Add your time complexity here, e.g. O(n) \n | kaushikp0908 | NORMAL | 2023-02-06T07:16:53.484396+00:00 | 2023-02-06T07:16:53.484436+00:00 | 152 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nPoop the loop\n\n# Complexity\n- Time complexity: O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n def countPrefixes(... | 2 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | Easy solution | easy-solution-by-shubhanjalisingh_001-o4nk | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ShubhanjaliSingh_001 | NORMAL | 2022-12-02T04:38:44.407793+00:00 | 2022-12-02T04:38:44.407830+00:00 | 307 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | C++|| Easy Approach || 93% Faster | c-easy-approach-93-faster-by-spects-acul-jdd8 | \nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) { \n int ans = 0;\n for(int i = 0 ; i<words.size() ; i++){\n | spects-acular | NORMAL | 2022-10-14T06:51:36.663268+00:00 | 2022-10-14T06:51:36.663310+00:00 | 309 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) { \n int ans = 0;\n for(int i = 0 ; i<words.size() ; i++){\n string element = words[i];\n int len = element.length();\n \n for(int j = 0 ; j<len ; j++){\n \n ... | 2 | 0 | [] | 1 |
count-prefixes-of-a-given-string | ✅ 'startsWith()' in Java | Java | startswith-in-java-java-by-sayakmondal-5z5x | \nif(you like)\n\tplease upvote;\n\n\n\n# Java Code\n\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int cnt = 0;\n | Sayakmondal | NORMAL | 2022-10-02T16:18:18.031088+00:00 | 2022-10-02T16:23:48.320824+00:00 | 311 | false | ```\nif(you like)\n\tplease upvote;\n```\n\n\n# Java Code\n```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int cnt = 0;\n for(String str:words) \n ... | 2 | 0 | ['String', 'Java'] | 0 |
count-prefixes-of-a-given-string | python solution | python-solution-by-dugu0607-mfk9 | \tclass Solution:\n\t\tdef countPrefixes(self, words: List[str], s: str) -> int:\n\t\t\tres = 0\n\t\t\tfor word in words:\n\t\t\t\tif s[:len(word)] == word: res | Dugu0607 | NORMAL | 2022-10-02T06:20:02.825221+00:00 | 2022-10-02T06:20:02.825265+00:00 | 371 | false | \tclass Solution:\n\t\tdef countPrefixes(self, words: List[str], s: str) -> int:\n\t\t\tres = 0\n\t\t\tfor word in words:\n\t\t\t\tif s[:len(word)] == word: res += 1\n\t\t\treturn res | 2 | 0 | [] | 0 |
count-prefixes-of-a-given-string | python ||O(N) | python-on-by-sneh713-aiva | Time Complexcity O(N)\nspace Complexcity O(1)\n\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count=0\n for wo | Sneh713 | NORMAL | 2022-09-26T12:53:29.158577+00:00 | 2022-09-26T12:53:29.158614+00:00 | 905 | false | Time Complexcity O(N)\nspace Complexcity O(1)\n```\nclass Solution:\n def countPrefixes(self, words: List[str], s: str) -> int:\n count=0\n for word in words:\n n=len(word)\n if s[0:n]==word:\n count+=1\n return count\n \n``` | 2 | 0 | ['String', 'Python3'] | 1 |
count-prefixes-of-a-given-string | Python straight forward one-liner | python-straight-forward-one-liner-by-trp-k5e4 | Here is what I did:\n\npython\n def countPrefixes(self, words: List[str], s: str) -> int:\n return sum(s.startswith(word) for word in words)\n | trpaslik | NORMAL | 2022-09-12T10:52:52.174389+00:00 | 2022-09-12T10:52:52.174434+00:00 | 330 | false | Here is what I did:\n\n```python\n def countPrefixes(self, words: List[str], s: str) -> int:\n return sum(s.startswith(word) for word in words)\n``` | 2 | 0 | ['Python'] | 1 |
count-prefixes-of-a-given-string | ✅ [JavaScript] Straightforward Solution || Simple & Understandable || Fast | javascript-straightforward-solution-simp-wt1s | Runtime: 68 ms, faster than 91.85% of JavaScript online submissions for Count Prefixes of a Given String.\nMemory Usage: 43 MB, less than 40.00% of JavaScript o | PoeticIvy302543 | NORMAL | 2022-07-03T00:48:15.488151+00:00 | 2022-07-25T21:32:49.484404+00:00 | 121 | false | **Runtime: 68 ms, faster than 91.85% of JavaScript online submissions for Count Prefixes of a Given String.\nMemory Usage: 43 MB, less than 40.00% of JavaScript online submissions for Count Prefixes of a Given String.**\n\n```\nvar countPrefixes = function(words, s) {\n let count = 0\n for (i of words) {\n ... | 2 | 0 | ['String', 'JavaScript'] | 0 |
count-prefixes-of-a-given-string | C# LINQ One-line | c-linq-one-line-by-aip-s5vd | \n\tpublic int CountPrefixes(string[] words, string s) \n => words.Count(w => s.StartsWith(w)); \n | AIP | NORMAL | 2022-05-04T08:59:50.653397+00:00 | 2022-05-04T08:59:50.653429+00:00 | 76 | false | ```\n\tpublic int CountPrefixes(string[] words, string s) \n => words.Count(w => s.StartsWith(w)); \n``` | 2 | 0 | [] | 0 |
count-prefixes-of-a-given-string | Rust solutions | rust-solutions-by-bigmih-3tuw | Simle use of starts_with:\n\nimpl Solution {\n pub fn count_prefixes(words: Vec<String>, s: String) -> i32 {\n words.iter().filter(|w| s.starts_with(w | BigMih | NORMAL | 2022-05-01T09:25:16.219701+00:00 | 2022-05-01T09:25:16.219732+00:00 | 64 | false | 1. Simle use of `starts_with`:\n```\nimpl Solution {\n pub fn count_prefixes(words: Vec<String>, s: String) -> i32 {\n words.iter().filter(|w| s.starts_with(w.as_str())).count() as _\n }\n}\n```\n2. Slice comparison\n```\nimpl Solution {\n pub fn count_prefixes(words: Vec<String>, s: String) -> i32 {\n ... | 2 | 0 | ['Rust'] | 0 |
count-prefixes-of-a-given-string | Go golang solution | go-golang-solution-by-leaf_peng-rxt2 | go\nfunc countPrefixes(words []string, s string) int {\n ans := 0\n for _, word := range words {\n n := len(word)\n if n > len(s) { continue | leaf_peng | NORMAL | 2022-05-01T01:08:25.469918+00:00 | 2022-05-01T01:08:50.166341+00:00 | 73 | false | ```go\nfunc countPrefixes(words []string, s string) int {\n ans := 0\n for _, word := range words {\n n := len(word)\n if n > len(s) { continue }\n if word == s[:n] { ans++ }\n }\n return ans\n}\n``` | 2 | 0 | ['Go'] | 0 |
count-prefixes-of-a-given-string | C++ very easy to Understand | solution using unordered set | c-very-easy-to-understand-solution-using-du23 | ```\nclass Solution {\npublic:\n int countPrefixes(vector& words, string s) {\n unordered_set st;\n int size = s.size();\n for(int i = 0 | Ashishks0211 | NORMAL | 2022-04-30T18:46:16.772904+00:00 | 2022-04-30T18:46:16.772935+00:00 | 55 | false | ```\nclass Solution {\npublic:\n int countPrefixes(vector<string>& words, string s) {\n unordered_set<string> st;\n int size = s.size();\n for(int i = 0;i<size;i++){\n string temp = "";\n for(int j = 0;j<=i;j++){\n temp+=s[j];\n }\n st.i... | 2 | 0 | ['C'] | 0 |
count-prefixes-of-a-given-string | 📌 [JAVA] | Simple | java-simple-by-kanojiyakaran-y91t | ```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n \n int preFixCount=0;\n for(String curr:words){\n int | kanojiyakaran | NORMAL | 2022-04-30T16:00:56.377246+00:00 | 2022-04-30T16:03:22.896520+00:00 | 345 | false | ```\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n \n int preFixCount=0;\n for(String curr:words){\n int i=0;\n \n if(curr.length()>s.length()) // if current string length greater than s length invalid prefix\n continue; \n ... | 2 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | Optimized Set-Based Solution | O(n + m) Time | Python 3 | optimized-set-based-solution-on-m-time-p-hkk1 | IntuitionWe need to count how many words in words are prefixes of the string s. A prefix is any substring that starts from the beginning of s. Instead of checki | NRSingh | NORMAL | 2025-04-03T19:23:59.393721+00:00 | 2025-04-03T19:23:59.393721+00:00 | 19 | false | # Intuition
We need to count how many words in words are prefixes of the string s. A prefix is any substring that starts from the beginning of s. Instead of checking each word against s directly (which would be inefficient), we can precompute all valid prefixes of s and store them in a set for quick lookup.
<!-- Descri... | 1 | 0 | ['Array', 'String', 'Python3'] | 0 |
count-prefixes-of-a-given-string | Count Prefixes of a Given String (Optimal Approach) | count-prefixes-of-a-given-string-optimal-mi26 | IntuitionApproachGo through the array and check if the each element starting matches the prefix word.Complexity
Time complexity:
O(n)
Space complexity:
O(1)Code | Vinil07 | NORMAL | 2025-04-01T05:22:03.798635+00:00 | 2025-04-01T05:22:03.798635+00:00 | 19 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
Go through the array and check if the each element starting matches the prefix word.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
... | 1 | 0 | ['Array', 'String', 'C++'] | 0 |
count-prefixes-of-a-given-string | 0MS || Beats 100% || Easy and Simple Java Solution | 0ms-beats-100-easy-and-simple-java-solut-g2d7 | 🔍 Intuition:
We need to count how many strings in thewordsarray areprefixesof strings.
Aprefixmeans the string appears at thebeginningofs.
Example Understanding | Yashvendra | NORMAL | 2025-02-20T16:13:35.467778+00:00 | 2025-02-20T16:13:35.467778+00:00 | 100 | false | ## 🔍 Intuition:
- We need to count how many strings in the `words` array are **prefixes** of string `s`.
- A **prefix** means the string appears at the **beginning** of `s`.
- **Example Understanding:**
- **Input:** `words = ["a", "b", "ab", "abc"]`, `s = "abc"`
- `"a"`, `"ab"`, and `"abc"` are **prefixes** of `... | 1 | 0 | ['Array', 'String', 'Java'] | 0 |
count-prefixes-of-a-given-string | 0ms, Beats 100%, Hopefully my solution will help everyone with an easy problem solving | 0ms-beats-100-hopefully-my-solution-will-vn5n | IntuitionApproachComplexity
Time complexity:O(n)
Space complexity:O(1)
Code | bvc01654 | NORMAL | 2025-02-05T08:58:45.397939+00:00 | 2025-02-05T08:58:45.397939+00:00 | 66 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:O(n)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
... | 1 | 0 | ['String', 'Iterator', 'C++'] | 0 |
count-prefixes-of-a-given-string | [C++] Simple Solution | c-simple-solution-by-samuel3shin-8lai | Code | Samuel3Shin | NORMAL | 2025-01-23T16:11:58.057030+00:00 | 2025-01-23T16:11:58.057030+00:00 | 35 | false | # Code
```cpp []
class Solution {
public:
int countPrefixes(vector<string>& words, string s) {
unordered_set<string> hs;
for(int i=1; i<=s.size(); i++) {
hs.insert(s.substr(0, i));
}
int ans = 0;
for(auto w : words) {
if(hs.count(w)) ans++;
}
... | 1 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | 0MS 🥇|| BEATS 100% 🥂 || 5 LINE CODE ⛳ || 🌟JAVA ☕ | 0ms-beats-100-5-line-code-java-by-galani-giph | Code | Galani_jenis | NORMAL | 2025-01-18T05:30:24.151194+00:00 | 2025-01-18T05:30:24.151194+00:00 | 88 | false | 
# Code
```java []
class Solution {
public int countPrefixes(String[] words, String s) {
int ans = 0;
for (String word : words) {
if (s.startsWith(wo... | 1 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | 4 lines of python code beats 100% | 4-lines-of-python-code-beats-100-by-vsab-mr5x | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | vsabari146 | NORMAL | 2025-01-10T07:35:59.463227+00:00 | 2025-01-10T07:35:59.463227+00:00 | 113 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | Easy solution for begineers || 100% | easy-solution-for-begineers-100-by-kunal-lzjm | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | kunal44885 | NORMAL | 2025-01-09T15:03:26.643077+00:00 | 2025-01-09T15:03:26.643077+00:00 | 112 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
`... | 1 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | Built-in Function | built-in-function-by-khushi_1502-7kuc | Complexity
Time complexity:O(N)
Space complexity:O(1)
Code | khushi_1502 | NORMAL | 2025-01-09T06:42:51.795312+00:00 | 2025-01-09T06:42:51.795312+00:00 | 124 | false |
# Complexity
- Time complexity:O(N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int countPrefixes(String[] words, String s) {
int count=0;
for(int i=0;i<words.length;... | 1 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | 🔍 Count Prefixes in a String – Clean and Concise Python Solution!🚀 | count-prefixes-in-a-string-clean-and-con-whmg | 🧠 Intuition💡 The task is to count how many words in the list words are prefixes of the string s.
A word is a prefix of s if it matches the initial characters of | Ramanand98 | NORMAL | 2025-01-09T06:32:51.300285+00:00 | 2025-01-09T06:32:51.300285+00:00 | 95 | false | 
# 🧠 **Intuition**
💡 The task is to count how many words in the list `words` are prefixes of the string `s`.
- A word is a prefix of `s` if it matches the initial characters of `s`.
- We can check... | 1 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | Easy Solution | Beat 100% | c++ Solution | easy-solution-beat-100-c-solution-by-nit-5pl7 | IntuitionThe problem asks us to count how many words in the given list words are prefixes of the string s. The simplest approach is to check each word in the li | NITian_Dilip | NORMAL | 2025-01-09T05:41:53.359594+00:00 | 2025-01-09T05:45:54.721417+00:00 | 82 | false | # Intuition
The problem asks us to count how many words in the given list `words` are prefixes of the string `s`. The simplest approach is to check each word in the list and see if it matches the prefix of `s`. If the word is a prefix, we increment the count.
---
# Approach
1. **Initialize a count variable** to k... | 1 | 0 | ['Array', 'String', 'String Matching', 'C++'] | 0 |
count-prefixes-of-a-given-string | 2255. Count Prefixes of a Given String | 2255-count-prefixes-of-a-given-string-by-ky4l | Complexity
Time complexity: O(M*N)
Space complexity: O(1)
Code | Adarshjha | NORMAL | 2025-01-09T05:16:29.130771+00:00 | 2025-01-09T05:16:29.130771+00:00 | 7 | false | # Complexity
- Time complexity: O(M*N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```c []
int countPrefixes(char** words, int wordsSize, char* s) {
int t, count = 0;
for (int i = 0; i < wordsSize; i++) {
... | 1 | 0 | ['C'] | 0 |
count-prefixes-of-a-given-string | Easy solution with explanation beats 100% | easy-solution-with-explanation-beats-100-3u05 | IntuitionThe goal is to count how many words in the array words are prefixes of the string s. The Java String class provides the startsWith() method, which make | Shubhash_Singh | NORMAL | 2025-01-09T04:24:21.809558+00:00 | 2025-01-09T04:24:21.809558+00:00 | 21 | false | # Intuition
The goal is to count how many words in the array `words` are prefixes of the string `s`. The Java String class provides the `startsWith()` method, which makes it straightforward to check if one string is a prefix of another.
# Approach
1. Iterate Through the Words: Loop through each word in the words array... | 1 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | 🔥BEATS 💯 % 🎯 |✨SUPER EASY FOR BEGINNERS 👏 | beats-super-easy-for-beginners-by-vignes-z7nb | Code | vigneshvaran0101 | NORMAL | 2025-01-09T04:00:08.915637+00:00 | 2025-01-09T04:00:08.915637+00:00 | 40 | false | 
# Code
```python3 []
class Solution:
def countPrefixes(self, words: List[str], s: str) -> int:
c = 0
for i in words:
if s.startswith(i):
c += 1
... | 1 | 0 | ['Python3'] | 0 |
count-prefixes-of-a-given-string | ads; | ads-by-abhaydeep_24-b04j | Complexity
Time complexity:
O(N*M)
N=size of words array
M=max length of word in words array
Space complexity:
O(1)
Code | abhaydeep_24 | NORMAL | 2025-01-09T01:25:47.749151+00:00 | 2025-01-09T01:25:47.749151+00:00 | 30 | false | 
# Complexity
- Time complexity:
O(N*M)
N=size of words array
M=max length of word in words array
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
O(1)
<!-- Add your space complex... | 1 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | Simplest answer in 1 loop | simplest-answer-in-1-loop-by-saksham_gup-a8bs | Complexity
Time complexity: O(n.m)
(n for strings in array and m for length of each string)
Space complexity: O(1)
Code | Saksham_Gupta_ | NORMAL | 2025-01-04T11:19:52.844737+00:00 | 2025-01-04T11:19:52.844737+00:00 | 116 | false |
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity: ***O(n.m)***
*(n for strings in array and m for length of each string)*
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity: ***O(1)***
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java [... | 1 | 0 | ['Array', 'String', 'Java'] | 0 |
count-prefixes-of-a-given-string | leetcodedaybyday - Beats 100% with C++ and Beats 100% with Python3 | leetcodedaybyday-beats-100-with-c-and-be-3j3i | IntuitionThe task is to count how many words in the words list are prefixes of the string s. A prefix of a string is defined as a substring that starts at the b | tuanlong1106 | NORMAL | 2024-12-28T08:31:04.567395+00:00 | 2024-12-28T08:31:04.567395+00:00 | 82 | false | # Intuition
The task is to count how many words in the `words` list are prefixes of the string `s`. A prefix of a string is defined as a substring that starts at the beginning of the string. To solve this, we can compare each word in the `words` list with the corresponding prefix of `s` of the same length.
# Approach
... | 1 | 0 | ['Array', 'String', 'C++', 'Python3'] | 0 |
count-prefixes-of-a-given-string | Easy and 100% fast solution using string method | easy-and-100-fast-solution-using-string-qmloz | IntuitionTo solve the problem of finding how many strings in the words array are a prefix of string s, let us break this problem :-
Understanding a Prefix :- | code_space | NORMAL | 2024-12-17T09:33:19.897625+00:00 | 2024-12-17T09:33:19.897625+00:00 | 142 | false | \n\n# Intuition\nTo solve the problem of finding how many strings in the `words` array are a prefix of string `s`, let us break this problem :-\n1. **Understanding a Prefix** :- A string w is said to be a ... | 1 | 0 | ['Array', 'String', 'C++', 'Java', 'JavaScript'] | 0 |
count-prefixes-of-a-given-string | Count Prefixes of a Given String | count-prefixes-of-a-given-string-by-tejd-3o8v | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem is to determine how many strings in an array are prefixes of a given string | tejdekiwadiya | NORMAL | 2024-06-12T17:21:54.940920+00:00 | 2024-06-12T17:21:54.940943+00:00 | 21 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem is to determine how many strings in an array are prefixes of a given string. A prefix of a string is a substring that starts at the beginning of the string and can be of any length up to the length of the string itself. To sol... | 1 | 0 | ['Array', 'String', 'Java'] | 0 |
count-prefixes-of-a-given-string | 2 Line Solution || 100% faster || Single Pass | 2-line-solution-100-faster-single-pass-b-v43p | Intuition\nIf given words are the substring at index 0, they are the desired string count\n Describe your first thoughts on how to solve this problem. \n\n# App | Nitesh_Gupta_NG | NORMAL | 2024-06-07T18:25:42.451243+00:00 | 2024-06-07T18:25:42.451279+00:00 | 102 | false | # Intuition\nIf given words are the substring at index 0, they are the desired string count\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nUse indexOf function to validate, words are coming at first index or not\n<!-- Describe your approach to solving the problem. -->\n\n# Complexit... | 1 | 0 | ['Java'] | 1 |
count-prefixes-of-a-given-string | [Java] Best solution in design scenario - Using Trie Tree | java-best-solution-in-design-scenario-us-kshz | java\nclass Solution {\n public int countPrefixes(final String[] words, final String s) {\n final TrieNode root = new TrieNode();\n\n // Insert | YTchouar | NORMAL | 2024-05-08T06:59:02.029667+00:00 | 2024-05-08T06:59:13.506925+00:00 | 385 | false | ```java\nclass Solution {\n public int countPrefixes(final String[] words, final String s) {\n final TrieNode root = new TrieNode();\n\n // Insert all words in Trie\n\n for(final String word : words) {\n TrieNode curr = root;\n\n for(int i = 0; i < word.length(); ++i) {\n ... | 1 | 0 | ['Java'] | 0 |
count-prefixes-of-a-given-string | Easiest C || C++ || Java Solutions -- 100% beats Easy[Simple] | easiest-c-c-java-solutions-100-beats-eas-qlco | Intuition\n\nC []\nint countPrefixes(char** words, int wordsSize, char* s) {\n int c = 0;\n for (int i = 0; i < wordsSize; i++) {\n strncmp(words[i | Edwards310 | NORMAL | 2024-02-03T06:39:31.690431+00:00 | 2024-02-03T06:39:31.690462+00:00 | 3 | false | # Intuition\n\n```C []\nint countPrefixes(char** words, int wordsSize, char* s) {\n int c = 0;\n for (int i = 0; i < wordsSize; i++) {\n strncmp(words[i], s, strlen(words... | 1 | 0 | ['C', 'C++', 'Java'] | 0 |
count-prefixes-of-a-given-string | Java || Simple Solution || 100% Beats | java-simple-solution-100-beats-by-akash_-3r88 | \n# Code\n\nimport java.util.ArrayList;\nimport java.util.List;\n\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count | akash_kce | NORMAL | 2024-02-01T09:51:01.589296+00:00 | 2024-02-01T09:51:01.589330+00:00 | 379 | false | \n# Code\n```\nimport java.util.ArrayList;\nimport java.util.List;\n\nclass Solution {\n public int countPrefixes(String[] words, String s) {\n int count = 0;\n for (String i : words)\n if (s.startsWith(i)) count++;\n return count;\n }\n}\n\n``` | 1 | 0 | ['Array', 'String', 'Java'] | 0 |
count-prefixes-of-a-given-string | Easy one liner solution | easy-one-liner-solution-by-elangoram1998-95c9 | \n# Code\n\nfunction countPrefixes(words: string[], s: string): number {\n return words.filter((word) => s.startsWith(word)).length;\n};\n | elangoram1998 | NORMAL | 2023-12-01T10:46:23.724286+00:00 | 2023-12-01T10:46:23.724312+00:00 | 45 | false | \n# Code\n```\nfunction countPrefixes(words: string[], s: string): number {\n return words.filter((word) => s.startsWith(word)).length;\n};\n``` | 1 | 0 | ['TypeScript', 'JavaScript'] | 0 |
count-prefixes-of-a-given-string | Count Prefixes of a Given String Solution in C++ | count-prefixes-of-a-given-string-solutio-humn | 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_Kunal_Singh | NORMAL | 2023-03-14T18:03:58.544456+00:00 | 2023-03-14T18:03:58.544490+00:00 | 51 | 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)$$ -->\nO(n*m)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n... | 1 | 0 | ['C++'] | 0 |
count-prefixes-of-a-given-string | Ruby solution. 1-line | ruby-solution-1-line-by-guswhitten-rdez | \n# @param {String[]} words\n# @param {String} s\n# @return {Integer}\ndef count_prefixes(words, s)\n words.count { |word| s.start_with?(word) }\nend\n | guswhitten | NORMAL | 2023-02-12T14:37:56.157570+00:00 | 2023-02-12T14:37:56.157600+00:00 | 7 | false | ```\n# @param {String[]} words\n# @param {String} s\n# @return {Integer}\ndef count_prefixes(words, s)\n words.count { |word| s.start_with?(word) }\nend\n``` | 1 | 0 | ['Ruby'] | 0 |
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