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minimum-number-of-people-to-teach | Is the expected I/O dataset wrong for this problem? | is-the-expected-io-dataset-wrong-for-thi-rn0a | Can anybody help me understand why is the answer for the following input be 4 instead of 3? \n\n\n17\n[[4,7,2,14,6],[15,13,6,3,2,7,10,8,12,4,9],[16],[10],[10,3 | bridge_four | NORMAL | 2021-01-23T16:39:56.504502+00:00 | 2021-01-23T16:39:56.504543+00:00 | 83 | false | Can anybody help me understand why is the answer for the following input be `4` instead of `3`? \n\n```\n17\n[[4,7,2,14,6],[15,13,6,3,2,7,10,8,12,4,9],[16],[10],[10,3],[4,12,8,1,16,5,15,17,13],[4,13,15,8,17,3,6,14,5,10],[11,4,13,8,3,14,5,7,15,6,9,17,2,16,12],[4,14,6],[16,17,9,3,11,14,10,12,1,8,13,4,5,6],[14],[7,14],[1... | 1 | 0 | [] | 1 |
minimum-number-of-people-to-teach | Java Solution | java-solution-by-abstracted_0507-710e | \nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n int m = languages.length;\n boolean[][] kno | abstracted_0507 | NORMAL | 2021-01-23T16:34:26.424002+00:00 | 2021-01-23T16:34:26.424029+00:00 | 108 | false | ```\nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n int m = languages.length;\n boolean[][] knowsLang = new boolean[m][n];\n for(int i = 0; i < m; i++) {\n for(int j = 0; j < languages[i].length; j++) {\n knowsLang[i][l... | 1 | 0 | [] | 1 |
minimum-number-of-people-to-teach | C++ O(n2logn) with tutorial | c-on2logn-with-tutorial-by-hawkings9999-z4gy | The Idea in this problem is the following, we have n langauges and there are m people and l pairs of people,\nwe need to find which langauge we need to make all | hawkings9999 | NORMAL | 2021-01-23T16:30:12.219175+00:00 | 2021-01-23T16:48:36.322661+00:00 | 210 | false | The Idea in this problem is the following, we have n langauges and there are m people and l pairs of people,\nwe need to find which langauge we need to make all of them be able to talk to all the people that they are paired with. \nSome Observations,\nWe dont need to consider pair of people that have at least one langu... | 1 | 1 | [] | 0 |
minimum-number-of-people-to-teach | C# Beats 100% Space/Time | c-beats-100-spacetime-by-rameshthaleia-b3qn | IntuitionUse bool arrays to simplify comparisons.Don't double-count a users language needs.Accumulate wanted languages when the user's can't speak.ApproachEach | rameshthaleia | NORMAL | 2025-04-06T16:56:54.535866+00:00 | 2025-04-06T16:56:54.535866+00:00 | 1 | false | # Intuition
Use bool arrays to simplify comparisons.
Don't double-count a users language needs.
Accumulate wanted languages when the user's can't speak.
# Approach
Each user has a bool array indicating what languages they speak.
Each user has a bool array indicating whether they have already requested a language or... | 0 | 0 | ['C#'] | 0 |
minimum-number-of-people-to-teach | C# Beats 100% Space/Time | c-beats-100-spacetime-by-rameshthaleia-td2y | IntuitionUse dictionaries to simplify comparisons.
Don't double-count a users language needs.
Accumulate wanted languages when the user's can't speak.ApproachEa | rameshthaleia | NORMAL | 2025-04-06T16:42:42.104831+00:00 | 2025-04-06T16:42:42.104831+00:00 | 2 | false | # Intuition
Use dictionaries to simplify comparisons.
Don't double-count a users language needs.
Accumulate wanted languages when the user's can't speak.
# Approach
Each user has a bool[n] indicating what languages they speak.
Keep track of a user's wanted languages so we only count them once.
Accumulates the wanted l... | 0 | 0 | ['C#'] | 0 |
minimum-number-of-people-to-teach | While doing the program of minimum of teach two friends setup chars to use in the program | while-doing-the-program-of-minimum-of-te-gclq | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Girsha | NORMAL | 2025-02-14T07:57:37.658081+00:00 | 2025-02-14T07:57:37.658081+00:00 | 4 | 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
`... | 0 | 0 | ['Python'] | 0 |
minimum-number-of-people-to-teach | easy to understand, good for novice | easy-to-understand-good-for-novice-by-ji-cdty | Intuitionbrute force each language, see which one has the least people to teach.Complexity
Time complexity:
O(n*m)Code | jiayi9063 | NORMAL | 2025-02-06T21:06:02.883900+00:00 | 2025-02-06T21:06:02.883900+00:00 | 4 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
brute force each language, see which one has the least people to teach.
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
O(n*m)
# Code
```python3 []
class Solution:
def minimumTeachings(self, n: int,... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | 1733. Minimum Number of People to Teach | 1733-minimum-number-of-people-to-teach-b-zwtu | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-15T01:25:03.323300+00:00 | 2025-01-15T01:25:03.323300+00:00 | 4 | 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
`... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | One language to rule them all | one-language-to-rule-them-all-by-tonitan-w2rl | IntuitionApproachComplexity
Time complexity:
O(n^2)
Space complexity:
Code | tonitannoury01 | NORMAL | 2024-12-19T18:56:37.024373+00:00 | 2024-12-19T18:56:37.024373+00:00 | 5 | 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^2)
- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```javascript []
/**
* @param {number} n
* @pa... | 0 | 0 | ['JavaScript'] | 0 |
minimum-number-of-people-to-teach | brutforce solution with Sets | brutforce-solution-with-sets-by-debareto-zfnd | 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 | debareto | NORMAL | 2024-10-31T14:58:20.356158+00:00 | 2024-10-31T14:58:20.356187+00:00 | 3 | 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. $$O(n)$$ --... | 0 | 0 | ['Python'] | 0 |
minimum-number-of-people-to-teach | Minimum Number Of People To Teach!!! | minimum-number-of-people-to-teach-by-sin-7n1o | \n\n# Code\njava []\nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n //Step 1: Creating a map \n | sindhujadid_1804 | NORMAL | 2024-09-16T17:43:59.657825+00:00 | 2024-09-16T17:43:59.657853+00:00 | 28 | false | \n\n# Code\n```java []\nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n //Step 1: Creating a map \n // language --> person speaking\n HashMap<Integer, ArrayList<Integer>> map = new HashMap<>();\n for(int i = 0; i < languages.length; i++){\... | 0 | 0 | ['Array', 'Hash Table', 'Java'] | 0 |
minimum-number-of-people-to-teach | Very Easy Approach and Short Code in C++ | very-easy-approach-and-short-code-in-c-b-i45g | Intuition and Approach\nstep1. find the total friends who can\'t comm. each other\nstep2.find the most comman lang btwn them\nstep3. ans=totalfriendsWhoCantComm | Akki2910 | NORMAL | 2024-09-02T05:05:56.157787+00:00 | 2024-09-02T05:05:56.157805+00:00 | 32 | false | # Intuition and Approach\nstep1. find the total friends who can\'t comm. each other\nstep2.find the most comman lang btwn them\nstep3. ans=totalfriendsWhoCantComm-countOfMaximumComman Lang\n\n\nintuition:- if we know there are 10 people who cant comm. with each and maximum 5 of them know a comman language then minimum ... | 0 | 0 | ['Greedy', 'C++'] | 0 |
minimum-number-of-people-to-teach | Hashing || C++ | hashing-c-by-lotus18-xhbd | Code\n\nclass Solution \n{\npublic:\n int minimumTeachings(int lang, vector<vector<int>>& languages, vector<vector<int>>& friendships) \n {\n map<i | lotus18 | NORMAL | 2024-08-17T05:55:19.044081+00:00 | 2024-08-17T05:55:19.044113+00:00 | 8 | false | # Code\n```\nclass Solution \n{\npublic:\n int minimumTeachings(int lang, vector<vector<int>>& languages, vector<vector<int>>& friendships) \n {\n map<int,set<int>> m;\n int n=languages.size();\n for(int x=0; x<n; x++)\n {\n set<int> st;\n for(auto it: languages[x... | 0 | 0 | ['Hash Table', 'C++'] | 0 |
minimum-number-of-people-to-teach | Solution Minimum Number of People to Teach | solution-minimum-number-of-people-to-tea-2sqm | 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 | Suyono-Sukorame | NORMAL | 2024-07-02T06:45:28.968399+00:00 | 2024-07-02T06:45:28.968432+00:00 | 3 | 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. $$O(n)$$ --... | 0 | 0 | ['PHP'] | 0 |
minimum-number-of-people-to-teach | 1733. Minimum Number of People to Teach.cpp | 1733-minimum-number-of-people-to-teachcp-kyo3 | Code\n\nclass Solution {\npublic:\n bool findCommon(vector<int>&a, vector<int>&b) {\n for(int i=0;i<a.size();i++) {\n for(int j=0;j<b.size( | 202021ganesh | NORMAL | 2024-06-10T09:05:34.467140+00:00 | 2024-06-10T09:05:34.467173+00:00 | 0 | false | **Code**\n```\nclass Solution {\npublic:\n bool findCommon(vector<int>&a, vector<int>&b) {\n for(int i=0;i<a.size();i++) {\n for(int j=0;j<b.size();j++) {\n if(a[i]==b[j])\n return 1;\n }\n }\n return 0;\n } \n int minimumTeachings... | 0 | 0 | ['C'] | 0 |
minimum-number-of-people-to-teach | Java HashMap | java-hashmap-by-abhinvsinh-l4r5 | Intuition\n1. We need to find out which friends can already communicate so that we don\'t need to calculate the language teaching for them.\n2. For every langua | abhinvsinh | NORMAL | 2024-06-01T06:49:05.415786+00:00 | 2024-06-01T06:49:05.415818+00:00 | 21 | false | # Intuition\n1. We need to find out which friends can already communicate so that we don\'t need to calculate the language teaching for them.\n2. For every language , iterate through the friendship array and check if we need to teach the language to both the friends to be able to communicate.\n3. At the end , compare w... | 0 | 0 | ['Java'] | 0 |
minimum-number-of-people-to-teach | O(M*N) Greedy solution using Set | omn-greedy-solution-using-set-by-vgvishe-k6q0 | \n# Complexity\n- Time complexity:\nO(M*N)\n\n- Space complexity:\nO(M)\n\n# Code\n\nfunction minimumTeachings(n: number, languages: number[][], friendships: nu | vgvishesh | NORMAL | 2024-05-19T04:14:20.446781+00:00 | 2024-05-19T04:14:20.446809+00:00 | 1 | false | \n# Complexity\n- Time complexity:\nO(M*N)\n\n- Space complexity:\nO(M)\n\n# Code\n```\nfunction minimumTeachings(n: number, languages: number[][], friendships: number[][]): number {\n function createFriendsLanguageSets() {\n let friendsLanguageSets: Set<number>[] = new Array(1);\n for (let i = 0; i < lang... | 0 | 0 | ['TypeScript'] | 0 |
minimum-number-of-people-to-teach | 3lines: | 3lines-by-qulinxao-nnez | \n# Code\n\nclass Solution:\n def minimumTeachings(_, n: int, l: List[List[int]], f: List[List[int]]) -> int:\n q=set()\n for a,b in f: not set | qulinxao | NORMAL | 2024-04-23T08:31:37.257521+00:00 | 2024-04-23T08:33:36.318259+00:00 | 19 | false | \n# Code\n```\nclass Solution:\n def minimumTeachings(_, n: int, l: List[List[int]], f: List[List[int]]) -> int:\n q=set()\n for a,b in f: not set(l[a-1]).intersection(set(l[b-1]))and(q.add(a),q.add(b))\n return len(q)-(max(c.values())if ((c:=Counter()).update(chain.from_iterable((l[h-1]for h in... | 0 | 0 | ['Union Find', 'Python3'] | 0 |
minimum-number-of-people-to-teach | Simple Intuitive Greedy (Beats 97% TE and 77% ME) | simple-intuitive-greedy-beats-97-te-and-pd276 | Intuition\nIf people in a friendship are able to communicate with each other with at least 1 language, we do not care about them here.\nBut if they are not able | anushree97 | NORMAL | 2024-04-10T07:26:17.942832+00:00 | 2024-04-10T07:26:17.942864+00:00 | 23 | false | # Intuition\nIf people in a friendship are able to communicate with each other with at least 1 language, we do not care about them here.\nBut if they are not able to communicate, we need to find the language which is most popular amoung this set and the number of speakers of that language. We will need to teach the oth... | 0 | 0 | ['Greedy', 'Python3'] | 1 |
minimum-number-of-people-to-teach | Sol in Kotlin | sol-in-kotlin-by-arista_agarwal-r7f7 | 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 | Arista_agarwal | NORMAL | 2024-03-02T04:59:10.409828+00:00 | 2024-03-02T04:59:10.409854+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. $$O(n)$$ --... | 0 | 0 | ['Kotlin'] | 0 |
minimum-number-of-people-to-teach | Beats 100%, Hash Table, O(n + m) | beats-100-hash-table-on-m-by-igor0-cfkc | Intuition\nCreate a list of users\' pairs which need to learn a new language.\n\n# Approach\nAt first create new dictionary dicLang with a user as a key and a s | igor0 | NORMAL | 2024-02-28T22:30:23.694059+00:00 | 2024-02-28T22:37:27.850246+00:00 | 18 | false | # Intuition\nCreate a list of users\' pairs which need to learn a new language.\n\n# Approach\nAt first create new dictionary `dicLang` with a user as a key and a set of user\'s languages as a value:\n```\nvar dicLang = CreateDicOfLanguages(languages);\n```\nThe create a list of users who need to learn a new language:\... | 0 | 0 | ['Hash Table', 'C#'] | 0 |
minimum-number-of-people-to-teach | 1733. Minimum Number of People to Teach | 1733-minimum-number-of-people-to-teach-b-2idn | 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 | pgmreddy | NORMAL | 2024-01-15T13:43:31.825001+00:00 | 2024-01-15T13:43:31.825039+00:00 | 14 | 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. $$O(n)$$ --... | 0 | 0 | ['JavaScript'] | 0 |
minimum-number-of-people-to-teach | Easy to understand JavaScript solution (Greedy) | easy-to-understand-javascript-solution-g-eyft | Complexity\n- Time complexity:\nO(mn)\n\n- Space complexity:\nO(mn)\n\n# Code\n\nvar minimumTeachings = function(n, languages, friendships) {\n const size = | tzuyi0817 | NORMAL | 2023-10-22T06:29:56.362543+00:00 | 2023-10-22T06:29:56.362564+00:00 | 14 | false | # Complexity\n- Time complexity:\n$$O(mn)$$\n\n- Space complexity:\n$$O(mn)$$\n\n# Code\n```\nvar minimumTeachings = function(n, languages, friendships) {\n const size = languages.length;\n const communicate = Array(size).fill(\'\').map(_ => Array(n).fill(false));\n const users = new Set();\n let result = N... | 0 | 0 | ['JavaScript'] | 0 |
minimum-number-of-people-to-teach | C++ | Commented | c-commented-by-jiahangli-slfn | 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 | JiahangLi | NORMAL | 2023-10-11T15:21:10.112524+00:00 | 2023-10-11T15:21:10.112542+00:00 | 33 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-number-of-people-to-teach | C++ | c-by-user5976fh-06em | \nclass Solution {\npublic:\n bool same(int a, int b, vector<vector<int>>& l){\n int i = 0, y = 0;\n while (i < l[a].size() && y < l[b].size()) | user5976fh | NORMAL | 2023-09-24T02:31:53.073038+00:00 | 2023-09-24T02:31:53.073057+00:00 | 6 | false | ```\nclass Solution {\npublic:\n bool same(int a, int b, vector<vector<int>>& l){\n int i = 0, y = 0;\n while (i < l[a].size() && y < l[b].size()){\n if (l[a][i] == l[b][y]) return true;\n else if (l[a][i] < l[b][y]) ++i;\n else ++y;\n }\n return false;\n ... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | C++ Brute Force Solution | c-brute-force-solution-by-ranjan_him212-tmy7 | \nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int ans = INT_MAX;\n | ranjan_him212 | NORMAL | 2023-09-20T08:04:11.751462+00:00 | 2023-09-20T08:04:11.751493+00:00 | 6 | false | ```\nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int ans = INT_MAX;\n vector<set<int>> lang;\n set<int> stt;\n lang.push_back(stt);\n for(auto &x : languages)\n {\n set<int> st;\n ... | 0 | 0 | ['C'] | 0 |
minimum-number-of-people-to-teach | C++ solution | c-solution-by-pejmantheory-2j25 | 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 | pejmantheory | NORMAL | 2023-09-01T04:00:20.209998+00:00 | 2023-09-01T04:00:20.210019+00:00 | 27 | 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. $$O(n)$$ --... | 0 | 0 | ['C++'] | 0 |
minimum-number-of-people-to-teach | good one | good-one-by-guventuncay-cf20 | 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 | guventuncay | NORMAL | 2023-06-25T16:05:12.063834+00:00 | 2023-06-25T16:05:12.063873+00:00 | 61 | 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. $$O(n)$$ --... | 0 | 0 | ['Java'] | 0 |
minimum-number-of-people-to-teach | Space: O(m*n), Time: O(m*n) | space-omn-time-omn-by-iitjsagar-m80u | class Solution(object):\n def minimumTeachings(self, n, languages, friendships):\n """\n :type n: int\n :type languages: List[List[int]] | iitjsagar | NORMAL | 2023-06-20T14:02:46.801555+00:00 | 2023-06-20T14:02:46.801579+00:00 | 35 | false | class Solution(object):\n def minimumTeachings(self, n, languages, friendships):\n """\n :type n: int\n :type languages: List[List[int]]\n :type friendships: List[List[int]]\n :rtype: int\n """\n \n people = len(languages)+1\n \n lang_matrix = [[F... | 0 | 0 | ['Graph'] | 0 |
minimum-number-of-people-to-teach | Java solution 70% | java-solution-70-by-valushaprogramist-1a94 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe most complicated part was to understand that it must be one language to solve every | ValushaProgramist | NORMAL | 2023-06-19T21:44:49.659523+00:00 | 2023-06-19T21:44:49.659540+00:00 | 79 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe most complicated part was to understand that it must be one language to solve everything. \n# Approach\n<!-- Describe your approach to solving the problem. -->\nI decided to create a map of zeros and ones to catch all teach events and... | 0 | 0 | ['Array', 'Java'] | 0 |
minimum-number-of-people-to-teach | Python3 solution: logic + brute force | python3-solution-logic-brute-force-by-hu-o2ll | Intuition\n Describe your first thoughts on how to solve this problem. \nFirst intuition was graph, because the problem was framed like a graph problem\n# Appro | huikinglam02 | NORMAL | 2023-05-31T05:41:34.091529+00:00 | 2023-05-31T05:41:34.091564+00:00 | 87 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nFirst intuition was graph, because the problem was framed like a graph problem\n# Approach\n<!-- Describe your approach to solving the problem. -->\nSince constraints are rather small, brute force trial and error would suffice\n# Complexi... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | C# Solution using Set || Easy To understand | c-solution-using-set-easy-to-understand-799q3 | \n\npublic class Solution {\n public int MinimumTeachings(int n, int[][] languages, int[][] friendships) {\n HashSet<int> langsUnion = new HashSet<int | mohamedAbdelety | NORMAL | 2023-04-29T14:09:54.576168+00:00 | 2023-04-29T14:09:54.576211+00:00 | 50 | false | \n```\npublic class Solution {\n public int MinimumTeachings(int n, int[][] languages, int[][] friendships) {\n HashSet<int> langsUnion = new HashSet<int>();\n HashSet<int> diffSet = new HashSet<int>();\n foreach(var friend in friendships){\n int u = friend[0] - 1, v = friend[1] - 1;\... | 0 | 0 | ['C#'] | 0 |
minimum-number-of-people-to-teach | Go clean code | go-clean-code-by-ythosa-ovx1 | \n\nfunc minimumTeachings(n int, languages [][]int, friendships [][]int) int {\n\tvar cantSpeekPersons = make(map[int]struct{})\n\tfor _, friendship := range fr | ythosa | NORMAL | 2023-04-28T00:12:28.488576+00:00 | 2023-04-28T00:12:28.488608+00:00 | 50 | false | \n```\nfunc minimumTeachings(n int, languages [][]int, friendships [][]int) int {\n\tvar cantSpeekPersons = make(map[int]struct{})\n\tfor _, friendship := range friendships {\n\t\tfst := friendship[0] - 1\n\t\tsnd := friendship[1] - 1\n\t\tif !isLanguagesIntersects(languages, fst, snd) {\n\t\t\tcantSpeekPersons[fst] = ... | 0 | 0 | ['Go'] | 0 |
minimum-number-of-people-to-teach | clean commented code | clean-commented-code-by-grumpyg-y8z8 | the question is tricky to understand but easy to implement when you overcome that intial hurdle. \n\n\nclass Solution:\n def minimumTeachings(self, n: int, l | grumpyG | NORMAL | 2023-03-29T21:33:05.011838+00:00 | 2023-03-29T21:33:05.011868+00:00 | 92 | false | the question is tricky to understand but easy to implement when you overcome that intial hurdle. \n\n```\nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n # modify languages to be a set for quick intersection \n languages = {person... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | Just a runnable solution | just-a-runnable-solution-by-ssrlive-qavx | Code\n\nimpl Solution {\n pub fn minimum_teachings(n: i32, languages: Vec<Vec<i32>>, friendships: Vec<Vec<i32>>) -> i32 {\n fn check(a: &[Vec<i32>], u | ssrlive | NORMAL | 2023-02-20T06:02:46.032580+00:00 | 2023-02-20T06:02:46.032625+00:00 | 50 | false | # Code\n```\nimpl Solution {\n pub fn minimum_teachings(n: i32, languages: Vec<Vec<i32>>, friendships: Vec<Vec<i32>>) -> i32 {\n fn check(a: &[Vec<i32>], u: usize, v: usize, mem: &mut [Vec<i32>]) -> bool {\n if mem[u][v] != 0 {\n return mem[u][v] == 1;\n }\n for... | 0 | 0 | ['Rust'] | 0 |
minimum-number-of-people-to-teach | Solution | solution-by-deleted_user-gamb | C++ []\nclass Solution {\nprivate:\n bool intersects(vector<int>& a, vector<int>& b, vector<bool>& vis){\n for(int elem: a){\n vis[elem] = | deleted_user | NORMAL | 2023-02-03T07:28:57.463158+00:00 | 2023-03-08T08:15:42.190052+00:00 | 225 | false | ```C++ []\nclass Solution {\nprivate:\n bool intersects(vector<int>& a, vector<int>& b, vector<bool>& vis){\n for(int elem: a){\n vis[elem] = true;\n }\n\n bool intersectionFound = false;\n for(int elem: b){\n if(vis[elem]){\n intersectionFound = true;... | 0 | 0 | ['C++', 'Java', 'Python3'] | 0 |
minimum-number-of-people-to-teach | Python short 3-line solution | python-short-3-line-solution-by-vincent_-ivlh | \ndef minimumTeachings(self, n: int, ls: List[List[int]], es: List[List[int]]) -> int:\n\tls = [{0}]+[set(i) for i in ls]\n\tns = reduce(ior, [{i, j} for i, j i | vincent_great | NORMAL | 2023-01-22T13:17:23.287948+00:00 | 2023-01-22T13:19:20.952936+00:00 | 57 | false | ```\ndef minimumTeachings(self, n: int, ls: List[List[int]], es: List[List[int]]) -> int:\n\tls = [{0}]+[set(i) for i in ls]\n\tns = reduce(ior, [{i, j} for i, j in es if not ls[i]&ls[j]] or [{0}]) # ns=={0} means no teaching is needed\n\treturn min(sum(i not in ls[k] for k in ns) for i in range(1, n+1)) if len(ns)>1 e... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | O(friendships * languages) time, O(number_of_people space) | ofriendships-languages-time-onumber_of_p-xmh1 | Approach\nFirst check who are the people that have some unsatisfying friendship (have a friend that they can\'t speak with).\nnow among those people check what | sroninio | NORMAL | 2023-01-10T12:04:36.435861+00:00 | 2023-01-10T12:04:36.435892+00:00 | 84 | false | # Approach\nFirst check who are the $$people$$ that have some unsatisfying friendship (have a friend that they can\'t speak with).\nnow among those $$people$$ check what is the most popular language.\nthis is the language that needs to be taught to all the $$people$$ that dont know it.\n\n\n# Code\n```\nclass Solution(... | 0 | 0 | ['Python'] | 0 |
minimum-number-of-people-to-teach | Python(Simple Maths) | pythonsimple-maths-by-rnotappl-ekgg | 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 | rnotappl | NORMAL | 2022-12-19T15:45:50.346432+00:00 | 2022-12-19T15:45:50.346471+00:00 | 171 | 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. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | Java solution intuitive | java-solution-intuitive-by-jeshupatelg37-t4ns | \n\n# Code\n\nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n ArrayList<HashSet<Integer>> lan = new | jeshupatelg3774 | NORMAL | 2022-12-18T21:15:58.874460+00:00 | 2022-12-18T21:15:58.874501+00:00 | 190 | false | \n\n# Code\n```\nclass Solution {\n public int minimumTeachings(int n, int[][] languages, int[][] friendships) {\n ArrayList<HashSet<Integer>> lan = new ArrayList<>();\n for(int i =0;i<languages.length;i++){//converting languages array to list of set for easy search\n lan.add(new HashSet<>()... | 0 | 0 | ['Java'] | 0 |
minimum-number-of-people-to-teach | [JavaScript] Clean Solution | javascript-clean-solution-by-cucla101-ru1l | Code\n\n\n\nlet minimumTeachings = function(n, languages, friendships) {\n let mostSpokenLanguages = new Array(n + 1).fill(0);\n let PPL = new Map | cucla101 | NORMAL | 2022-12-07T20:42:57.881922+00:00 | 2022-12-07T20:44:43.025561+00:00 | 127 | false | # Code\n\n```\n\nlet minimumTeachings = function(n, languages, friendships) {\n let mostSpokenLanguages = new Array(n + 1).fill(0);\n let PPL = new Map();\n\n friendships.forEach( ( element ) => \n {\n let LANG = new Set( languages[element[0] - 1] );\n\n let shareLang =... | 0 | 0 | ['JavaScript'] | 0 |
minimum-number-of-people-to-teach | [Python] - Commented and Simple Python Solution | python-commented-and-simple-python-solut-a0ho | Intuition\n Describe your first thoughts on how to solve this problem. \nI think this problem is in the harder range of medium problems.\n\nThere are some thing | Lucew | NORMAL | 2022-12-01T11:10:09.141381+00:00 | 2022-12-01T11:10:09.141407+00:00 | 114 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nI think this problem is in the harder range of medium problems.\n\nThere are some things you need to consider:\n1) We only deal with users that are disconnected from other users\n2) We need to keep track of the most common known language ... | 0 | 0 | ['Python3'] | 0 |
minimum-number-of-people-to-teach | C++ solution with explanation | c-solution-with-explanation-by-slothalan-4108 | So, the question statement shall be paraphrased as follows:\n\nFind the minimum number of users to learn a new language L, such that for each pair of immediate | slothalan | NORMAL | 2022-11-29T13:53:51.982272+00:00 | 2022-11-29T13:53:51.982317+00:00 | 54 | false | So, the question statement shall be paraphrased as follows:\n\nFind the minimum number of users to learn a new language L, such that *for each pair of immediate friends (**no** nested relationships...), they don\'t have a common language, but each of them may or may not have learned L.*\n\n(-_-)\n\n```cpp\nclass Soluti... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | python greedy solution | python-greedy-solution-by-alston16-e7nh | \nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n counter = [set() for i in r | ALSTON16 | NORMAL | 2022-11-12T04:15:52.259794+00:00 | 2022-11-12T04:15:52.259847+00:00 | 31 | false | ```\nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n counter = [set() for i in range(len(languages))]\n memo = dict()\n for i in range(len(languages)):\n for j in languages[i]:\n counter[i].add(j)\... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | [C++] |Simple|Easy to read and Unerstand | c-simpleeasy-to-read-and-unerstand-by-en-pe9c | \nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int m=languages.size() | endless_mercury | NORMAL | 2022-11-02T16:56:04.891779+00:00 | 2022-11-02T16:56:04.891820+00:00 | 74 | false | ```\nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int m=languages.size();\n //keep track if person can speak language already\n vector<vector<bool>> can_speak(m+1,vector<bool>(n+1,false));\n for(int i=0;i<m;i... | 0 | 0 | ['C'] | 0 |
minimum-number-of-people-to-teach | Python 3 Solution | python-3-solution-by-aryaman73-nluj | \nIndexing is a bit annoying, since everything is 1-indexed, so keep that in mind. \n\npy\nclass Solution:\n def minimumTeachings(self, n: int, languages: Li | aryaman73 | NORMAL | 2022-10-17T03:30:41.589871+00:00 | 2022-10-17T03:30:41.589917+00:00 | 110 | false | \nIndexing is a bit annoying, since everything is 1-indexed, so keep that in mind. \n\n```py\nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n \n # Find which friendships can not communicate, i.e. which users need to be taught\n ... | 0 | 0 | ['Python'] | 0 |
minimum-number-of-people-to-teach | Use sets to get friends who can't communicate | use-sets-to-get-friends-who-cant-communi-dy9x | \nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n hmap = {}\n for i, l | rktayal | NORMAL | 2022-09-27T02:25:31.241925+00:00 | 2022-09-27T02:25:31.241966+00:00 | 40 | false | ```\nclass Solution:\n def minimumTeachings(self, n: int, languages: List[List[int]], friendships: List[List[int]]) -> int:\n hmap = {}\n for i, l_list in enumerate(languages):\n hmap[i+1] = set()\n for item in l_list:\n hmap[i+1].add(item)\n \n mis_fr... | 0 | 0 | ['Ordered Set', 'Python'] | 0 |
minimum-number-of-people-to-teach | Confusion | confusion-by-zsun273-6bch | For this example, why is the answer 3?\n\n4\n[[1],[2],[3],[4]]\n[[1,2],[3,4]]\n\nwe can simply teach 1 language 2 and teach 3 language 4, which leads to answer | zsun273 | NORMAL | 2022-09-20T19:25:15.429202+00:00 | 2022-09-20T19:25:15.429241+00:00 | 22 | false | For this example, why is the answer 3?\n\n4\n[[1],[2],[3],[4]]\n[[1,2],[3,4]]\n\nwe can simply teach 1 language 2 and teach 3 language 4, which leads to answer 2. | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | [Time & Memory 100%] Try all languages and find minimum | time-memory-100-try-all-languages-and-fi-rpov | Complexity\nTime: O(m * n)\nSpace: O(m * n)\n\ntypescript\nfunction minimumTeachings(n: number, languages: number[][], friendships: number[][]): number {\n / | jhw123 | NORMAL | 2022-08-30T09:11:51.520803+00:00 | 2022-08-30T09:11:51.520849+00:00 | 70 | false | ### Complexity\nTime: `O(m * n)`\nSpace: `O(m * n)`\n\n```typescript\nfunction minimumTeachings(n: number, languages: number[][], friendships: number[][]): number {\n // filter the friends who do not have a common language and need to be taught\n\tconst friendshipsWithProblem = []\n for (const [f1, f2] of friends... | 0 | 0 | ['Ordered Set', 'TypeScript', 'JavaScript'] | 0 |
minimum-number-of-people-to-teach | C++, filter friendships, 100% | c-filter-friendships-100-by-cx3129-001a | \nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int m=languages.size() | cx3129 | NORMAL | 2022-06-05T10:48:40.935142+00:00 | 2022-06-05T10:59:54.473877+00:00 | 112 | false | ```\nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& languages, vector<vector<int>>& friendships) {\n int m=languages.size();\n vector<vector<bool>> hasLan(m+1,vector<bool>(n+1,false)); //cache for which languages each user known\n for(int i=1;i<=m;++i) \n ... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | [C++] || Set,Map | c-setmap-by-rahul921-x45n | \nclass Solution {\npublic:\n set<int> cantTalk ;\n bool InCommon(vector<int>& lang1 , vector<int>&lang2 ){\n //checks if two friends can speak an | rahul921 | NORMAL | 2022-05-27T08:30:09.095518+00:00 | 2022-05-27T08:30:09.095559+00:00 | 180 | false | ```\nclass Solution {\npublic:\n set<int> cantTalk ;\n bool InCommon(vector<int>& lang1 , vector<int>&lang2 ){\n //checks if two friends can speak any language in common ?\n for(auto &x : lang1)\n for(auto &y : lang2)\n if(x == y) return true ;\n \n \n ... | 0 | 0 | ['C'] | 0 |
minimum-number-of-people-to-teach | Intersection of common languages | intersection-of-common-languages-by-ttn6-yiq8 | Very unclear description, I must say...\nBut not a bad question after all.\n\ncpp\n// check if user u and v share a common language.\n// sorted array and two-po | ttn628826 | NORMAL | 2022-05-21T12:46:50.598397+00:00 | 2022-05-21T12:51:19.985851+00:00 | 108 | false | Very unclear description, I must say...\nBut not a bad question after all.\n\n```cpp\n// check if user u and v share a common language.\n// sorted array and two-pointer, \n// binary search might be even better.\nbool haveCommon(vector<vector<int>>& lan, int u, int v)\n{\n\tint i = 0;\n\tint j = 0;\n\t\n\twhile (i < lan... | 0 | 0 | ['C'] | 0 |
minimum-number-of-people-to-teach | 93% Time and 89% Space, 277 ms Simple Logic Explained | 93-time-and-89-space-277-ms-simple-logic-2jj0 | ```\nclass Solution {\npublic:\n #define pii pair\n #define f first\n #define s second\n int minimumTeachings(int n, vector>& langs, vector>& conn) | ms2000 | NORMAL | 2022-05-21T04:35:58.870046+00:00 | 2022-05-21T04:39:11.032535+00:00 | 74 | false | ```\nclass Solution {\npublic:\n #define pii pair<int,int>\n #define f first\n #define s second\n int minimumTeachings(int n, vector<vector<int>>& langs, vector<vector<int>>& conn) {\n \n vector<int> allLang(n+1,0);\n int m = conn.size();\n vector<pii> graph;\n int mx = 0;... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | Scala | scala-by-fairgrieve-q2nc | \nimport scala.collection.immutable.BitSet\nimport scala.util.chaining.scalaUtilChainingOps\n\nobject Solution {\n def minimumTeachings(n: Int, languages: Arra | fairgrieve | NORMAL | 2022-04-04T05:17:42.591734+00:00 | 2022-04-04T05:17:42.591771+00:00 | 38 | false | ```\nimport scala.collection.immutable.BitSet\nimport scala.util.chaining.scalaUtilChainingOps\n\nobject Solution {\n def minimumTeachings(n: Int, languages: Array[Array[Int]], friendships: Array[Array[Int]]): Int = {\n languages\n .map(_.to(BitSet))\n .pipe { languages =>\n friendships\n ... | 0 | 0 | ['Scala'] | 0 |
minimum-number-of-people-to-teach | count for friends without same language c++ | count-for-friends-without-same-language-67atg | \nclass Solution {\n bool hassame(const vector<int>& v1, const vector<int>& v2)\n {\n int i1=0, i2=0;\n while(i1<v1.size() && i2<v2.size())\ | dolaamon2 | NORMAL | 2022-04-02T13:47:03.670331+00:00 | 2022-04-02T13:49:45.537020+00:00 | 78 | false | ```\nclass Solution {\n bool hassame(const vector<int>& v1, const vector<int>& v2)\n {\n int i1=0, i2=0;\n while(i1<v1.size() && i2<v2.size())\n {\n if (v1[i1] == v2[i2])\n return true;\n else if (v1[i1] < v2[i2])\n i1++;\n else\n... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | Java solution with explanation | java-solution-with-explanation-by-ptdao-jdn5 | \n\nclass Solution {\n \n /**\n Algorithm:\n\n 1. We will initialize the variable \'minimumUsers\' = total users, and it will store the minimum | ptdao | NORMAL | 2022-03-22T00:05:14.364621+00:00 | 2022-03-22T00:05:14.364721+00:00 | 112 | false | \n```\nclass Solution {\n \n /**\n Algorithm:\n\n 1. We will initialize the variable \'minimumUsers\' = total users, and it will store the minimum number of users that needs to learn the common language.\n 2. Maintain a user Map, which stores users who cannot communicate with their friends.\n ... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | Golang | O(N * Len(FriendShip) | Bitset, SIMD, Loop Unroll | Beat 100% | golang-on-lenfriendship-bitset-simd-loop-hesr | \nfunc minimumTeachings(n int, langs [][]int, friends [][]int) int {\n m := len(langs)\n \n set := make([][8]uint64, m)\n for i, lang := range langs | linhduong | NORMAL | 2021-12-13T13:45:54.918861+00:00 | 2021-12-13T13:46:17.193654+00:00 | 242 | false | ```\nfunc minimumTeachings(n int, langs [][]int, friends [][]int) int {\n m := len(langs)\n \n set := make([][8]uint64, m)\n for i, lang := range langs {\n for _, l := range lang {\n set[i] = turnOn(set[i], l - 1)\n } \n }\n \n added := make([]bool, m)\n \n min := m+1\... | 0 | 0 | ['Go'] | 0 |
minimum-number-of-people-to-teach | [Golang] Two steps brute force solution | golang-two-steps-brute-force-solution-by-tut9 | Get the people who need learn other language then store in a map.\n2. Check the number of people need to learn if teaching them from language "1" to "n". \nThe | genius52 | NORMAL | 2021-12-03T05:25:01.966280+00:00 | 2021-12-03T05:25:36.537438+00:00 | 95 | false | 1. Get the people who need learn other language then store in a map.\n2. Check the number of people need to learn if teaching them from language "1" to "n". \nThe minimum number is we need.\n```\nfunc minimumTeachings(n int, languages [][]int, friendships [][]int) int {\n\tvar need_learn_people map[int]bool = make(map[... | 0 | 0 | ['Go'] | 0 |
minimum-number-of-people-to-teach | C# 100% | c-100-by-rahulvn389-rwnx | public class Solution {\n \n private bool CanCommunicate(int[][] languages, int a, int b)\n {\n var languagesA = languages[a-1];\n var la | rahulvn389 | NORMAL | 2021-11-02T19:02:38.919863+00:00 | 2021-11-02T19:02:38.919899+00:00 | 58 | false | public class Solution {\n \n private bool CanCommunicate(int[][] languages, int a, int b)\n {\n var languagesA = languages[a-1];\n var languagesB = languages[b-1];\n \n int i=0;\n int j=0;\n while(i<languagesA.Length && j<languagesB.Length)\n {\n if(l... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | c++ | bitset | c-bitset-by-_seg_fault-pueq | Idea\nFor every person who can\'t talk with its friend, store which language he/she needs to be taught and store the frequency of that language in an array toTe | _seg_fault | NORMAL | 2021-09-08T09:27:39.655978+00:00 | 2021-09-08T09:27:55.897171+00:00 | 115 | false | <b>Idea</b>\nFor every person who can\'t talk with its friend, store which language he/she needs to be taught and store the frequency of that language in an array `toTeach`. The language that needs to be taught to minimum number of people is the desired one. \nNote: In order to ensure that the same person to be taught ... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | [100% Solution] Concise Solution speeding up with bitmap + Cpp Code | 100-solution-concise-solution-speeding-u-3oab | \nRuntime: 120 ms, faster than 100.00% of C++ online submissions \nMemory Usage: 59.2 MB, less than 98.82% of C++ online submissions\n\nWe have N languages and | mkyang | NORMAL | 2021-08-31T10:49:05.158840+00:00 | 2021-08-31T10:50:46.300189+00:00 | 181 | false | ```\nRuntime: 120 ms, faster than 100.00% of C++ online submissions \nMemory Usage: 59.2 MB, less than 98.82% of C++ online submissions\n```\nWe have N languages and M persons. Let us image that we have a MxN table "L".\n```\nL[m][n] = 1 means the person m knows the languag n, and 0 means opposite.\n```\nIf two people ... | 0 | 0 | [] | 0 |
minimum-number-of-people-to-teach | C++ | unordered_set | c-unordered_set-by-sonusharmahbllhb-9q44 | \nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& lng, vector<vector<int>>& frnd) {\n unordered_set<int> unc;\n in | sonusharmahbllhb | NORMAL | 2021-08-28T07:09:35.944265+00:00 | 2021-08-28T07:09:35.944307+00:00 | 336 | false | ```\nclass Solution {\npublic:\n int minimumTeachings(int n, vector<vector<int>>& lng, vector<vector<int>>& frnd) {\n unordered_set<int> unc;\n int lsz=lng.size();\n int fsz=frnd.size();\n vector<unordered_set<int>> v(lsz+1);\n for(int i=0;i<lsz;i++){\n unordered_set<int... | 0 | 0 | ['C', 'C++'] | 0 |
minimum-number-of-people-to-teach | JavaScript faster than 100% | javascript-faster-than-100-by-lilongxue-wouw | \n/**\n * @param {number} n\n * @param {number[][]} languages\n * @param {number[][]} friendships\n * @return {number}\n */\nvar minimumTeachings = function(n, | lilongxue | NORMAL | 2021-07-17T12:03:57.245319+00:00 | 2021-07-17T12:03:57.245366+00:00 | 110 | false | ```\n/**\n * @param {number} n\n * @param {number[][]} languages\n * @param {number[][]} friendships\n * @return {number}\n */\nvar minimumTeachings = function(n, languages, friendships) {\n // user count\n const len = languages.length\n function Node(val) {\n \n }\n Node.table = new Array(1 + len... | 0 | 0 | [] | 0 |
sum-of-prefix-scores-of-strings | [C++, Java, Python3] Easy Trie Explained with diagram | c-java-python3-easy-trie-explained-with-52svp | \n\nWe construct a trie to find the number of times each node has been visited.\n\nIterate over ["abc","ab","bc","b"] and the trie looks like:\n\n\n After const | tojuna | NORMAL | 2022-09-18T04:01:13.157566+00:00 | 2022-09-18T04:23:38.390015+00:00 | 9,979 | false | \n\nWe construct a trie to find the number of times each node has been visited.\n\nIterate over `["abc","ab","bc","b"]` and the trie looks like:\n\n\n* After constructing the `trie` we just iterate over the giv... | 105 | 1 | [] | 13 |
sum-of-prefix-scores-of-strings | Simple step by step solution using Trie data structure | simple-step-by-step-solution-using-trie-nyto0 | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem can be effectively solved using a Trie (prefix tree) data structure. The Tr | Reddaiah12345 | NORMAL | 2024-09-25T02:44:19.606444+00:00 | 2024-09-25T02:44:19.606488+00:00 | 14,796 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem can be effectively solved using a Trie (prefix tree) data structure. The Trie structure allows us to efficiently store and traverse the prefixes of words. Each node in the Trie keeps track of how many times a particular prefix... | 88 | 3 | ['Array', 'String', 'Trie', 'Counting', 'C++', 'Java', 'Python3'] | 7 |
sum-of-prefix-scores-of-strings | C++ | Trie | Related Problems | c-trie-related-problems-by-kiranpalsingh-kdf6 | Store each string in trie and add 1 to each prefix of string while inserting.\n- Then, for each string, we sum up the count for all its prefixes.\n\n\n\ncpp\nst | kiranpalsingh1806 | NORMAL | 2022-09-18T04:01:16.143848+00:00 | 2022-09-19T05:21:10.853388+00:00 | 5,459 | false | - Store each string in trie and **add 1 to each prefix of string while inserting**.\n- Then, for each string, **we sum up the count for all its prefixes**.\n\n\n\n```cpp\nstruct TrieNode {\n TrieNode *next[2... | 68 | 2 | ['Trie', 'C'] | 13 |
sum-of-prefix-scores-of-strings | [Python] Explanation with pictures, 2 solutions | python-explanation-with-pictures-2-solut-f21p | Solution 1. Counter of prefix\n\nUse counter C to collect every prefix of every word. \nThen sum up C[pre] for every prefix pre of word word, this is the score | Bakerston | NORMAL | 2022-09-18T04:01:58.958795+00:00 | 2022-09-18T04:40:03.144501+00:00 | 2,676 | false | ### Solution 1. Counter of prefix\n\nUse counter `C` to collect every prefix of every word. \nThen sum up `C[pre]` for every prefix `pre` of word `word`, this is the score of `word`.\n\n**python**\n\n```\nclass Solution:\n def sumPrefixScores(self, W: List[str]) -> List[int]:\n C = collections.defaultdict(int... | 52 | 0 | [] | 11 |
sum-of-prefix-scores-of-strings | Full Dry Run + Brute to Trie Explained, illustrations || Let's Go !! | full-dry-run-brute-to-trie-explained-ill-buzx | Youtube Explanation\nSoon to be on Channel : https://www.youtube.com/@Intuit_and_Code\nEdited:\nhttps://youtu.be/Sh0ca33JjMM?si=9tzS3AuxGtHSgkew\n\n# Intuition | Rarma | NORMAL | 2024-09-25T02:11:13.740855+00:00 | 2024-09-25T11:53:02.876888+00:00 | 6,309 | false | # Youtube Explanation\nSoon to be on Channel : https://www.youtube.com/@Intuit_and_Code\nEdited:\nhttps://youtu.be/Sh0ca33JjMM?si=9tzS3AuxGtHSgkew\n\n# Intuition & Approach\n\n> If you don\'t know trie, don\'t worry this article will help you get an idea about how it works.\n\n with line by line comments. | python3-simple-trie-o-nk-with-line-by-li-5moe | The idea is to use trie. When initializing the trie, instead of storing the end of the word, we are tracking the number of times this prefix has occurred.\nOnce | MeidaChen | NORMAL | 2022-09-18T04:02:09.029154+00:00 | 2024-01-04T19:26:05.006558+00:00 | 1,211 | false | The idea is to use trie. When initializing the trie, instead of storing the end of the word, we are tracking the number of times this prefix has occurred.\nOnce we have the trie, we will go through all the words one more time and count how many times its prefix has occurred to build our result.\n\n```\nclass Solution:\... | 24 | 0 | [] | 3 |
sum-of-prefix-scores-of-strings | Trie | trie-by-votrubac-75g6 | \nFirst, we add all strings into a Trie, incrementing the count for each prefix.\n \nThen, for each string, we aggregate the count for all its prefixes. \n | votrubac | NORMAL | 2022-09-18T04:01:59.754053+00:00 | 2022-09-18T04:01:59.754098+00:00 | 2,401 | false | \nFirst, we add all strings into a Trie, incrementing the count for each prefix.\n \nThen, for each string, we aggregate the count for all its prefixes. \n \n**C++**\n```cpp\nstruct Trie {\n Trie* ch[26] = {};\n int cnt = 0;\n void insert(string &w, int i = 0) {\n auto n = this;\n for (auto... | 23 | 0 | [] | 7 |
sum-of-prefix-scores-of-strings | Trie Solution || O(N*k) || Beats 97% || C++ || Python | trie-solution-onk-beats-97-c-python-by-r-yy7a | Intuition\n Describe your first thoughts on how to solve this problem. \nUse a datastructure that allows quick prefix count\n\n# Approach\n Describe your approa | Rzhek | NORMAL | 2024-09-25T00:10:46.634229+00:00 | 2024-09-25T00:24:56.447120+00:00 | 11,220 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse a datastructure that allows quick prefix count\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n- Add each word to a trie\n- Increment score of nodes while adding words (except root node)\n- Run DFS for each wor... | 22 | 0 | ['String', 'Trie', 'C++', 'Python3'] | 8 |
sum-of-prefix-scores-of-strings | trie, java | trie-java-by-clasiqh-vtsy | Code:\n\n Node root = new Node(); // Trie root.\n class Node {\n int score = 0;\n Node[] child = new Node[26];\n }\n \n public int | clasiqh | NORMAL | 2022-09-18T04:00:32.785885+00:00 | 2022-09-18T04:22:45.741052+00:00 | 1,668 | false | **Code:**\n\n Node root = new Node(); // Trie root.\n class Node {\n int score = 0;\n Node[] child = new Node[26];\n }\n \n public int[] sumPrefixScores(String[] words) {\n for(String word : words) add(word); // make trie.\n \n int [] res = new int[words.length];\n ... | 21 | 2 | ['Trie', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | Trie again | trie-again-by-anwendeng-qesq | Intuition\n Describe your first thoughts on how to solve this problem. \nUse Trie with memeber variable len\n# Approach\n Describe your approach to solving the | anwendeng | NORMAL | 2024-09-25T02:56:44.580955+00:00 | 2024-09-25T02:56:44.580973+00:00 | 1,171 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nUse Trie with memeber variable len\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. Define Trie struct with construtor, insert methods.\n2. In solution, let `trie` be the memeber variable.\n3. Define a member metho... | 16 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 7 |
sum-of-prefix-scores-of-strings | [C++] Detailed Explanation w/ Diagram | Easy and commented code | Simple language | c-detailed-explanation-w-diagram-easy-an-z9cd | IDEA:\n We store each word of the words array in the trie.\n Every Node of the trie contains two arrays of size 26.\n First array denotes the characters of engl | aryanttripathi | NORMAL | 2022-09-18T07:51:37.586428+00:00 | 2022-09-18T07:51:37.586473+00:00 | 1,206 | false | **IDEA:**\n* We store each word of the words array in the trie.\n* Every Node of the trie contains two arrays of size 26.\n* First array denotes the characters of english alphabets i.e index 0 denotes character \'a\', index 1 denotes character \'b\'... like this.\n* Second array number denotes how many number of times ... | 16 | 2 | ['C'] | 2 |
sum-of-prefix-scores-of-strings | Python | Trie-based Pattern | python-trie-based-pattern-by-khosiyat-s5cx | see the Successfully Accepted Submission\n\n# Code\npython3 []\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.prefix_count | Khosiyat | NORMAL | 2024-09-25T01:42:27.827759+00:00 | 2024-09-25T01:42:27.827791+00:00 | 1,175 | false | [see the Successfully Accepted Submission](https://leetcode.com/problems/sum-of-prefix-scores-of-strings/submissions/1401354774/?envType=daily-question&envId=2024-09-25)\n\n# Code\n```python3 []\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.prefix_count = 0\n\nclass Solution:\n ... | 13 | 0 | ['Python3'] | 3 |
sum-of-prefix-scores-of-strings | Trie || Beats 97% || C++ || Python || Java | trie-beats-97-c-python-java-by-as_313-mxrh | Intuition\nThe goal of the problem is to calculate the total score of every non-empty prefix of each word in a given list of strings. The score of a prefix is d | Baslik69 | NORMAL | 2024-09-25T00:33:38.060780+00:00 | 2024-09-25T00:33:38.060814+00:00 | 3,920 | false | # Intuition\nThe goal of the problem is to calculate the total score of every non-empty prefix of each word in a given list of strings. The score of a prefix is defined as the number of strings in the list that start with that prefix.\n\nTo solve this efficiently, we can use a Trie (prefix tree) data structure. The Tri... | 13 | 1 | ['Array', 'String', 'Trie', 'Counting', 'Python', 'C++', 'Java', 'Python3'] | 1 |
sum-of-prefix-scores-of-strings | Thought process explained | Trie | For beginner | thought-process-explained-trie-for-begin-pigq | So First question?\n How I got to know that Trie will work.\n\t So let\'s thing :\n\t We have 1000 words of 1000 length each. Also we have to get the value for | faltu_admi | NORMAL | 2022-09-18T04:02:38.666712+00:00 | 2022-09-18T04:02:38.666754+00:00 | 578 | false | ## So First question?\n* How I got to know that Trie will work.\n\t* So let\'s thing :\n\t* We have 1000 words of 1000 length each. Also we have to get the value for substring also.\n\t\t* Will HashMap works? No because for 1k word of 1k length, going to store all substring that would cause TLE.\n\t* So let\'s think so... | 12 | 0 | [] | 2 |
sum-of-prefix-scores-of-strings | ⚙ C++ || NO TRIE || NLOGN SOLUTION | c-no-trie-nlogn-solution-by-venom-xd-frvr | \t\tclass Solution {\n\t\tpublic:\n\t\t\tvector sumPrefixScores(vector& words) {\n\t\t\t\tint n = words.size();\n\t\t\t\tvector> words2;\n\t\t\t\tfor (int i = 0 | venom-xd | NORMAL | 2022-09-18T04:31:12.435990+00:00 | 2022-09-18T04:31:12.436041+00:00 | 1,302 | false | \t\tclass Solution {\n\t\tpublic:\n\t\t\tvector<int> sumPrefixScores(vector<string>& words) {\n\t\t\t\tint n = words.size();\n\t\t\t\tvector<pair<string, int>> words2;\n\t\t\t\tfor (int i = 0; i < n; ++i) {\n\t\t\t\t\twords2.push_back(make_pair(words[i], i));\n\t\t\t\t}\n\t\t\t\tsort(words2.begin(), words2.end());\n\n\... | 10 | 0 | ['Trie', 'C', 'C++', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | C# Solution for Sum of Prefix Scores of Strings Problem | c-solution-for-sum-of-prefix-scores-of-s-26iq | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires us to compute the score of each word in an array, where the score | Aman_Raj_Sinha | NORMAL | 2024-09-25T02:52:36.987948+00:00 | 2024-09-25T02:52:36.987984+00:00 | 235 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires us to compute the score of each word in an array, where the score is the sum of how many words have each prefix of the word. A brute force approach (checking each prefix of each word against all other words) would be ... | 9 | 0 | ['C#'] | 0 |
sum-of-prefix-scores-of-strings | 🌟 [Explained]🔍 From Brute Force to Tries: 🚀 Mastering Sum Of Prefix Scores Optimization 🌟 | explained-from-brute-force-to-tries-mast-ub2r | \n# Approach 1: Brute Force (Memory Limit Exceeded - MLE) \uD83D\uDED1\nThis approach uses a brute-force method by storing each word\'s prefixes in a map with v | AlgoArtisan | NORMAL | 2024-09-25T02:24:05.494606+00:00 | 2024-09-25T04:47:36.811546+00:00 | 1,436 | false | \n# Approach 1: Brute Force (Memory Limit Exceeded - MLE) \uD83D\uDED1\nThis approach uses a brute-force method by storing each word\'s prefixes in a map with vector<string>. For every prefix, the corresponding vector stores all occurrences of the prefix, leading to an excessive memory footprint.\n\n# Time Complexity:\... | 9 | 0 | ['Hash Table', 'String', 'Trie', 'C++'] | 3 |
sum-of-prefix-scores-of-strings | [Python] Trie Solution | python-trie-solution-by-lee215-ceji | Explanation\nUse Trie to build up a tree.\nadd word and add 1 point to each node on the word path.\nsocre will sum up the point on the word path.\n\n\n# Complex | lee215 | NORMAL | 2022-09-18T04:06:02.757623+00:00 | 2022-09-19T16:26:38.776881+00:00 | 555 | false | # **Explanation**\nUse Trie to build up a tree.\n`add` word and add `1` point to each node on the word path.\n`socre` will sum up the point on the word path.\n<br>\n\n# **Complexity**\nTime `O(words)`\nSpace `O(words)`\n<br>\n\n\n**Python**\n```py\nclass Trie(object):\n\n def __init__(self):\n T = lambda: def... | 9 | 0 | [] | 6 |
sum-of-prefix-scores-of-strings | [Java/Python 3] Trie build/search prefix w/ brief explanation and analysis. | javapython-3-trie-buildsearch-prefix-w-b-ns8l | Build Trie and accumulate the frequencies of each pefix at the same time; then search each word and compute the corresponding score.\n\njava\nclass Trie {\n | rock | NORMAL | 2022-09-18T04:20:04.183372+00:00 | 2022-11-10T12:02:39.116374+00:00 | 502 | false | Build Trie and accumulate the frequencies of each pefix at the same time; then search each word and compute the corresponding score.\n\n```java\nclass Trie {\n int cnt = 0;\n Map<Character, Trie> kids = new HashMap<>();\n public void add(String word) {\n Trie t = this;\n for (char c : word.toChar... | 8 | 0 | [] | 1 |
sum-of-prefix-scores-of-strings | ✅ One Line Solution | one-line-solution-by-mikposp-ev44 | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1 - Brute Force Counting\npython3\nclass | MikPosp | NORMAL | 2024-09-25T10:55:20.179735+00:00 | 2024-09-25T16:43:57.557965+00:00 | 750 | false | (Disclaimer: this is not an example to follow in a real project - it is written for fun and training mostly)\n\n# Code #1 - Brute Force Counting\n```python3\nclass Solution:\n def sumPrefixScores(self, a: List[str]) -> List[int]:\n return (z:=Counter(s[:i+1] for s in a for i in range(len(s)))) and [sum(z[s[:i... | 7 | 0 | ['Array', 'String', 'Trie', 'Counting', 'Python', 'Python3'] | 1 |
sum-of-prefix-scores-of-strings | String hashing | No trie needed | Python3 | string-hashing-no-trie-needed-python3-by-mvdt | Intuition\nWe can use a rolling hash to cumulatively calculate the hash of all the prefixes in O(len(word)).\nThus, we can store the occurences of this hash in | reas0ner | NORMAL | 2024-09-25T09:30:05.796482+00:00 | 2024-09-25T09:30:05.796514+00:00 | 827 | false | # Intuition\nWe can use a rolling hash to cumulatively calculate the hash of all the prefixes in O(len(word)).\nThus, we can store the occurences of this hash in various strings and lookup the counts in a second loop to calculate the score.\n\n# Complexity\n- Time complexity:\nO(len(words) * len(word))\n\n- Space compl... | 7 | 0 | ['Hash Table', 'Trie', 'Rolling Hash', 'Python3'] | 3 |
sum-of-prefix-scores-of-strings | C++ || Trie || Easy to Understand || Visualization | c-trie-easy-to-understand-visualization-pted9 | Introduction to Tries\n###### A Trie, also known as a prefix tree, is an efficient tree-like data structure used for storing and searching strings. It\'s partic | khalidalam980 | NORMAL | 2024-09-25T01:57:33.538219+00:00 | 2024-09-25T23:46:14.700082+00:00 | 1,029 | false | # Introduction to Tries\n###### A Trie, also known as a prefix tree, is an efficient tree-like data structure used for storing and searching strings. It\'s particularly useful for problems involving prefixes, which makes it ideal for our current challenge.\n\n**In a Trie:**\n- Each node represents a character in a stri... | 7 | 0 | ['String', 'Trie', 'Counting', 'C++'] | 2 |
sum-of-prefix-scores-of-strings | Trie | Java | Beginner friendly🔥 | trie-java-beginner-friendly-by-someone_c-i65q | class Solution {\n \n private class TrieNode{\n TrieNode[]children;\n boolean isEnd;\n int pre_count=0;\n \n TrieNode() | someone_codes | NORMAL | 2022-09-18T12:06:27.408795+00:00 | 2022-09-18T13:33:32.097017+00:00 | 629 | false | class Solution {\n \n private class TrieNode{\n TrieNode[]children;\n boolean isEnd;\n int pre_count=0;\n \n TrieNode(){\n children=new TrieNode[26];\n isEnd=false;\n }\n }\n \n private TrieNode root;\n \n private void insert_word(Stri... | 7 | 0 | ['Trie', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | Don't Trie | Simple hashing! | dont-trie-simple-hashing-by-rac101ran-coxd | \nclass Solution {\npublic:\n unordered_map<long long,int> mp;\n long long base = 31 , pw = 1 , mod = 1011001110001111;\n vector<int> sumPrefixScores(v | rac101ran | NORMAL | 2022-09-18T04:13:09.000335+00:00 | 2022-09-18T04:21:40.334824+00:00 | 1,301 | false | ```\nclass Solution {\npublic:\n unordered_map<long long,int> mp;\n long long base = 31 , pw = 1 , mod = 1011001110001111;\n vector<int> sumPrefixScores(vector<string>& words) {\n vector<int> ans;\n for(string s : words) {\n long long hash = 0 , pw = 1;\n for(int j=0; j<... | 7 | 0 | ['C', 'Rolling Hash'] | 6 |
sum-of-prefix-scores-of-strings | C++ | Trie | Searching | c-trie-searching-by-surajpatel-5fhj | \nclass Solution {\npublic:\n class Trie {\n public:\n Trie() {\n cnt = 0;\n for(int i=0;i<26;i++){\n this->ne | surajpatel | NORMAL | 2022-09-18T04:06:08.934164+00:00 | 2022-09-18T04:06:08.934204+00:00 | 770 | false | ```\nclass Solution {\npublic:\n class Trie {\n public:\n Trie() {\n cnt = 0;\n for(int i=0;i<26;i++){\n this->next[i] = NULL;\n }\n }\n void insert(string word, Trie* root) {\n Trie* node = root;\n int i = 0;\n ... | 7 | 0 | ['String', 'Trie', 'C'] | 4 |
sum-of-prefix-scores-of-strings | TLE Disaster | both hashmap and Trie giving TLE | tle-disaster-both-hashmap-and-trie-givin-w2oo | HashMap\n\nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n \n \n\n int n=words.length;\n int ans[]=new int[n] | deleted_user | NORMAL | 2022-09-18T04:01:14.547715+00:00 | 2022-09-18T04:20:13.883940+00:00 | 945 | false | 1. HashMap\n```\nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n \n \n\n int n=words.length;\n int ans[]=new int[n];\n \n HashMap<String,Integer> map=new HashMap<>();\n HashMap<Integer,HashSet<String>> sub_str=new HashMap<>();\n for(int i=0;... | 7 | 2 | ['Trie'] | 2 |
sum-of-prefix-scores-of-strings | String Hashing || Java | string-hashing-java-by-virendra115-hla6 | \npublic int[] sumPrefixScores(String[] words) {\n int n = words.length, p = 31, m = (int)1e9 + 9;\n int[] ans = new int[n];\n long[] hash = new long[n | virendra115 | NORMAL | 2022-09-18T04:11:45.663418+00:00 | 2022-09-18T04:23:51.725873+00:00 | 734 | false | ```\npublic int[] sumPrefixScores(String[] words) {\n int n = words.length, p = 31, m = (int)1e9 + 9;\n int[] ans = new int[n];\n long[] hash = new long[n], p_pow = new long[n];\n Arrays.fill(p_pow,1);\n for(int i = 0; i < 1000; i++){\n Map<Long,Integer> map = new HashMap<>();\n for(int j =... | 6 | 1 | [] | 3 |
sum-of-prefix-scores-of-strings | Optimized Solution Using a Custom Trie for Prefix Scores | Java | C++ | [Video Explanation] | optimized-solution-using-a-custom-trie-f-djsu | Intuition, approach, and complexity discussed in detail in the video solution\nhttps://youtu.be/HymJeL_ERlg\n# Code\njava []\nclass TrieNode{\n TrieNode chil | Lazy_Potato_ | NORMAL | 2024-09-25T11:30:53.165075+00:00 | 2024-09-25T11:30:53.165120+00:00 | 471 | false | # Intuition, approach, and complexity discussed in detail in the video solution\nhttps://youtu.be/HymJeL_ERlg\n# Code\n``` java []\nclass TrieNode{\n TrieNode children[] = new TrieNode[26];\n int nodeFreq = 0;\n void insert(String key){\n TrieNode curr = this;\n for(var chr : key.toCharArray()){\... | 5 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | 🔥BEATS 💯 % 🎯 |✨SUPER EASY BEGINNERS 👏 | beats-super-easy-beginners-by-codewithsp-j2vs | \n\n\n---\n\n## Intuition\nThe task involves calculating the "prefix score" for each word in a list, where the prefix score is the sum of the counts of all pref | CodeWithSparsh | NORMAL | 2024-09-25T05:51:19.848217+00:00 | 2024-09-25T05:51:19.848287+00:00 | 517 | false | \n\n\n---\n\n## Intuition\nThe task involves calculating the "prefix score" for each word in a list, where the prefix score is the sum of the counts of all prefixes of the word in the given word list. The Tr... | 5 | 0 | ['Trie', 'C', 'C++', 'Java', 'Go', 'Python3', 'JavaScript', 'Dart'] | 0 |
sum-of-prefix-scores-of-strings | 💡 O(n⋅m) | Easy Solution | C++ 462ms Beats 87.78% | Java Py3 | onm-easy-solution-c-462ms-beats-8778-jav-xvcp | \n#\n---\n\n# Intuition\nWe can efficiently solve this problem using a Trie (prefix tree). A Trie helps in storing prefixes and counting their occurrences while | user4612MW | NORMAL | 2024-09-25T03:02:45.557215+00:00 | 2024-09-25T03:08:23.334455+00:00 | 579 | false | \n#\n---\n\n# **Intuition**\nWe can efficiently solve this problem using a **Trie** (prefix tree). A **Trie** helps in storing prefixes and counting their occurrences while reducing redundant comparisons. Each node in the Trie represents a character, and we increment the count for each character as we insert the string... | 5 | 0 | ['Trie', 'C++', 'Java', 'Python3'] | 2 |
sum-of-prefix-scores-of-strings | 🌟 Multi-Language Solution for Prefix Score Mastery Made Easy 🔥💯 | multi-language-solution-for-prefix-score-xokw | \n\nExplore a collection of solutions to LeetCode problems in multiple programming languages. Each solution includes a detailed explanation and step-by-step app | withaarzoo | NORMAL | 2024-09-25T02:00:06.001079+00:00 | 2024-09-25T02:00:06.001099+00:00 | 1,126 | false | \n\nExplore a collection of solutions to LeetCode problems in multiple programming languages. Each solution includes a detailed explanation and step-by-step approach to solving the pr... | 5 | 0 | ['Array', 'String', 'Trie', 'Counting', 'Python', 'C++', 'Java', 'Go', 'Python3', 'JavaScript'] | 2 |
sum-of-prefix-scores-of-strings | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 100 | easiestfaster-lesser-cpython3javacpython-vnqu | Intuition\n\n Describe your first thoughts on how to solve this problem. \n- JavaScript Code --> https://leetcode.com/problems/sum-of-prefix-scores-of-strings/s | Edwards310 | NORMAL | 2024-09-25T01:10:19.347447+00:00 | 2024-09-25T01:10:19.347482+00:00 | 546 | false | # Intuition\n\n<!-- Describe your first thoughts on how to solve this problem. -->\n- ***JavaScript Code -->*** https://leetcode.com/problems/sum-of-prefix-scores-of-strings/submissions/1401331474?... | 5 | 0 | ['Array', 'String', 'Trie', 'C', 'Python', 'C++', 'Java', 'Python3', 'JavaScript', 'C#'] | 0 |
sum-of-prefix-scores-of-strings | Hashing C++ easy to understand | hashing-c-easy-to-understand-by-virujtha-fu19 | The hashing technique is simple. The algorithm states that\n1) Assign a hash code to all prefixes of a word ( Using any hash function or Rabin Karp algorithm). | virujthakur01 | NORMAL | 2022-10-04T08:50:33.236092+00:00 | 2022-10-04T08:50:33.236138+00:00 | 702 | false | The hashing technique is simple. The algorithm states that\n1) Assign a hash code to all prefixes of a word ( Using any hash function or Rabin Karp algorithm). Repeat this for every word in the list of words.\n2) Alongside maintain a frequency map for the hash code, or the prefix, which maintains the number of occurren... | 5 | 0 | ['C'] | 1 |
sum-of-prefix-scores-of-strings | C++ Simple DP solution without Trie (Beats 99% runtime, 97% Space) | c-simple-dp-solution-without-trie-beats-ov5kw | Firstly Sort the array of strings without losing original indices (eg: by using vector>).\n\nThis solution takes advantage of the fact that after sorting the st | Sadid_005 | NORMAL | 2022-09-21T15:44:40.241986+00:00 | 2022-09-21T17:50:02.069792+00:00 | 365 | false | Firstly Sort the array of strings without losing original indices (eg: by using vector<pair<string,int>>).\n\nThis solution takes advantage of the fact that after sorting the strings with identical prefixes will form a consecutive segment.\n\nExample: \nBefore Sorting => "abc" "bc" "ab" "b"\nAfter Sorting => *... | 5 | 0 | ['Dynamic Programming', 'Sorting'] | 0 |
sum-of-prefix-scores-of-strings | Concise Python Tire Implementation | concise-python-tire-implementation-by-51-0hcb | \nclass TrieNode:\n def __init__(self):\n self.next = {}\n self.score = 0\n\n\nclass Solution:\n def sumPrefixScores(self, words: List[str]) | 515384231 | NORMAL | 2022-09-18T04:32:32.377861+00:00 | 2022-09-18T04:32:32.377892+00:00 | 292 | false | ```\nclass TrieNode:\n def __init__(self):\n self.next = {}\n self.score = 0\n\n\nclass Solution:\n def sumPrefixScores(self, words: List[str]) -> List[int]:\n result = []\n \n root = TrieNode()\n \n # build the trie\n for i, word in enumerate(words):\n ... | 5 | 0 | ['Trie', 'Python'] | 3 |
sum-of-prefix-scores-of-strings | C++ Explained Solution | Using Trie | Time - O(NM) | | c-explained-solution-using-trie-time-onm-24d5 | Approach -\nStep 1 - Create a node class that stores characters, a pointer to its next characters \nand count how many times this character appears in every pre | abhinandan__22 | NORMAL | 2022-09-18T04:00:41.335302+00:00 | 2022-09-18T04:13:00.278526+00:00 | 222 | false | **Approach -**\n**Step 1** - Create a node class that stores characters, a pointer to its next characters \nand count how many times this character appears in every prefix of every string of a given vector\n**Step 2** - Create a trie class that has to member function\ni) Push - To push the string in trie\nii) Score - T... | 5 | 0 | [] | 0 |
sum-of-prefix-scores-of-strings | Dcc 25/09/2024 || Trie 🔥 | dcc-25092024-trie-by-sajaltiwari007-tn1u | Intuition:\nThe problem involves calculating the sum of prefix scores for each word in an array of words. The prefix score of a word is determined by the number | sajaltiwari007 | NORMAL | 2024-09-25T09:24:24.553515+00:00 | 2024-09-25T09:24:24.553554+00:00 | 100 | false | ### Intuition:\nThe problem involves calculating the sum of prefix scores for each word in an array of words. The prefix score of a word is determined by the number of times its prefixes appear across all the words in the list.\n\nA **Trie** (prefix tree) is well-suited for problems involving prefixes because it allows... | 4 | 0 | ['Trie', 'Java'] | 0 |
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