question_slug stringlengths 3 77 | title stringlengths 1 183 | slug stringlengths 12 45 | summary stringlengths 1 160 ⌀ | author stringlengths 2 30 | certification stringclasses 2
values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
classes | content stringlengths 4 576k | upvotes int64 0 11.5k | downvotes int64 0 358 | tags stringlengths 2 193 | comments int64 0 2.56k |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
sum-of-prefix-scores-of-strings | Solution By Dare2Solve | Detailed Explanation | Clean Code | solution-by-dare2solve-detailed-explanat-gxhd | Exlanation []\nauthorslog.com/blog/FxtvdgRwjW\n\n# Code\n\ncpp []\nclass TrieNode {\npublic:\n unordered_map<char, TrieNode*> children;\n int prefixCount; | Dare2Solve | NORMAL | 2024-09-25T06:51:42.615867+00:00 | 2024-09-25T06:51:42.615900+00:00 | 787 | false | ```Exlanation []\nauthorslog.com/blog/FxtvdgRwjW\n```\n# Code\n\n```cpp []\nclass TrieNode {\npublic:\n unordered_map<char, TrieNode*> children;\n int prefixCount;\n \n TrieNode() {\n prefixCount = 0;\n }\n};\n\nclass Trie {\npublic:\n TrieNode* root;\n \n Trie() {\n root = new Tri... | 4 | 0 | ['String', 'Trie', 'Python', 'C++', 'Java', 'Python3', 'JavaScript'] | 0 |
sum-of-prefix-scores-of-strings | C++ || Easy Tries Implementation ✅✅ || Beats 82.29% | c-easy-tries-implementation-beats-8229-b-67ym | Intuition\nThe problem requires calculating prefix scores for a list of words. The prefix score of a word is defined as the sum of the counts of occurrences of | arunk_leetcode | NORMAL | 2024-09-25T05:52:44.613774+00:00 | 2024-09-25T05:52:44.613804+00:00 | 288 | false | # Intuition\nThe problem requires calculating prefix scores for a list of words. The prefix score of a word is defined as the sum of the counts of occurrences of all its prefixes in a trie structure. The trie allows efficient insertion and retrieval of prefix counts, making it a suitable data structure for this task.\n... | 4 | 0 | ['String', 'Trie', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | simple and easy Python solution || Trie | simple-and-easy-python-solution-trie-by-ypbpa | if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: ww | shishirRsiam | NORMAL | 2024-09-25T05:05:59.695746+00:00 | 2024-09-25T05:05:59.695776+00:00 | 368 | false | # if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam\n\n\n\n# Complexity\n- Time complexity: O(n * l)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n\n# Code\n```python3 []\nclas... | 4 | 0 | ['Array', 'String', 'Trie', 'Counting', 'Python', 'Python3'] | 7 |
sum-of-prefix-scores-of-strings | Simple Solution Using Trie Data Structure | Java | simple-solution-using-trie-data-structur-rj75 | \n# Code\njava []\nclass Trie {\n Trie[] arr = new Trie[26];\n int count = 0;\n}\nclass Solution {\n\n Trie root = new Trie();\n\n void add_word(Str | eshwaraprasad | NORMAL | 2024-09-25T04:59:02.526614+00:00 | 2024-09-25T04:59:02.526647+00:00 | 247 | false | \n# Code\n```java []\nclass Trie {\n Trie[] arr = new Trie[26];\n int count = 0;\n}\nclass Solution {\n\n Trie root = new Trie();\n\n void add_word(String str) {\n Trie curr = root;\n int ind;\n for(char ch : str.toCharArray()) {\n ind = ch - \'a\';\n if(curr.arr[i... | 4 | 0 | ['Java'] | 0 |
sum-of-prefix-scores-of-strings | simple and easy C++ solution || Trie | simple-and-easy-c-solution-trie-by-shish-dfqt | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook | shishirRsiam | NORMAL | 2024-09-25T04:54:03.126625+00:00 | 2024-09-25T04:54:03.126665+00:00 | 514 | false | \n# if it\'s help, please up \u2B06 vote! \u2764\uFE0F\n\n###### Let\'s Connect on LinkedIn: www.linkedin.com/in/shishirrsiam\n###### Let\'s Connect on Facebook: www.fb.com/shishirrsiam\n\n# Complexity\n- Time complexity: O(n * l)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```cpp []\nclass TrieNo... | 4 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 6 |
sum-of-prefix-scores-of-strings | 2 Approaches || Trie & Hashing | 2-approaches-trie-hashing-by-imdotrahul-z4up | Intuition\n Describe your first thoughts on how to solve this problem. \nTrie:\nTrie approach leverages the tree structure to store and count the frequency of e | imdotrahul | NORMAL | 2024-09-25T03:35:42.553575+00:00 | 2024-09-25T03:35:42.553617+00:00 | 393 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n**Trie:**\nTrie approach leverages the tree structure to store and count the frequency of each prefix. Each node in the Trie represents a prefix formed by the characters from the root to that node. As words are inserted into the Trie, we ... | 4 | 0 | ['Array', 'Hash Table', 'String', 'Trie', 'Counting', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | All solutions | all-solutions-by-dixon_n-t85c | 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 | Dixon_N | NORMAL | 2024-09-25T03:01:15.368206+00:00 | 2024-09-25T03:11:57.990356+00:00 | 28 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 4 | 0 | ['Java'] | 4 |
sum-of-prefix-scores-of-strings | Sum of Prefix Scores of Strings || DCC 25/09/24 || Trie Solution ✅✅ | sum-of-prefix-scores-of-strings-dcc-2509-d1sz | Intuition\nThe problem revolves around finding the sum of scores for all prefixes of each word in the list. A Trie (prefix tree) is particularly suited for this | Ayush_Singh2004 | NORMAL | 2024-09-25T02:36:02.165959+00:00 | 2024-09-25T02:36:02.165991+00:00 | 161 | false | # Intuition\nThe problem revolves around finding the sum of scores for all prefixes of each word in the list. A Trie (prefix tree) is particularly suited for this task, as it efficiently stores and retrieves prefixes of words. By traversing the Trie, we can track how many times each prefix appears, which allows us to c... | 4 | 0 | ['Array', 'Trie', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | Trie, just 3 steps | trie-just-3-steps-by-raviteja_29-v20w | Intuition\n Describe your first thoughts on how to solve this problem. \nTo solve this problem, you need to efficiently calculate the number of times each prefi | raviteja_29 | NORMAL | 2024-07-24T17:12:09.056160+00:00 | 2024-07-24T17:12:09.056191+00:00 | 176 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTo solve this problem, you need to efficiently calculate the number of times each prefix of a given string appears in the list of words. A direct approach could involve checking each prefix of each string against all words, but this would... | 4 | 0 | ['Python3'] | 2 |
sum-of-prefix-scores-of-strings | c++ two solution trie || hashing | c-two-solution-trie-hashing-by-dilipsuth-ylai | \nclass Solution\n{\n public:\n struct node\n {\n node *child[26] = { NULL\n };\n int count = 0;\n };\n | dilipsuthar17 | NORMAL | 2022-09-18T04:01:48.861504+00:00 | 2022-09-18T07:00:46.630347+00:00 | 361 | false | ```\nclass Solution\n{\n public:\n struct node\n {\n node *child[26] = { NULL\n };\n int count = 0;\n };\n node *root = new node();\n void insert(string & s)\n {\n int n = s.size();\n node *curr = root;\n for (int i = 0; i < n; i++)\... | 4 | 0 | ['Trie', 'C', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | Trie || striver Templete | trie-striver-templete-by-mayank670-hene | 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 | Mayank670 | NORMAL | 2024-09-25T13:25:25.359060+00:00 | 2024-09-25T13:25:25.359082+00:00 | 22 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['String', 'Trie', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | JAVA | Trie Data Structure | java-trie-data-structure-by-sudhikshagha-q9dl | Code\njava []\nclass TrieNode {\n TrieNode [] children = new TrieNode[26];\n int count = 0;\n}\nclass Solution {\n TrieNode root = new TrieNode();\n | SudhikshaGhanathe | NORMAL | 2024-09-25T12:57:31.022161+00:00 | 2024-09-25T12:57:31.022197+00:00 | 24 | false | # Code\n```java []\nclass TrieNode {\n TrieNode [] children = new TrieNode[26];\n int count = 0;\n}\nclass Solution {\n TrieNode root = new TrieNode();\n public void add(String word) {\n TrieNode curr = root;\n for (char c : word.toCharArray()) {\n int idx = c - \'a\';\n ... | 3 | 0 | ['Trie', 'Counting', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | [EASY] Simplified explantion of logic (using Trie Data Structure): | easy-simplified-explantion-of-logic-usin-h15g | Intuition\nThe problem asks us to compute the sum of prefix scores for each word in the given list. A prefix score is the total number of words in the list that | rinsane | NORMAL | 2024-09-25T11:25:27.674667+00:00 | 2024-09-25T11:25:27.674700+00:00 | 40 | false | # Intuition\nThe problem asks us to compute the sum of prefix scores for each word in the given list. A prefix score is the total number of words in the list that share the same prefix, up to every letter in the word.\n\nTo efficiently solve this problem, we can leverage a **Trie** (prefix tree). Tries are well-suited ... | 3 | 0 | ['Trie', 'Python3'] | 1 |
sum-of-prefix-scores-of-strings | Sum of Prefix Scores of Strings | sum-of-prefix-scores-of-strings-by-pushp-viai | \n# Approach\n Describe your approach to solving the problem. \nHashMap and Trie\n\n\n\n# Time complexity:\n Add your time complexity here, e.g. O(n) \n- Hashma | Pushparaj_Shetty | NORMAL | 2024-09-25T05:41:08.626147+00:00 | 2024-09-25T05:41:08.626186+00:00 | 89 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\n**HashMap** and **Trie**\n\n\n\n# Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n- **Hashmap**: O(n * k\xB2)\n \n\n- **Trie**: O(n * k)\n\n\n\n# Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\... | 3 | 0 | ['Trie', 'Counting', 'Python3'] | 1 |
sum-of-prefix-scores-of-strings | Easy Solution | Explained with Example & Visual Walkthrough | Beginner Friendly | | easy-solution-explained-with-example-vis-iw3s | Approach\nBuilding the Trie:\nEvery time a letter is inserted into the Trie, we update the count for that letter\u2019s node. This count keeps track of how many | Guna01 | NORMAL | 2024-09-25T01:56:07.257494+00:00 | 2024-09-25T01:57:35.709746+00:00 | 93 | false | # Approach\nBuilding the Trie:\nEvery time a letter is inserted into the Trie, we update the count for that letter\u2019s node. This count keeps track of how many words pass through that node.\n\nExample Words: ["abc", "ab", "bc", "b"]\n\nInsert "abc":\n\'a\' goes into the Trie (count is now 1).\n\'b\' goes into the Tr... | 3 | 0 | ['Array', 'String', 'Trie', 'Counting', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | Rust Solution | rust-solution-by-evanchun-txf6 | Intuition\n\n# Approach\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n\n#[derive(Default)]\nstruct Trie {\n cnt: usize,\n | evanchun | NORMAL | 2024-06-05T06:39:39.949211+00:00 | 2024-06-05T06:39:39.949240+00:00 | 47 | false | # Intuition\n\n# Approach\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(n)\n\n# Code\n```\n#[derive(Default)]\nstruct Trie {\n cnt: usize,\n children: [Option<Box<Trie>>; 27],\n}\n\nimpl Solution {\n pub fn sum_prefix_scores(words: Vec<String>) -> Vec<i32> {\n let mut root = Trie::de... | 3 | 0 | ['Rust'] | 0 |
sum-of-prefix-scores-of-strings | [Python3] Trie - Simple Solution + Detailed Explanation | python3-trie-simple-solution-detailed-ex-rylt | Intuition\n Describe your first thoughts on how to solve this problem. \n- Check for Common Prefix for string -> Trie\n\n# Approach\n Describe your approach to | dolong2110 | NORMAL | 2024-02-18T10:37:47.966641+00:00 | 2024-09-25T07:24:31.004457+00:00 | 143 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n- Check for Common Prefix for string -> Trie\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n**1. Node Class:**\n\n- Represents a node in the Trie data structure.\n- `cnt_prf` stores the count of prefixes that en... | 3 | 0 | ['String', 'Trie', 'Counting', 'Python3'] | 0 |
sum-of-prefix-scores-of-strings | C++ || EASY || TRIE | c-easy-trie-by-ganeshkumawat8740-em2l | Code\n\nclass trie{\n public:\n int x;\n trie *v[26];\n};\nvoid maketrie(string str,trie* node){\n for(auto &i: str){\n if(node->v[i- | ganeshkumawat8740 | NORMAL | 2023-05-26T09:10:27.874682+00:00 | 2023-05-26T09:10:27.874711+00:00 | 1,373 | false | # Code\n```\nclass trie{\n public:\n int x;\n trie *v[26];\n};\nvoid maketrie(string str,trie* node){\n for(auto &i: str){\n if(node->v[i-\'a\'] == NULL){\n node->v[i-\'a\'] = new trie();\n node = node->v[i-\'a\'];\n node->x = node->x +1 ;\n }else{\n ... | 3 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | C++ || TRIE || EASY TO UNDERSTAND || SHORT & SWEET CODE | c-trie-easy-to-understand-short-sweet-co-35ai | \nclass trie{//make trie class\n public:\n int cnt;\n trie* v[26];\n trie(){\n cnt = 0;//no of common prefix\n for | yash___sharma_ | NORMAL | 2023-04-12T12:57:32.046538+00:00 | 2023-04-12T12:59:16.023159+00:00 | 205 | false | ````\nclass trie{//make trie class\n public:\n int cnt;\n trie* v[26];\n trie(){\n cnt = 0;//no of common prefix\n for(int i = 0; i < 26; i++){\n v[i] = NULL;\n }\n }\n};\nclass Solution {\npublic:\n vector<int> sumPrefixScores(vector<str... | 3 | 0 | ['Trie', 'C', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | C++ Fully optimized Trie 99% faster 87% less memory | c-fully-optimized-trie-99-faster-87-less-3ykr | This is based on the Trie approach but with significantly less memory usage.\nMy original Trie was 779 ms\t700.9 MB\nThis one is 353 ms\t186.5 MB\nhttps://leetc | flowac | NORMAL | 2022-10-17T07:10:06.129557+00:00 | 2022-10-17T07:10:06.129592+00:00 | 369 | false | This is based on the Trie approach but with significantly less memory usage.\nMy original Trie was 779 ms\t700.9 MB\nThis one is 353 ms\t186.5 MB\nhttps://leetcode.com/submissions/detail/824255922/\n\nI don\'t remember what this method is called. Please comment if you know the name.\n\nTake "abcd", "abef" for example:\... | 3 | 0 | ['C', 'C++'] | 1 |
sum-of-prefix-scores-of-strings | [Javascript] get rid of memory allocation problem | javascript-get-rid-of-memory-allocation-36ybg | I found that in javascript, you will face memory problem in some testcase\nSo i\'m posting my solution here\n\nyou have to know that words with different inital | haocherhong | NORMAL | 2022-09-19T16:13:03.051066+00:00 | 2022-09-19T16:13:03.051115+00:00 | 154 | false | I found that in javascript, you will face memory problem in some testcase\nSo i\'m posting my solution here\n\nyou have to know that words with different inital characters wont affect each others\' score\nfor example, `abcd, aaab, acab` wont affect the scores of `cbcd, bacd, bbab, ccccd`.\nKnowing the fact, you know yo... | 3 | 0 | [] | 2 |
sum-of-prefix-scores-of-strings | C++ Rabin Karp Algorithm || Easy Implementation | c-rabin-karp-algorithm-easy-implementati-e443 | \ntypedef long long ll;\nclass Solution {\npublic:\n ll mod = 100000000007;\n ll dp[1005],pa[1005];\n void preProcess(string s){\n ll p = 29; / | manavmajithia6 | NORMAL | 2022-09-19T03:18:40.755823+00:00 | 2022-09-19T03:18:40.755956+00:00 | 98 | false | ```\ntypedef long long ll;\nclass Solution {\npublic:\n ll mod = 100000000007;\n ll dp[1005],pa[1005];\n void preProcess(string s){\n ll p = 29; // any prime no.\n ll pow = p;\n\n dp[0] = s[0] - \'a\' + 1;\n pa[0] = 1;\n for(ll i = 1;i < s.length();i++){\n dp[i] =... | 3 | 0 | ['Dynamic Programming', 'C'] | 0 |
sum-of-prefix-scores-of-strings | Easy with Trie | easy-with-trie-by-deleted_user-m2h4 | \nclass TrieNode{\n constructor(){\n this.ch = {};\n this.score = 0;\n }\n}\n\nclass Trie{\n constructor(){\n this.root = new TrieNode();\n }\n ad | deleted_user | NORMAL | 2022-09-18T16:46:30.724748+00:00 | 2022-09-18T16:46:30.724794+00:00 | 166 | false | ```\nclass TrieNode{\n constructor(){\n this.ch = {};\n this.score = 0;\n }\n}\n\nclass Trie{\n constructor(){\n this.root = new TrieNode();\n }\n add(word){\n let cur = this.root;\n for(let c of word){\n if (!cur.ch.hasOwnProperty(c)){\n cur.ch[c] = new TrieNode();\n }\n cur =... | 3 | 0 | ['Trie', 'JavaScript'] | 1 |
sum-of-prefix-scores-of-strings | ✅✅Faster || Easy To Understand || C++ Code | faster-easy-to-understand-c-code-by-__kr-ycjg | Trie\n\n Time Complexity :- O(N * M)\n\n Space Complexity :- O(N)\n\n\nclass Solution {\npublic:\n \n // structure for TrieNode\n \n struct TrieNode | __KR_SHANU_IITG | NORMAL | 2022-09-18T05:38:48.648195+00:00 | 2022-09-18T05:38:48.648232+00:00 | 289 | false | * ***Trie***\n\n* ***Time Complexity :- O(N * M)***\n\n* ***Space Complexity :- O(N)***\n\n```\nclass Solution {\npublic:\n \n // structure for TrieNode\n \n struct TrieNode\n {\n TrieNode* child[26];\n \n int count;\n \n TrieNode()\n {\n count = 0;\n ... | 3 | 0 | ['Trie', 'C', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | Easy peasy || Trie | easy-peasy-trie-by-kageyama09-21o4 | Using normal trie implementation with a branch property which at any node wil tell us the total number of branches that are merged with the current path i.e, th | kageyama09 | NORMAL | 2022-09-18T04:10:15.945624+00:00 | 2022-09-18T04:10:15.945650+00:00 | 115 | false | Using normal trie implementation with a **branch** property which at any node wil tell us the total number of branches that are merged with the current path i.e, the total number of prefixes ending at the current node.\nImplementing this and storing all the string in that trie will be a key to that answer, because afte... | 3 | 0 | ['Trie'] | 0 |
sum-of-prefix-scores-of-strings | Trie [Count and fetch] | C++ | trie-count-and-fetch-c-by-xxvvpp-k098 | C++\n struct trie {\n trie *child[26] = {};\n int count=0;\n }; \n \n vector sumPrefixScores(vector& A) {\n trie td;\n \n | xxvvpp | NORMAL | 2022-09-18T04:02:12.182150+00:00 | 2022-09-18T04:18:59.036580+00:00 | 143 | false | # C++\n struct trie {\n trie *child[26] = {};\n int count=0;\n }; \n \n vector<int> sumPrefixScores(vector<string>& A) {\n trie td;\n \n //put all words\n for(auto w:A){\n //insert word in trie\n auto root= &td;\n for(auto j:w){\n ... | 3 | 0 | ['Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | trie| c++ 🔥 the very fundamental of trie data structure Easy & Intutive | trie-c-the-very-fundamental-of-trie-data-b4kc | first of all how we can identify that this problem is related to trie data structure?\n\n\ntime complexcity total: Sum (size of each words)<=10^6\nspace complex | demon_code | NORMAL | 2024-09-25T17:37:19.233104+00:00 | 2024-09-26T17:11:54.734508+00:00 | 7 | false | first of all how we can identify that this problem is related to trie data structure?\n\n```\ntime complexcity total: Sum (size of each words)<=10^6\nspace complexcity total : Sum (size of each words) <=10^6\n```\n\nfirst when we see what they are asking is calculate sum of ( frequencies of prefixes (prifix which is g... | 2 | 0 | ['Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | Fastest Solution in Java with Explanation | fastest-solution-in-java-with-explanatio-w1qu | Intuition\n Describe your first thoughts on how to solve this problem. \nThe goal of this problem is to calculate a score for each word, where the score is the | heyysankalp | NORMAL | 2024-09-25T16:30:23.839791+00:00 | 2024-09-28T12:16:21.388542+00:00 | 40 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe goal of this problem is to calculate a score for each word, where the score is the sum of lengths of common prefixes between the word and every other word in the input array. To efficiently compute this, sorting the words and processi... | 2 | 0 | ['Array', 'String', 'Trie', 'Counting', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | EASY AND BEGINNER FRIENDLY PYTHON SOLUTION🔥🔥🔥 | easy-and-beginner-friendly-python-soluti-1xld | Intuition\nThe problem requires calculating the prefix score of each word, which depends on how many other words share the same prefix at each step. A Trie (pre | FAHAD_26 | NORMAL | 2024-09-25T14:28:10.271245+00:00 | 2024-09-25T14:28:10.271287+00:00 | 10 | false | # Intuition\nThe problem requires calculating the prefix score of each word, which depends on how many other words share the same prefix at each step. A Trie (prefix tree) is ideal for such tasks because it allows efficient prefix matching. By building a Trie with all the words and keeping track of how many words pass ... | 2 | 0 | ['Python'] | 0 |
sum-of-prefix-scores-of-strings | Concise and beautiful Trie | Cracking Hard into Easy | concise-and-beautiful-trie-cracking-hard-tyvz | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nJust use a custom Trie, | sulerhy | NORMAL | 2024-09-25T12:02:45.722178+00:00 | 2024-09-25T12:02:45.722218+00:00 | 23 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nJust use a custom Trie, with infomation of `count` as the number of words in `words` which also has this character\n\n# Complexity\n- Time complexity: O(n * k)\n<!-- A... | 2 | 0 | ['String', 'Trie', 'Python3'] | 0 |
sum-of-prefix-scores-of-strings | Small & Simple Graph Count Traversal in Kotlin (Beats 100%) | small-simple-graph-count-traversal-in-ko-jmml | \n\n# Intuition\nThe question that was asked; to sum up all the prefixes and occurences for every seperate word, can be achieved by simply counting how many tim | mdekaste | NORMAL | 2024-09-25T11:08:19.176137+00:00 | 2024-09-25T11:14:18.498386+00:00 | 36 | false | \n\n# Intuition\nThe question that was asked; to sum up all the prefixes and occurences for every seperate word, can be achieved by simply counting how many times a character reaches a node in a graph.\n\n#... | 2 | 0 | ['Trie', 'Kotlin'] | 1 |
sum-of-prefix-scores-of-strings | JAVA EASY SOLUTION | java-easy-solution-by-007harsh-qc0o | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires counting how many strings in the given list start with each prefix | 007Harsh- | NORMAL | 2024-09-25T09:37:22.152039+00:00 | 2024-09-25T09:37:22.152073+00:00 | 16 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires counting how many strings in the given list start with each prefix of every word. A naive approach would be to check each prefix of every word against all other words, but this would be inefficient. Using a Trie `(Pre... | 2 | 0 | ['Trie', 'Prefix Sum', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | Easy Solution || C++ || Tries || TC- O(N*L) | easy-solution-c-tries-tc-onl-by-manavtor-ke9p | Intuition\nWhen solving this problem, the main challenge is calculating the prefix scores efficiently for multiple words. A brute force approach of calculating | manavtore | NORMAL | 2024-09-25T08:40:31.502404+00:00 | 2024-09-25T08:40:31.502460+00:00 | 52 | false | # Intuition\nWhen solving this problem, the main challenge is calculating the prefix scores efficiently for multiple words. A brute force approach of calculating prefix scores individually for each word would be inefficient, especially with longer word lists. The Trie (prefix tree) data structure is well-suited for pro... | 2 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 1 |
sum-of-prefix-scores-of-strings | C++ Simple Trie / Prefix Tree implementation | c-simple-trie-prefix-tree-implementation-5ibo | Intuition\n- In this problem we need to find the number of strings that has a prefix in the current string.\n- Thus when it comes to counting prefixes of string | Codesmith28 | NORMAL | 2024-09-25T06:47:41.969002+00:00 | 2024-09-25T12:50:45.244243+00:00 | 23 | false | # Intuition\n- In this problem we need to find the number of strings that has a prefix in the current string.\n- Thus when it comes to counting prefixes of strings and counting occurrences, we can rely on Trie / prefix tree data structure to efficiently store the strings and traverse through them.\n# What is Trie ?\n- ... | 2 | 0 | ['String', 'Trie', 'Counting', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | C++ Solution || Trie | c-solution-trie-by-rohit_raj01-t8z9 | Intuition\nThe problem is about calculating a score for each word based on how many times its prefixes appear across all the words in the input list. A Trie (pr | Rohit_Raj01 | NORMAL | 2024-09-25T06:33:47.863491+00:00 | 2024-09-25T06:33:47.863522+00:00 | 47 | false | # Intuition\nThe problem is about calculating a score for each word based on how many times its prefixes appear across all the words in the input list. A Trie (prefix tree) is an efficient data structure for handling prefix-based problems because it allows for fast insertion and traversal of strings. By storing prefix ... | 2 | 0 | ['String', 'Trie', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | Simple C++ solution | TRIE | simple-c-solution-trie-by-saurabhdamle11-p6qt | Intuition\nFor every word, we need to find all the prefixes and get the count of all such prefixes which will be the score of that word.\n\nSince we need to fin | saurabhdamle11 | NORMAL | 2024-09-25T05:26:22.391583+00:00 | 2024-09-25T05:26:22.391620+00:00 | 71 | false | # Intuition\nFor every word, we need to find all the prefixes and get the count of all such prefixes which will be the score of that word.\n\nSince we need to find and store all the prefixes, a Trie will be the best Data Structure to use here.\n\n# Approach\n1. Initialize a trie and insert all the words into the trie.\... | 2 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 2 |
sum-of-prefix-scores-of-strings | [C++] ✅|| 2 Approach | HashMap & Trie | Easy to understand 🚀 | c-2-approach-hashmap-trie-easy-to-unders-3941 | Approach - 1: (Using Hashmap) \n\nclass Solution {\npublic:\n vector<int> sumPrefixScores(vector<string>& words) \n {\n unordered_map<string, int>m | RIOLOG | NORMAL | 2024-09-25T05:06:59.978248+00:00 | 2024-09-25T05:07:35.685520+00:00 | 27 | false | Approach - 1: (Using Hashmap) \n```\nclass Solution {\npublic:\n vector<int> sumPrefixScores(vector<string>& words) \n {\n unordered_map<string, int>mp;\n for (int i=0;i<words.size();i++)\n {\n string temp = "";\n string curr = words[i];\n \n for (i... | 2 | 0 | ['Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | HashMap Solution | hashmap-solution-by-jastegsingh007-y1ql | Soch\nFollowing up for yesterday\'s POTD, store all subsets with there subset and count and just add them\n\n# Code\npython3 []\nclass Solution:\n def sumPre | jastegsingh007 | NORMAL | 2024-09-25T04:20:33.876354+00:00 | 2024-09-25T04:20:33.876403+00:00 | 243 | false | # Soch\nFollowing up for yesterday\'s POTD, store all subsets with there subset and count and just add them\n\n# Code\n```python3 []\nclass Solution:\n def sumPrefixScores(self, words: List[str]) -> List[int]:\n k=defaultdict(int)\n ans=[]\n for i in words:\n for j in range(len(i)):\n... | 2 | 0 | ['Python3'] | 2 |
sum-of-prefix-scores-of-strings | Java: Trie solution with comments and explanation. Easy to understand. | java-trie-solution-with-comments-and-exp-jmlc | Intuition\nWe are given a array of words. For each word, we need to compute prefix score. Any word of length n has n prefixes. For example, word abcd has prefix | cg7293 | NORMAL | 2024-09-25T04:09:51.117128+00:00 | 2024-09-25T04:09:51.117192+00:00 | 65 | false | # Intuition\nWe are given a array of words. For each word, we need to compute $$prefix score.$$ Any word of length `n` has `n` prefixes. For example, word `abcd` has prefixes `a`, `ab`, `abc` and `abcd`. For each prefix we need to find number of times it occurs in the words array.\n\n# Approach\nFor each word, we add t... | 2 | 0 | ['Trie', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | Simple trie | simple-trie-by-vortexz-sf3i | 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 | vortexz | NORMAL | 2024-09-25T03:10:53.576254+00:00 | 2024-09-25T03:10:53.576302+00:00 | 39 | 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 | ['Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | Python 3 solution - clean use of trie data structure and hash map soluton. | python-3-solution-clean-use-of-trie-data-xn6b | Code\npython [Prefix Trie]\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.count = 0\n\nclass Trie:\n def __init__(self): | iloabachie | NORMAL | 2024-09-25T01:51:51.114003+00:00 | 2024-09-25T03:00:15.836290+00:00 | 231 | false | # Code\n```python [Prefix Trie]\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.count = 0\n\nclass Trie:\n def __init__(self):\n self.root = TrieNode()\n \n def insert(self, words: List[str]) -> None:\n for word in words:\n current = self.root\n ... | 2 | 0 | ['Python3'] | 0 |
sum-of-prefix-scores-of-strings | TRIE makes it easy !! | trie-makes-it-easy-by-saurav_1928-qz1w | \n# Code\ncpp []\nclass Node {\npublic:\n unordered_map<char, pair<Node*, int>> arr;\n bool isEnd = false;\n};\n\nvoid insert(Node* root, string& word) {\ | saurav_1928 | NORMAL | 2024-09-25T01:18:51.249115+00:00 | 2024-09-25T01:18:51.249195+00:00 | 61 | false | \n# Code\n```cpp []\nclass Node {\npublic:\n unordered_map<char, pair<Node*, int>> arr;\n bool isEnd = false;\n};\n\nvoid insert(Node* root, string& word) {\n Node* curr = root;\n for (auto ch : word) {\n if (curr->arr.find(ch) == curr->arr.end()) {\n curr->arr[ch] = {new Node(), 0};\n ... | 2 | 0 | ['Array', 'String', 'Trie', 'Counting', 'C++'] | 1 |
sum-of-prefix-scores-of-strings | Easy, well explained Javascript and Trie solution | easy-well-explained-javascript-and-trie-oheti | Intuition\nThe task is to compute the prefix score for each word by summing how many words share each prefix. It\'s possible to solve it using a HashMap, but it | balfin | NORMAL | 2024-09-25T00:40:20.205023+00:00 | 2024-09-25T00:41:56.490448+00:00 | 157 | false | # Intuition\nThe task is to compute the prefix score for each word by summing how many words share each prefix. It\'s possible to solve it using a HashMap, but it will fail by time limit on the last casess, so more efficient data structure is needed. Using a Trie allows us to efficiently store and count shared prefixes... | 2 | 0 | ['Trie', 'JavaScript'] | 0 |
sum-of-prefix-scores-of-strings | brain dead solution in python | brain-dead-solution-in-python-by-mostkno-veeq | Intuition\nUses the idea of tries but no tries hehe\n\nHonestly, just build the prefix of a word as you go.\n\n# Approach\nFor each word, I create a buffer and | mostknownunknown | NORMAL | 2023-02-06T02:09:20.985569+00:00 | 2023-02-06T02:09:20.985614+00:00 | 263 | false | # Intuition\nUses the idea of tries but no tries hehe\n\nHonestly, just build the prefix of a word as you go.\n\n# Approach\nFor each word, I create a `buffer` and build a prefix for each character of the word.\n\nSo I continuously build a `buffer` character at a time without doing some weird indexing calculation.\n\nO... | 2 | 0 | ['Python3'] | 0 |
sum-of-prefix-scores-of-strings | Count all Prefixes in a Hash Map | count-all-prefixes-in-a-hash-map-by-jana-zbky | Intuition\nIterate over all the workds x prefixes and create a hashmap of all prefix counts. Then just use that to count the score\n\n# Approach\niterate over a | janakagoon | NORMAL | 2023-01-15T05:09:12.802485+00:00 | 2023-01-15T05:09:12.802533+00:00 | 57 | false | # Intuition\nIterate over all the workds x prefixes and create a hashmap of all prefix counts. Then just use that to count the score\n\n# Approach\niterate over all the workds x prefixes and create a hashmap of all prefix counts. Then just use that to count the score\n\n# Complexity\n- Time complexity:\nO(W x L)\n\nW: ... | 2 | 0 | ['Go'] | 0 |
sum-of-prefix-scores-of-strings | C++ | Trie + search Prefixes | c-trie-search-prefixes-by-ajay_5097-i7zi | cpp\nclass TrieNode {\npublic:\n vector<TrieNode*> child;\n int availablePrefix;\n};\n\nconst int CHILD_COUNT = 26;\nconst int nax = 1e7 + 9;\nTrieNode po | ajay_5097 | NORMAL | 2022-11-21T11:08:02.135159+00:00 | 2022-11-21T11:08:02.135206+00:00 | 1,085 | false | ```cpp\nclass TrieNode {\npublic:\n vector<TrieNode*> child;\n int availablePrefix;\n};\n\nconst int CHILD_COUNT = 26;\nconst int nax = 1e7 + 9;\nTrieNode pool[nax];\nint cnt;\nTrieNode *getNode() {\n auto t = &pool[cnt++];\n t->child.resize(CHILD_COUNT);\n for (int i = 0; i < CHILD_COUNT; ++i) {\n ... | 2 | 0 | [] | 0 |
sum-of-prefix-scores-of-strings | [Javascript] Simple Trie implementation | javascript-simple-trie-implementation-by-mkk3 | It\'s a pretty simple and understandable solution, written in Javascript\n\n```\nvar sumPrefixScores = function(words) {\n const n = words.length;\n const | vik_13 | NORMAL | 2022-11-17T10:25:56.537869+00:00 | 2022-11-18T23:30:52.905921+00:00 | 152 | false | It\'s a pretty simple and understandable solution, written in Javascript\n\n```\nvar sumPrefixScores = function(words) {\n const n = words.length;\n const trie = {_count: 0};\n const result = [];\n \n\t// Create our own custom trie with _count property.\n // We are storing how many time we passed current... | 2 | 0 | ['Trie', 'JavaScript'] | 0 |
sum-of-prefix-scores-of-strings | Python | Trie | Modular clean code | python-trie-modular-clean-code-by-jnaik-3b4l | Standard Trie implmentation, which I follow to solve all the Trie problems with just slight variation in search and insert method as per problem statement. Maki | jnaik | NORMAL | 2022-10-21T17:17:45.347485+00:00 | 2022-10-21T17:31:55.054512+00:00 | 121 | false | Standard Trie implmentation, which I follow to solve all the Trie problems with just slight variation in search and insert method as per problem statement. Making a habit to follow a template makes you fast and well organized and it leaves the best impression on interviewer.\n\nSolution inspired from: https://leetcode.... | 2 | 0 | ['String', 'Trie', 'Python', 'Python3'] | 0 |
sum-of-prefix-scores-of-strings | Java | Trie | Detailed Explanation and Comments | java-trie-detailed-explanation-and-comme-7t41 | Here I will explain the super detailed thought process to solve this problem. For a concise explanation, only read section Problem Approach and Final Approach I | zheng_four_oranges | NORMAL | 2022-09-24T02:53:05.425818+00:00 | 2022-09-24T02:58:20.659346+00:00 | 291 | false | Here I will explain the super detailed thought process to solve this problem. For a concise explanation, only read section **Problem Approach** and **Final Approach Implementation**.\n\n**Problem Approach:**\nFor each String **S** in words, for all **S\'s prefixes Sj** (**j denoting the length of Sj**), we essentially ... | 2 | 0 | ['Depth-First Search', 'Trie', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | Beginner Friendly Trie Implementation | C++ Solution | beginner-friendly-trie-implementation-c-t5xko | In addition to the general implementation of Trie, Storing an extra information how many prefixes end with a specific character in each node.\n\nstruct node{\n | wa_survivor | NORMAL | 2022-09-18T17:27:59.676905+00:00 | 2022-09-18T17:28:28.550459+00:00 | 211 | false | In addition to the general implementation of Trie, Storing an extra information how many prefixes end with a specific character in each node.\n```\nstruct node{\n node* mp[26];\n int cnt[26];\n bool isPresent(char c){\n if(mp[c-\'a\']==NULL) return false;\n return true;\n }\n node* getKey(c... | 2 | 0 | ['Trie', 'C', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | Trie Java Solution. | trie-java-solution-by-harshraj9988-jh3j | \nclass Solution {\n\tclass TrieNode {\n TrieNode[] next;\n int[] cnt;\n\n public TrieNode() {\n next = new TrieNode[26];\n | HarshRaj9988 | NORMAL | 2022-09-18T05:40:03.269899+00:00 | 2022-09-18T05:40:03.269946+00:00 | 39 | false | ```\nclass Solution {\n\tclass TrieNode {\n TrieNode[] next;\n int[] cnt;\n\n public TrieNode() {\n next = new TrieNode[26];\n cnt = new int[26];\n }\n\n \n }\n\n public void insert(String word, TrieNode root) {\n TrieNode curr = root;\n for (... | 2 | 0 | ['Depth-First Search', 'Trie'] | 0 |
sum-of-prefix-scores-of-strings | C++ Easy Solution | Trie | c-easy-solution-trie-by-b_rabbit007-slnc | I have created Trie data structure with array of pointers next to point all alphabets and wordCount which will store number of words having prefix string ending | B_Rabbit007 | NORMAL | 2022-09-18T05:06:07.081209+00:00 | 2022-09-23T01:45:51.031904+00:00 | 53 | false | I have created `Trie` data structure with array of pointers `next` to point all alphabets and `wordCount` which will store number of words having prefix string ending at current node.\n\nFor example - consider array of strings `["abc", "ab", "b"]`\nTrie will be like -\n```\n\t\t\t\t\t\t\t Root\n\t\t\t\t\t\t wordCount =... | 2 | 0 | ['Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | Python3 ✅ || Simple Trie | python3-simple-trie-by-husharbabu-dqbu | `.\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.word = False\n self.count = 1\n \nclass | HusharBabu | NORMAL | 2022-09-18T04:42:08.785353+00:00 | 2022-09-18T04:42:08.785391+00:00 | 32 | false | ```.\nclass TrieNode:\n def __init__(self):\n self.children = {}\n self.word = False\n self.count = 1\n \nclass Trie:\n def __init__(self):\n self.root = TrieNode()\n\n def insert(self, word):\n trav = self.root\n for c in word:\n ... | 2 | 0 | [] | 0 |
sum-of-prefix-scores-of-strings | Java | Roling Hash | java-roling-hash-by-aaveshk-dhk1 | \nclass Solution\n{\n final int m = 1023;\n public int[] sumPrefixScores(String[] words)\n {\n HashMap<Long,Integer> map = new HashMap<>();\n | aaveshk | NORMAL | 2022-09-18T04:22:19.628297+00:00 | 2022-09-18T04:22:30.890943+00:00 | 243 | false | ```\nclass Solution\n{\n final int m = 1023;\n public int[] sumPrefixScores(String[] words)\n {\n HashMap<Long,Integer> map = new HashMap<>();\n for(String S : words)\n {\n char[] s = S.toCharArray();\n long hash_so_far = 0L;\n int n = s.length;\n ... | 2 | 0 | ['Rolling Hash', 'Java'] | 2 |
sum-of-prefix-scores-of-strings | SIMPLE TRIE | C++ | BRUTE FORCE | simple-trie-c-brute-force-by-deepanshu_d-icx7 | TRIE PRACTICE PROBLEMS \n\n1. Implement Trie (Prefix Tree)\n2. Maximum XOR of Two Numbers in an Array\n3. Search Suggestions System\n\n// trie node \nstruct no | Deepanshu_Dhakate | NORMAL | 2022-09-18T04:19:23.567619+00:00 | 2022-09-18T04:19:23.567664+00:00 | 73 | false | **TRIE PRACTICE PROBLEMS **\n\n[1. Implement Trie (Prefix Tree)](https://leetcode.com/problems/implement-trie-prefix-tree/)\n[2. Maximum XOR of Two Numbers in an Array](https://leetcode.com/problems/maximum-xor-of-two-numbers-in-an-array/)\n[3. Search Suggestions System](https://leetcode.com/problems/search-suggestions... | 2 | 1 | ['Depth-First Search', 'Trie', 'C'] | 0 |
sum-of-prefix-scores-of-strings | C++ | Trie Solution | c-trie-solution-by-travanj05-hxai | cpp\nclass Node {\npublic:\n char ch;\n Node* children[26] = {nullptr};\n int prefix_cnt;\n};\n\nclass Solution {\nprivate:\n Node* root = new Node; | travanj05 | NORMAL | 2022-09-18T04:14:11.844955+00:00 | 2022-09-18T04:14:11.844986+00:00 | 43 | false | ```cpp\nclass Node {\npublic:\n char ch;\n Node* children[26] = {nullptr};\n int prefix_cnt;\n};\n\nclass Solution {\nprivate:\n Node* root = new Node;\n int prefixCount(string &prefix) {\n Node* tmp = root;\n int cnt = 0;\n for (char &ch : prefix) {\n int index = ch - \'a... | 2 | 0 | ['Trie', 'C', 'C++'] | 0 |
sum-of-prefix-scores-of-strings | [Java] Trie | O(NM) Time | Easy to Understand | java-trie-onm-time-easy-to-understand-by-17es | java\nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n if (words.length == 1) {\n return new int[] { words[0].length() } | jjeeffff | NORMAL | 2022-09-18T04:12:10.387993+00:00 | 2022-09-21T08:09:55.883935+00:00 | 227 | false | ``` java\nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n if (words.length == 1) {\n return new int[] { words[0].length() };\n }\n \n Trie trie = new Trie();\n for (String word : words) {\n trie.addWord(word);\n }\n int[] ans ... | 2 | 0 | ['Trie', 'Java'] | 0 |
sum-of-prefix-scores-of-strings | Java || Trie Solution | java-trie-solution-by-yog_esh-361o | \nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n \n Tries trie=new Tries();\n \n \n \n for(Str | yog_esh | NORMAL | 2022-09-18T04:07:00.240820+00:00 | 2022-09-18T04:07:00.240846+00:00 | 397 | false | ```\nclass Solution {\n public int[] sumPrefixScores(String[] words) {\n \n Tries trie=new Tries();\n \n \n \n for(String word:words){\n \n add(word,trie);\n }\n \n int[] res=new int[words.length];\n \n \n for(i... | 2 | 0 | ['Trie', 'Java'] | 1 |
sum-of-prefix-scores-of-strings | Very simple Python trie code | very-simple-python-trie-code-by-rjmcmc-xked | ```\nclass Solution:\n def sumPrefixScores(self, words: List[str]) -> List[int]:\n trie = {}\n \n for word in words:\n t = tr | rjmcmc | NORMAL | 2022-09-18T04:03:19.731203+00:00 | 2022-09-18T04:03:19.731241+00:00 | 67 | false | ```\nclass Solution:\n def sumPrefixScores(self, words: List[str]) -> List[int]:\n trie = {}\n \n for word in words:\n t = trie\n for char in word:\n if char not in t:\n t[char] = {}\n t = t[char]\n if \'$\' no... | 2 | 0 | [] | 0 |
sum-of-prefix-scores-of-strings | Simple Solution Using Trie Tree | simple-solution-using-trie-tree-by-pbara-ycv3 | The question gives Intution of Trie tree as storing all the prefixes will cause TLE\n\nWith the standard implementation of trie tree we can keep a array of coun | pbarak6 | NORMAL | 2022-09-18T04:02:07.624350+00:00 | 2022-09-18T04:02:07.624389+00:00 | 124 | false | The question gives Intution of Trie tree as storing all the prefixes will cause TLE\n\nWith the standard implementation of trie tree we can keep a array of count that will count the frequency of that charcater\n\n=>> Below is the simple CPP implementation\n\n```\nstruct Node{\n bool flag=false;\n int count[26] = ... | 2 | 0 | ['Tree', 'Trie', 'C++'] | 1 |
sum-of-prefix-scores-of-strings | [JavaScript] Trie | javascript-trie-by-jialihan-cv7c | \nclass TrieNode {\n constructor() {\n this.children = {};\n this.cnt = 0;\n }\n}\n\n\nvar sumPrefixScores = function(words) {\n const ro | jialihan | NORMAL | 2022-09-18T04:01:40.170494+00:00 | 2022-09-18T04:01:40.170546+00:00 | 97 | false | ```\nclass TrieNode {\n constructor() {\n this.children = {};\n this.cnt = 0;\n }\n}\n\n\nvar sumPrefixScores = function(words) {\n const root = new TrieNode();\n function add(w) {\n let cur = root;\n for(let i = 0; i<w.length; i++){\n const c = w[i];\n if(!cur... | 2 | 1 | ['Trie', 'JavaScript'] | 0 |
sum-of-prefix-scores-of-strings | trie cpp | trie-cpp-by-sreyash_11-1hc5 | space complexity here, e.g. O(n) -->Code | sreyash_11 | NORMAL | 2025-03-17T06:50:17.580238+00:00 | 2025-03-17T06:50:17.580238+00:00 | 7 | false | space complexity here, e.g. $$O(n)$$ -->
# Code
```cpp []
struct Node {
Node *links[26];
int precnt = 0;
Node() {
for (int i = 0; i < 26; i++) links[i] = nullptr;
}
bool containsKey(char ch) {
return links[ch - 'a'] != nullptr;
}
Node* get(char ch) {
return links[ch... | 1 | 0 | ['String', 'Trie', 'String Matching', 'C++'] | 0 |
type-of-triangle | ✅THE BEST EXPLANATION||Beats 100% Users | the-best-explanationbeats-100-users-by-h-00os | ---\n# PLEASE UPVOTE IF YOU LIKE IT \u2705\n---\n\n\n---\n# Intuition\n Describe your first thoughts on how to solve this problem. \n\nThe goal is to determine | harshal2024 | NORMAL | 2024-02-03T16:16:02.600254+00:00 | 2024-02-03T16:16:02.600277+00:00 | 1,830 | false | ---\n# PLEASE UPVOTE IF YOU LIKE IT \u2705\n---\n\n\n---\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\nThe goal is to determine the type of triangle that c... | 13 | 0 | ['Python', 'C++', 'Java', 'Python3'] | 1 |
type-of-triangle | Sort | sort-by-votrubac-q5jc | We sort the sides first to simplify the logic.\n\nC++\ncpp\nstring triangleType(vector<int>& n) {\n sort(begin(n), end(n));\n if (n[2] >= n[0] + n[1])\n | votrubac | NORMAL | 2024-02-03T16:04:32.494865+00:00 | 2024-02-03T16:04:32.494895+00:00 | 1,259 | false | We sort the sides first to simplify the logic.\n\n**C++**\n```cpp\nstring triangleType(vector<int>& n) {\n sort(begin(n), end(n));\n if (n[2] >= n[0] + n[1])\n return "none";\n return n[0] == n[2] ? "equilateral" : n[0] == n[1] || n[1] == n[2] ? "isosceles" : "scalene";\n}\n``` | 13 | 1 | ['C'] | 5 |
type-of-triangle | Simple java solution | simple-java-solution-by-siddhant_1602-0hto | Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\n\nclass Solution {\n public String triangleType(int[] nums) {\n if(nums[0]+ | Siddhant_1602 | NORMAL | 2024-02-03T16:01:16.030489+00:00 | 2024-02-03T16:06:45.022384+00:00 | 783 | false | # Complexity\n- Time complexity: $$O(1)$$\n\n- Space complexity: $$O(1)$$\n\n# Code\n```\nclass Solution {\n public String triangleType(int[] nums) {\n if(nums[0]+nums[1]>nums[2] && nums[1]+nums[2]>nums[0] && nums[2]+nums[0]>nums[1])\n {\n if(nums[0]==nums[1] && nums[1]==nums[2])\n ... | 12 | 0 | ['Java'] | 0 |
type-of-triangle | Without if-else spaghetti | without-if-else-spaghetti-by-almostmonda-teqg | Intuition\n Describe your first thoughts on how to solve this problem. \nThe \space Triangle \space Inequality \space Theorem says:\n> The sum of two sides of a | almostmonday | NORMAL | 2024-02-03T20:53:35.920274+00:00 | 2024-02-03T21:37:46.827111+00:00 | 379 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n$$The \\space Triangle \\space Inequality \\space Theorem$$ says:\n> The sum of two sides of a triangle must be greater than the third side.\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nFind the right type of a ... | 8 | 1 | ['Geometry', 'Python', 'Python3'] | 3 |
type-of-triangle | Sort & 3 conditions | sort-3-conditions-by-kreakemp-frlb | \n# C++\n\nstring triangleType(vector<int>& nums) {\n sort(nums.begin(), nums.end());\n if(nums[0] + nums[1] <= nums[2]) return "none";\n if(nums[0] == | kreakEmp | NORMAL | 2024-02-03T16:42:41.170212+00:00 | 2024-02-03T16:42:50.220112+00:00 | 2,567 | false | \n# C++\n```\nstring triangleType(vector<int>& nums) {\n sort(nums.begin(), nums.end());\n if(nums[0] + nums[1] <= nums[2]) return "none";\n if(nums[0] == nums[1] && nums[1] == nums[2]) return "equilateral";\n if(nums[0] == nums[1] || nums[1] == nums[2]) return "isosceles";\n return "scalene";\n}\n```\n\... | 7 | 1 | ['C++', 'Java'] | 4 |
type-of-triangle | Type of Triangle📐🔍: Simple if-else ✨|| 0ms⏰ ||Beats 100%✅ || JAVA ☕️ | type-of-triangle-simple-if-else-0ms-beat-myd9 | \n# Intuition\nYou\'re checking the type of triangle based on its side lengths, following these rules:\n\n- If any side length is greater than or equal to the s | Megha_Mathur18 | NORMAL | 2024-06-23T14:16:52.692709+00:00 | 2024-06-23T14:18:41.489093+00:00 | 117 | false | \n# Intuition\nYou\'re checking the type of triangle based on its side lengths, following these rules:\n\n- If any side length is greater than or equal to the sum of the other two side... | 6 | 0 | ['Java'] | 0 |
type-of-triangle | [Java] ✅ ⬆️ | Simple 🔥🔥 | Clean | Self Explanatory | java-simple-clean-self-explanatory-by-sa-2nep | Complexity\nTime complexity: O(1)\nSpace complexity: O(1)\n\npublic String triangleType(int[] nums) {\n int a = nums[0];\n int b = nums[1];\n | sandeepk97 | NORMAL | 2024-02-03T16:35:16.785226+00:00 | 2024-02-03T16:37:54.391528+00:00 | 248 | false | **Complexity**\nTime complexity: O(1)\nSpace complexity: O(1)\n```\npublic String triangleType(int[] nums) {\n int a = nums[0];\n int b = nums[1];\n int c = nums[2];\n\n if (isValidTriangle(a, b, c)) {\n if (a == b && b == c) {\n return "equilateral";\n }... | 5 | 0 | ['Java'] | 1 |
type-of-triangle | 🔥BEATS 💯 % 🎯 |✨SUPER EASY FOR BEGINNERS 👏 | beats-super-easy-for-beginners-by-vignes-f5u7 | Code | vigneshvaran0101 | NORMAL | 2025-02-25T17:00:14.251211+00:00 | 2025-02-25T17:00:14.251211+00:00 | 136 | false | 
# Code
```python3 []
class Solution:
def triangleType(self, nums: List[int]) -> str:
if nums[0] >= nums[1] + nums[2] or nums[1] >= nums[0] + nums[2] or nums[2] >= nums[0] + nums[1]:
... | 3 | 0 | ['Python3'] | 0 |
type-of-triangle | Java Check None first | java-check-none-first-by-hobiter-u869 | See code. \n# Complexity\n- Time complexity:\n Add your time complexity here, e.g. O(n) \nO(1)\n- Space complexity:\n Add your space complexity here, e.g. O(n) | hobiter | NORMAL | 2024-02-04T21:57:58.178602+00:00 | 2024-02-04T21:57:58.178620+00:00 | 871 | false | See code. \n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(1)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(1)\n# Code\n```\nclass Solution {\n final String EQ = "equilateral";\n final String ISO = "isosceles";\n final String SCA = "... | 3 | 0 | ['Java'] | 0 |
type-of-triangle | ✅Beats 100% || Classifying triangles in Java, Python & C++ | beats-100-classifying-triangles-in-java-xl2tw | Intuition\nThe code checks the validity of a triangle based on its side lengths and classifies it as equilateral, isosceles, scalene, or none.\n\n# Approach\nTh | Mohit-P | NORMAL | 2024-02-03T16:32:57.937652+00:00 | 2024-02-03T16:32:57.937675+00:00 | 203 | false | # Intuition\nThe code checks the validity of a triangle based on its side lengths and classifies it as equilateral, isosceles, scalene, or none.\n\n# Approach\nThe approach involves using conditional statements to check the triangle\'s validity and then determining its type based on the side lengths.\n\n# Complexity\n-... | 3 | 0 | ['Python', 'C++', 'Java'] | 1 |
type-of-triangle | Easy to Understand | Beats 100% | O(1) | 0ms Runtime🔥🧯🦹🏻♀️ | easy-to-understand-beats-100-o1-0ms-runt-7ata | Intuition : FOR THE LACKWITS!!
equilateral if it has all sides of equal length.
isosceles if it has exactly two sides of equal length.
scalene if all its sides | tyagideepti9 | NORMAL | 2025-01-30T17:10:42.700338+00:00 | 2025-01-30T17:13:28.103898+00:00 | 373 | false | # Intuition : FOR THE LACKWITS!!
- equilateral if it has all sides of equal length.
- isosceles if it has exactly two sides of equal length.
- scalene if all its sides are of different lengths.
- sum of two sides is not greater than third one, then not a valid triangle
<!-- Describe your first thoughts on how to solve... | 2 | 0 | ['Array', 'Math', 'Sorting', 'Java', 'C#', 'MS SQL Server'] | 1 |
type-of-triangle | Easy To Understand | Beats 100% | O(1) | 0ms runtime 🔥🧯 | easy-to-understand-beats-100-o1-0ms-runt-kmnp | Intuition - FOR THE LACKWITS!!Two Side Sum < Third Side - Not a Triangle
All Sides Equal - Equilateral
Two Sides Equal - Isosceles
No Side Equal - ScaleneApproa | aman-sagar | NORMAL | 2025-01-30T17:09:46.780862+00:00 | 2025-01-30T17:10:34.887117+00:00 | 252 | false | # Intuition - FOR THE LACKWITS!!
<!-- Describe your first thoughts on how to solve this problem. -->
Two Side Sum < Third Side - Not a Triangle
All Sides Equal - Equilateral
Two Sides Equal - Isosceles
No Side Equal - Scalene
# Approach
<!-- Describe your approach to solving the problem. -->
If if elif elif else else
... | 2 | 0 | ['Array', 'Math', 'Sorting', 'C++', 'Python3', 'MySQL'] | 1 |
type-of-triangle | easy answer for the questions beats 100% | easy-answer-for-the-questions-beats-100-pzum5 | 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 | wfjvxY4Pch | NORMAL | 2024-08-26T17:02:14.322395+00:00 | 2024-08-26T17:02:14.322422+00:00 | 79 | 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 |
type-of-triangle | Easy Java Solution || Beginner Friendly | easy-java-solution-beginner-friendly-by-jq0ex | 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 | anikets101 | NORMAL | 2024-04-20T13:53:49.037239+00:00 | 2024-04-20T13:53:49.037288+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. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
type-of-triangle | THE EASIEST SOLUTION FOR YOU | Type of Triangle | 3024 | the-easiest-solution-for-you-type-of-tri-dbkp | \n# Code\n\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n ns = set(nums)\n l = len(ns)\n a, b, c = nums[0], nums[1 | JustChis100 | NORMAL | 2024-03-22T13:17:09.579457+00:00 | 2024-03-22T13:17:09.579489+00:00 | 172 | false | \n# Code\n```\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n ns = set(nums)\n l = len(ns)\n a, b, c = nums[0], nums[1], nums[2]\n if not (a < b + c and b < a + c and c < b + a) :\n return "none"\n if l == 2 :\n return "isosceles"\n ... | 2 | 0 | ['Array', 'Math', 'Python', 'Python3'] | 1 |
type-of-triangle | MATH || Solution of type of triangle problem | math-solution-of-type-of-triangle-proble-5y7c | Approach\n- Triangle Properties - The sum of the length of the two sides of a triangle is greater than the length of the third side.\n\n# Complexity\n- Time com | tiwafuj | NORMAL | 2024-03-19T15:26:42.387639+00:00 | 2024-04-06T08:53:15.214650+00:00 | 98 | false | # Approach\n- Triangle Properties - The sum of the length of the two sides of a triangle is greater than the length of the third side.\n\n# Complexity\n- Time complexity: O(1) - as all algorithms require constant time \n\n- Space complexity: O(1) - as no extra space is requried\n\n# Code\n```\nclass Solution:\n def ... | 2 | 0 | ['Array', 'Math', 'Sorting', 'Python3'] | 0 |
type-of-triangle | Simple If-else solution | Beginners Friendly | 💯% user's beat | simple-if-else-solution-beginners-friend-p76d | \n\n\n# Approach\nif nums[0]==nums[1]==nums[2]:\nThis line checks if all three sides of the triangle are equal, which would indicate an equilateral triangle.\nr | Pankaj_Tyagi | NORMAL | 2024-02-04T11:38:50.175149+00:00 | 2024-02-05T09:49:36.409483+00:00 | 225 | false | \n\n\n# Approach\nif nums[0]==nums[1]==nums[2]:\nThis line checks if all three sides of the triangle are equal, which would indicate an equilateral triangle.\nreturn "equilateral"\n\n... | 2 | 0 | ['Python', 'Python3'] | 1 |
type-of-triangle | ✅✔️Simple Solution using Triangle Property ✈️✈️✈️ | simple-solution-using-triangle-property-wztql | 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 | ajay_1134 | NORMAL | 2024-02-03T16:08:44.896827+00:00 | 2024-02-03T16:08:44.896922+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. $$O(n)$$ --... | 2 | 0 | ['Math', 'C++'] | 0 |
type-of-triangle | a few solutions | a-few-solutions-by-claytonjwong-h416 | Game Plan\n\n\uD83D\uDEAB We cannot form a triangle if the sum of the minimum two side lengths is less-than-or-equal-to the maximum side length.\n \'none\'\n\n\ | claytonjwong | NORMAL | 2024-02-03T16:02:04.901408+00:00 | 2024-02-03T16:56:01.381456+00:00 | 520 | false | # Game Plan\n\n\uD83D\uDEAB We cannot form a triangle if the sum of the minimum two side lengths is less-than-or-equal-to the maximum side length.\n* `\'none\'`\n\n\u2705 Otherwise we return the type of triangle based on the **cardinality of the set of side lengths:**\n\n1. `\'equilateral\'`\n2. `\'isosceles\'`\n3. `\'... | 2 | 0 | ['C++', 'Python3', 'Rust', 'Kotlin', 'JavaScript'] | 2 |
type-of-triangle | Easy Java Solution | easy-java-solution-by-krishn13-fu9z | IntuitionSchool Level Mathematics., if sum of any two sides is greater than the third one, then it is a valid triangle.Approach
nums contains 3 values, denoting | krishn13 | NORMAL | 2025-03-16T07:36:45.644402+00:00 | 2025-03-16T07:36:45.644402+00:00 | 137 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
School Level Mathematics., if sum of any two sides is greater than the third one, then it is a valid triangle.
# Approach
<!-- Describe your approach to solving the problem. -->
1. `nums` contains 3 values, denoting sides of the triangle.,... | 1 | 0 | ['Java'] | 0 |
type-of-triangle | Java, Beats 100% | java-beats-100-by-sarn1-4igd | Code | Sarn1 | NORMAL | 2025-03-13T15:49:36.827493+00:00 | 2025-03-13T15:49:36.827493+00:00 | 32 | false |
# Code
```java []
class Solution {
public String triangleType(int[] nums) {
if (nums[0] + nums[1] <= nums[2] || nums[0] + nums[2] <= nums[1] || nums[1] + nums[2] <= nums[0]) return "none";
else if (nums[0] == nums[1] && nums[0] == nums[2]) return "equilateral";
else if (nums[0] == nums[1] |... | 1 | 0 | ['Java'] | 0 |
type-of-triangle | Mastering Triangle Classification with C++ || Beats 100% || O(1) || Easy for beginners to understand | mastering-triangle-classification-with-c-3clb | IntuitionThe problem requires us to determine the type of triangle that can be formed from three given side lengths. The key conditions to check are:Triangle Va | GeekyRahul04 | NORMAL | 2025-02-26T15:09:05.910744+00:00 | 2025-02-26T15:09:05.910744+00:00 | 103 | false | # Intuition
The problem requires us to determine the type of triangle that can be formed from three given side lengths. The key conditions to check are:
Triangle Validity:
A triangle is only valid if the sum of any two sides is greater than the third side.
This ensures the given lengths satisfy the Triangle Inequalit... | 1 | 0 | ['Array', 'Math', 'Sorting', 'C++'] | 0 |
type-of-triangle | O(1) 🔥 c++ | o1-c-by-varuntyagig-yxu5 | Complexity
Time complexity:
O(1)
Space complexity:
O(1)
Code | varuntyagig | NORMAL | 2025-02-21T06:01:01.781593+00:00 | 2025-02-21T06:01:01.781593+00:00 | 93 | false | # Complexity
- Time complexity:
$$O(1)$$
- Space complexity:
$$O(1)$$
# Code
```cpp []
class Solution {
public:
string triangleType(vector<int>& nums) {
if (nums[0] + nums[1] <= nums[2])
return "none";
if (nums[0] + nums[2] <= nums[1])
return "none";
if (nums[1] +... | 1 | 0 | ['Ordered Set', 'C++'] | 0 |
type-of-triangle | Python soln with thorough explanation | python-soln-with-thorough-explaination-b-vint | IntuitionAfter reading the problem and to solve it efficiently. We can proceed with hashset.ApproachUpon deciding, We chose hashsets as they dont repeat the sam | 9chaitanya11 | NORMAL | 2025-02-06T08:26:04.311667+00:00 | 2025-02-06T08:47:28.489430+00:00 | 69 | false | # Intuition
After reading the problem and to solve it efficiently. We can proceed with hashset.
# Approach
Upon deciding, We chose hashsets as they dont repeat the same value again. So, if a triangle has all three sides as {3,3,3}, the hashset will only store one 3. By which, we can conclude that the given triangle i... | 1 | 0 | ['Python3'] | 0 |
type-of-triangle | 3 LINES JAVA CODE BEATING 100% | 3-lines-java-code-beating-100-by-arshi_b-zc23 | Complexity
Time complexity: O(1)
Space complexity: O(1)
Code | arshi_bansal | NORMAL | 2025-01-09T07:54:04.905663+00:00 | 2025-01-09T07:54:04.905663+00:00 | 208 | false | # Complexity
- Time complexity: O(1)
<!-- 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 String triangleType(int[] nums){
if(nums[0]+nums[1]<=nums[2]||nums[1]+nums[2]<=nums[0]||nums[0... | 1 | 0 | ['Java'] | 0 |
type-of-triangle | O(1) Python Solution Using Sets | o1-python-solution-using-sets-by-odysseu-7foa | IntuitionA set is list of distinct objects. So that can help observe how many sides are the same since a repeated side will be deleted and that could be observe | Odysseus_v01 | NORMAL | 2024-12-28T03:26:18.645132+00:00 | 2024-12-28T03:26:18.645132+00:00 | 116 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
A set is list of distinct objects. So that can help observe how many sides are the same since a repeated side will be deleted and that could be observed in the length of the set.
# Approach
<!-- Describe your approach to solving the proble... | 1 | 0 | ['Python3'] | 1 |
type-of-triangle | 0 ms || BEATS 100% | 0-ms-beats-100-by-deepakk2510-zxtc | Code | deepakk2510 | NORMAL | 2024-12-16T09:28:10.780815+00:00 | 2024-12-16T09:28:10.780815+00:00 | 197 | false | \n# Code\n```cpp []\nclass Solution {\npublic:\n string triangleType(vector<int>& nums) {\n if((nums[0]==nums[1]) && (nums[1]==nums[2])) return "equilateral";\n sort(nums.begin(),nums.end()); \n if((nums[0]+nums[1])>nums[2]){\n if((nums[0]==nums[1]) || (nums[1]==nums[2]) || (nums[0]==... | 1 | 0 | ['C++'] | 0 |
type-of-triangle | 0ms very easy java code | 0ms-very-easy-java-code-by-galani_jenis-zr9x | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Galani_jenis | NORMAL | 2024-12-15T04:06:36.552956+00:00 | 2024-12-15T04:06:36.552956+00:00 | 165 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
type-of-triangle | Very easy code 0ms 100% beat java code | very-easy-code-0ms-100-beat-java-code-by-ci2d | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | Galani_jenis | NORMAL | 2024-12-15T04:05:20.493141+00:00 | 2024-12-15T04:05:20.493141+00:00 | 88 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Java'] | 0 |
type-of-triangle | C++ Simple 4-line Solution | c-simple-4-line-solution-by-yehudisk-cw9a | Intuition\nSimply check all conditions.\n\n# Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\ncpp []\nclass Solution {\npublic:\n s | yehudisk | NORMAL | 2024-11-26T13:40:42.698245+00:00 | 2024-11-26T13:40:42.698328+00:00 | 70 | false | # Intuition\nSimply check all conditions.\n\n# Complexity\n- Time complexity: O(1)\n\n- Space complexity: O(1)\n\n# Code\n```cpp []\nclass Solution {\npublic:\n string triangleType(vector<int>& nums) {\n if ((nums[0] + nums[1] <= nums[2]) || (nums[0] + nums[2] <= nums[1]) || (nums[1] + nums[2] <= nums[0])) re... | 1 | 0 | ['C++'] | 0 |
type-of-triangle | Easy solution | 100% beats | easy-solution-100-beats-by-ravindran_s-t48c | 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 | RAVINDRAN_S | NORMAL | 2024-11-11T05:17:27.144598+00:00 | 2024-11-11T05:17:27.144637+00:00 | 8 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(1)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(1)\n<!-- Add your space complexity here, e.g. $$O... | 1 | 0 | ['C++'] | 0 |
type-of-triangle | Type of Triangle | type-of-triangle-by-tejdekiwadiya-sr9b | Intuition\nThe problem requires determining the type of a triangle based on the lengths of its sides. A triangle can be categorized as:\n- Equilateral: All thre | tejdekiwadiya | NORMAL | 2024-10-21T16:25:42.577915+00:00 | 2024-10-21T16:25:42.577940+00:00 | 74 | false | # Intuition\nThe problem requires determining the type of a triangle based on the lengths of its sides. A triangle can be categorized as:\n- **Equilateral**: All three sides are equal.\n- **Isosceles**: Two sides are equal.\n- **Scalene**: All sides are different.\n- **None**: If the sides don\'t form a valid triangle.... | 1 | 0 | ['Array', 'Math', 'Sorting', 'Java'] | 0 |
type-of-triangle | Simplest logic!! 0ms runtime :) | simplest-logic-0ms-runtime-by-raana_01-2yzt | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Raana_01 | NORMAL | 2024-10-20T07:09:49.882534+00:00 | 2024-10-20T07:09:49.882560+00:00 | 98 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['Python3'] | 0 |
type-of-triangle | Easy Solution | easy-solution-by-shamnad_skr-x3i1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | shamnad_skr | NORMAL | 2024-09-18T11:43:10.966855+00:00 | 2024-09-18T11:43:10.966884+00:00 | 51 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['TypeScript', 'JavaScript'] | 0 |
type-of-triangle | PYTHON | python-by-kanvapuri_sai_praneetha-8tsx | Code\n\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n if len(nums)==3 and nums[0]<nums[1]+nums[2] and nums[1]<nums[0]+nums[2] an | kanvapuri_sai_praneetha | NORMAL | 2024-08-05T18:51:05.716653+00:00 | 2024-08-05T18:51:05.716689+00:00 | 64 | false | # Code\n```\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n if len(nums)==3 and nums[0]<nums[1]+nums[2] and nums[1]<nums[0]+nums[2] and nums[2]<nums[1]+nums[0]:\n if len(set(nums))==1:\n return "equilateral"\n elif len(set(nums))==2:\n r... | 1 | 0 | ['Python3'] | 1 |
type-of-triangle | Python solution in easy steps | python-solution-in-easy-steps-by-shubhas-v07l | \n\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n \n if nums[0] + nums[1] > nums[2] and nums[0] + nums[2] > nums[1] and n | Shubhash_Singh | NORMAL | 2024-08-04T11:56:50.357419+00:00 | 2024-08-04T11:56:50.357453+00:00 | 14 | false | \n```\nclass Solution:\n def triangleType(self, nums: List[int]) -> str:\n \n if nums[0] + nums[1] > nums[2] and nums[0] + nums[2] > nums[1] and nums[1] + nums[2] > nums[0]:\n if nums[0]==nums[1] and nums[1] == nums[2]:\n return "equilateral"\n elif nums[0] != nums[... | 1 | 0 | ['Python3'] | 0 |
type-of-triangle | Easy Java / C++ Solution (😍100% Beats Runtime in Java) 😎😎 -> 😮😮 | easy-java-c-solution-100-beats-runtime-i-629p | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. For "equilateral" tr | jeeleej | NORMAL | 2024-07-26T18:30:04.573702+00:00 | 2024-07-26T18:30:04.573740+00:00 | 30 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n1. For `"equilateral"` triangle all size of triangle is same so **num[0]==nums[1]==nums[2]** is the condition for that.\n2. For `"isosceles"` triangle any two side of ... | 1 | 0 | ['Array', 'C++', 'Java'] | 0 |
type-of-triangle | Simple Easy to Understand Java Code | simple-easy-to-understand-java-code-by-s-8mz1 | Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n\nclass Solution {\n public String triangleType(int[] nums) {\n if(nums[0] == | Saurabh_Mishra06 | NORMAL | 2024-06-24T12:53:33.379685+00:00 | 2024-06-24T12:53:33.379719+00:00 | 136 | false | # Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n```\nclass Solution {\n public String triangleType(int[] nums) {\n if(nums[0] == nums[1] && nums[1] == nums[2]){\n return "equilateral";\n }\n if(nums[0] + nums[1] <= nums[2] || nums[1] + nums[2] <= nums[0] || ... | 1 | 0 | ['Java'] | 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.