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values | created_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | updated_at stringdate 2013-10-25 17:32:12 2025-04-12 09:38:24 | hit_count int64 0 10.6M | has_video bool 2
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maximum-product-of-the-length-of-two-palindromic-subsequences | C++ EASY RECURSION + BACKTRACKING | c-easy-recursion-backtracking-by-the_exp-yrzx | ```\nclass Solution {\npublic:\n int ans =-1;\n bool ispal(vector&a)\n {\n int n=a.size();\n for(int i=0;i&s1,vector\&s2,int i)\n {\n | the_expandable | NORMAL | 2022-02-28T09:07:24.662365+00:00 | 2022-02-28T09:07:24.662396+00:00 | 139 | false | ```\nclass Solution {\npublic:\n int ans =-1;\n bool ispal(vector<char>&a)\n {\n int n=a.size();\n for(int i=0;i<n/2;i++)\n {\n if(a[i] != a[n-1-i])return false;\n }\n return true;\n }\n void solve(string s,vector<char>&s1,vector<char>&s2,int i)\n {\n ... | 1 | 0 | ['Backtracking'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | C++ || BITMASK | c-bitmask-by-easy_coder-cm2d | ```\nclass Solution {\npublic:\n \n bool isPalindrome(string s){\n if(s.size() == 1) return true;\n string tmp = s;\n reverse(tmp.beg | Easy_coder | NORMAL | 2022-01-18T07:33:20.924362+00:00 | 2022-01-18T07:33:20.924394+00:00 | 269 | false | ```\nclass Solution {\npublic:\n \n bool isPalindrome(string s){\n if(s.size() == 1) return true;\n string tmp = s;\n reverse(tmp.begin(),tmp.end());\n return tmp == s;\n }\n \n int maxProduct(string s) {\n int n = s.size();\n map<int,int> mp;\n f... | 1 | 0 | ['Bit Manipulation', 'C', 'Bitmask'] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Maximum Product of the Length of Two Palindromic Subsequences | C++ | LPS | maximum-product-of-the-length-of-two-pal-kgnp | LPS : Longest Palindromic Subsequence\n\nclass Solution {\npublic:\n int LPS(string& s1) {\n string s2 = s1;\n reverse(s2.begin(), s2.end());\n | renu_apk | NORMAL | 2022-01-15T18:15:00.133151+00:00 | 2022-01-15T18:15:28.271072+00:00 | 216 | false | LPS : Longest Palindromic Subsequence\n```\nclass Solution {\npublic:\n int LPS(string& s1) {\n string s2 = s1;\n reverse(s2.begin(), s2.end());\n \n int n = s1.size();\n int dp[n + 1][n + 1];\n for(int i = 0; i < n; i++) dp[i][0] = 0, dp[0][i] = 0;\n \n for(in... | 1 | 0 | [] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | faster than 93.69%, less than 94.86% , C++, simple and concise | faster-than-9369-less-than-9486-c-simple-rqko | \nclass Solution {\npublic:\n int maxProduct(string s) {\n int n=s.size();\n int maskMax=1<<n-1;\n \n int ret=0;\n for(int | zzwolf1306489 | NORMAL | 2022-01-11T08:38:43.071031+00:00 | 2022-01-11T08:38:43.071074+00:00 | 132 | false | ```\nclass Solution {\npublic:\n int maxProduct(string s) {\n int n=s.size();\n int maskMax=1<<n-1;\n \n int ret=0;\n for(int mask=1;mask<maskMax;mask++){\n string s1,s2;\n for(int i=0;i<s.size();i++){\n if(mask & 1<<i)\n s1+=... | 1 | 0 | [] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Java | Simple | Backtracking | java-simple-backtracking-by-pagalpanda-9xpp | \npublic int maxProduct(String s) {\n solve(s, "", "", 0);\n return ans;\n }\n int ans;\n void solve(String s, String s1, String s2, int | pagalpanda | NORMAL | 2021-09-28T01:48:43.278101+00:00 | 2021-09-28T01:48:43.278132+00:00 | 370 | false | ```\npublic int maxProduct(String s) {\n solve(s, "", "", 0);\n return ans;\n }\n int ans;\n void solve(String s, String s1, String s2, int start) {\n if(start == s.length()) {\n if(isPal(s1) && isPal(s2)) {\n int val = s1.length()*s2.length();\n an... | 1 | 0 | ['Backtracking', 'Java'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | C++ creating all subsets and storing them | c-creating-all-subsets-and-storing-them-93o97 | \nclass Solution {\npublic:\n bool isPalindrome(string str){\n int i=0,j = str.length()-1;\n while(i<j){\n if(str[i]!=str[j])return | shvmkj | NORMAL | 2021-09-18T10:51:01.937979+00:00 | 2021-09-18T10:51:01.938013+00:00 | 149 | false | ```\nclass Solution {\npublic:\n bool isPalindrome(string str){\n int i=0,j = str.length()-1;\n while(i<j){\n if(str[i]!=str[j])return false;\n i++;\n j--;\n }\n return true;\n }\n int maxProduct(string s) {\n int n = s.length();\n vect... | 1 | 0 | [] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Disjoint subsequence | Bitmasking | dp | c++ | disjoint-subsequence-bitmasking-dp-c-by-3m3fl | \nint max_palindrome(string a)\n {\n int n = a.size();\n if(n==0)\n return 0;\n vector<vector<int>> dp(n, vector<int>(n,0)); | Dr_Strange_10 | NORMAL | 2021-09-15T10:25:40.590135+00:00 | 2021-09-15T10:25:40.590167+00:00 | 308 | false | ```\nint max_palindrome(string a)\n {\n int n = a.size();\n if(n==0)\n return 0;\n vector<vector<int>> dp(n, vector<int>(n,0));\n for(int i=0;i<n;i++)\n {\n dp[i][i] = 1;\n }\n\t\t\n\t\t// dist is the distance between i and j pointer...\n for(in... | 1 | 0 | ['Dynamic Programming', 'Bitmask'] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | recursion and bitset | recursion-and-bitset-by-michelusa-ehxd | This approach:\n identify all palindroms\n store found palindroms as a bitset of their character positions in an unordered set\n* enumerate all combinations to | michelusa | NORMAL | 2021-09-14T15:15:32.516917+00:00 | 2021-09-14T15:22:26.369930+00:00 | 53 | false | This approach:\n* identify all palindroms\n* store found palindroms as a bitset of their character positions in an unordered set\n* enumerate all combinations to find the max product (length is how many bits are set)\n\n\nExplaining a bit the recursion: we can either have a character or not in the sequence we are ... | 1 | 0 | ['C'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | (C++) 2002. Maximum Product of the Length of Two Palindromic Subsequences | c-2002-maximum-product-of-the-length-of-7l36g | \n\nclass Solution {\npublic:\n int maxProduct(string s) {\n int n = s.size(); \n vector<int> dp(1 << n); \n for (int mask = 1; mask < ( | qeetcode | NORMAL | 2021-09-13T17:58:55.476889+00:00 | 2021-09-13T18:09:13.031701+00:00 | 113 | false | \n```\nclass Solution {\npublic:\n int maxProduct(string s) {\n int n = s.size(); \n vector<int> dp(1 << n); \n for (int mask = 1; mask < (1 << n); ++mask) \n if ((mask & mask-1) == 0) dp[mask] = 1; \n else {\n int lo = log2(mask & ~(mask-1)), hi = log2(mask)... | 1 | 0 | ['C'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | [Python3] bitmask dp | python3-bitmask-dp-by-ye15-oqf9 | Please check out this commit for solutions of weekly 258. \n\nclass Solution:\n def maxProduct(self, s: str) -> int:\n \n @cache\n def l | ye15 | NORMAL | 2021-09-13T14:59:14.593590+00:00 | 2021-09-13T18:13:32.527286+00:00 | 187 | false | Please check out this [commit](https://github.com/gaosanyong/leetcode/commit/0de189059bd6af4dabe23d9066dde1655f3334fc) for solutions of weekly 258. \n```\nclass Solution:\n def maxProduct(self, s: str) -> int:\n \n @cache\n def lps(mask): \n """Return length of longest palindromic seq... | 1 | 0 | ['Python3'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | NOT 2^n. O(n^2) + O(p^2) where n is length of string and p is number of palindromes | not-2n-on2-op2-where-n-is-length-of-stri-8tup | \n/*\n\tO(n^2) + O(plogp) where p is number of palindromes\n\n\tAlgorithm: for each posible palindrome starting charactera look at each possible palindrome end | justacoder3 | NORMAL | 2021-09-12T18:14:54.819966+00:00 | 2021-09-12T20:21:10.539866+00:00 | 66 | false | ```\n/*\n\tO(n^2) + O(plogp) where p is number of palindromes\n\n\tAlgorithm: for each posible palindrome starting charactera look at each possible palindrome end character, and if valid, add it\'s length and mask of bits, set corresponding to which positions were used by the palindrome, to a list of palindromes \n\n\t... | 1 | 0 | [] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | much faster than bitmask 100ms 100% | much-faster-than-bitmask-100ms-100-by-ak-8dtm | \nclass Solution:\n def maxProduct(self, s: str) -> int:\n \n @lru_cache(None)\n def findBigPalLen(x): #O(n^2)\n if len(x) <= | akbc | NORMAL | 2021-09-12T13:59:04.155324+00:00 | 2021-09-12T14:00:57.177868+00:00 | 163 | false | ```\nclass Solution:\n def maxProduct(self, s: str) -> int:\n \n @lru_cache(None)\n def findBigPalLen(x): #O(n^2)\n if len(x) <= 1: return len(x)\n if x[0] == x[-1]: return 2 + findBigPalLen(x[1:-1])\n return max(findBigPalLen(x[1:]), findBigPalLen(x[:-1]))\n ... | 1 | 0 | ['Python3'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | C++ bottom-up DP mask | c-bottom-up-dp-mask-by-walter01-m7nw | Let dp[mask] denotes the longest length of palindromic subsequence in the string "mask" specifies. For example, mask = "11010", s = "abcad", mask has the sequen | walter01 | NORMAL | 2021-09-12T08:57:24.144886+00:00 | 2021-09-12T08:57:24.144915+00:00 | 42 | false | Let dp[mask] denotes the longest length of palindromic subsequence in the string "mask" specifies. For example, mask = "11010", s = "abcad", mask has the sequence of "a, b, a", and dp[mask] = 3; other code is exactly to find the longest length of palindromic. \n\n\n\'\'\'\n\n int maxProduct(string s) {\n\t\t\tin... | 1 | 0 | [] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Python DFS + Memorisation | python-dfs-memorisation-by-jamescandy-kudx | Intuition: similar to LCS, we loop through every possible combination of parlimdromic subsequences, but also introduce two extra parameter, exclude which is the | jamescandy | NORMAL | 2021-09-12T05:12:39.993494+00:00 | 2021-09-12T06:34:55.965147+00:00 | 261 | false | Intuition: similar to LCS, we loop through every possible combination of parlimdromic subsequences, but also introduce two extra parameter, `exclude` which is the indexes of first palindromic subsequences, `num` indicating the number of palindromic subsequences e.g. 0 means first subsequence, 1 means second subsequence... | 1 | 0 | ['Depth-First Search', 'Python'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Isn't this insane how a small change (pass by reference) allows TLE to become AC | isnt-this-insane-how-a-small-change-pass-k15p | Isn\'t this insane how this small change allows the same solution to pass? Shouldn\'t leetcode be able to discern this is not some major change and allow both?\ | ritam777 | NORMAL | 2021-09-12T04:59:59.395449+00:00 | 2021-09-12T05:02:11.382068+00:00 | 128 | false | Isn\'t this insane how this small change allows the same solution to pass? Shouldn\'t leetcode be able to discern this is not some major change and allow both?\n\nAccepted\n```\nclass Solution {\npublic:\n int res, n;\n string str;\n bool isPalin(string s, int n) {\n for (int i=0; i<n/2; i++) {\n ... | 1 | 0 | [] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | C++ || Check All subsequences | c-check-all-subsequences-by-debo01-25pd | Check All subsequences\nand check their lps.\nthen finally return the max possible ans.\n\nclass Solution {\npublic:\n int ans=1;\n unordered_map<string,i | debo01 | NORMAL | 2021-09-12T04:47:14.669064+00:00 | 2021-09-12T04:52:32.264208+00:00 | 143 | false | Check All subsequences\nand check their lps.\nthen finally return the max possible ans.\n```\nclass Solution {\npublic:\n int ans=1;\n unordered_map<string,int>um;\n void func(string s,string x,string y,int i){\n \n if(i==s.length()){\n ans=max(ans,functiontofindlps(x)*functiontofindlp... | 1 | 0 | ['Backtracking', 'C'] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | javascript bitmask brute force 1239ms | javascript-bitmask-brute-force-1239ms-by-4si3 | Main idea:\n(1) get all palindrome subsequence\n(2) parse all ways of picking two of them, if disjoint, max the product in result.\n\nconst isPalindrome = (s) = | henrychen222 | NORMAL | 2021-09-12T04:26:12.997897+00:00 | 2021-09-12T04:37:48.404352+00:00 | 316 | false | Main idea:\n(1) get all palindrome subsequence\n(2) parse all ways of picking two of them, if disjoint, max the product in result.\n```\nconst isPalindrome = (s) => { let n = s.length; let i = 0; let j = n - 1; while (i < j) { if (s[i++] != s[j--]) return false; } return true; };\n\nconst maxProduct = (s) => {\n let... | 1 | 0 | ['Bitmask', 'JavaScript'] | 1 |
maximum-product-of-the-length-of-two-palindromic-subsequences | [python] iterative based brute force solution | python-iterative-based-brute-force-solut-w0ay | things to notice :\n1.deep copy\n2. covering every start point\n```\nclass Solution:\n def maxProduct(self, s: str) -> int:\n q= deque()\n for | user4436h | NORMAL | 2021-09-12T04:13:20.698886+00:00 | 2021-09-12T04:45:35.706459+00:00 | 114 | false | things to notice :\n1.deep copy\n2. covering every start point\n```\nclass Solution:\n def maxProduct(self, s: str) -> int:\n q= deque()\n for i in range(len(s)):\n q.append((s[i],[i]))\n ml=0\n n=0\n while q:\n st,li=q.pop()\n if st==st[::-1]:\n ... | 1 | 0 | ['Iterator', 'Python'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Brute force: Bitmask + Longest Palindromic Subsequence | brute-force-bitmask-longest-palindromic-bw5yi | Generate all possible splits of s into two sequences of minimum 1 character (using bitmask).\nRun the longest palindromic subsequence algorithm on both parts an | prezes | NORMAL | 2021-09-12T04:10:35.395984+00:00 | 2021-09-12T04:13:40.818403+00:00 | 163 | false | Generate all possible splits of s into two sequences of minimum 1 character (using bitmask).\nRun the[ longest palindromic subsequence](https://leetcode.com/problems/longest-palindromic-subsequence/) algorithm on both parts and return the max encountered product.\n\n```\nclass Solution {\n public int maxProduct(Stri... | 1 | 0 | ['Bitmask'] | 0 |
maximum-product-of-the-length-of-two-palindromic-subsequences | Java | Backtracking | java-backtracking-by-ashish10comp-fzue | \n\nclass Solution {\n private int maxProduct;\n\n public int maxProduct(String s) {\n maxProduct = 0;\n populateMaxProduct(s, new StringBui | ashish10comp | NORMAL | 2021-09-12T04:07:18.257353+00:00 | 2021-09-12T04:07:18.257392+00:00 | 164 | false | ```\n\nclass Solution {\n private int maxProduct;\n\n public int maxProduct(String s) {\n maxProduct = 0;\n populateMaxProduct(s, new StringBuilder(), new StringBuilder(), 0);\n return maxProduct;\n }\n\n private void populateMaxProduct(String s, StringBuilder s1, StringBuilder s2, int ... | 1 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++/Java Two Sum O(n * m) | cjava-two-sum-on-m-by-votrubac-hz2s | This looks very similar to Two Sum, and we will use Two Sum: Approach 3.\n\nWe just need to call our function "Two Product" :)\n\n> Update: see below for the op | votrubac | NORMAL | 2020-09-06T04:00:32.959731+00:00 | 2020-09-06T19:26:29.343002+00:00 | 6,955 | false | This looks very similar to Two Sum, and we will use [Two Sum: Approach 3](https://leetcode.com/problems/two-sum/solution/).\n\nWe just need to call our function "Two Product" :)\n\n> Update: see below for the optimized version!\n\n**C++**\n```cpp\nint numTriplets(vector<int>& n1, vector<int>& n2) {\n return accumula... | 94 | 2 | [] | 9 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ O(MN) with explanation | c-omn-with-explanation-by-lzl124631x-0gmq | See my latest update in repo LeetCode\n\n## Solution 1.\n\nType 1 and Type 2 are symmetrical so we can define a function count(A, B) which returns the count of | lzl124631x | NORMAL | 2020-09-06T04:02:39.697855+00:00 | 2020-09-09T07:15:28.670931+00:00 | 3,369 | false | See my latest update in repo [LeetCode](https://github.com/lzl124631x/LeetCode)\n\n## Solution 1.\n\nType 1 and Type 2 are symmetrical so we can define a function `count(A, B)` which returns the count of the Type 1 triples. The answer is `count(A, B) + count(B, A)`.\n\nFor `count(A, B)`, we can use a `unordered_map<int... | 38 | 1 | [] | 7 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | python solution | python-solution-by-akaghosting-osfq | \tclass Solution:\n\t\tdef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\t\t\td1 = collections.defaultdict(int)\n\t\t\td2 = collections.defaul | akaghosting | NORMAL | 2020-09-06T04:24:16.273455+00:00 | 2022-07-17T16:10:12.570724+00:00 | 1,718 | false | \tclass Solution:\n\t\tdef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\t\t\td1 = collections.defaultdict(int)\n\t\t\td2 = collections.defaultdict(int)\n\t\t\tres = 0\n\t\t\tfor num1 in nums1:\n\t\t\t\td1[num1 * num1] += 1\n\t\t\tfor num2 in nums2:\n\t\t\t\td2[num2 * num2] += 1 \n\t\t\tfor i i... | 19 | 0 | [] | 5 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Simple sort | java-simple-sort-by-hobiter-0q55 | An easy way to sort and count:\n1, sort all arrays by non-decreasing order;\n2, for each i, shrink from left and right as j and k, then find each triples\n3, ha | hobiter | NORMAL | 2020-09-06T04:02:10.027572+00:00 | 2020-09-13T20:39:55.335942+00:00 | 1,624 | false | An easy way to sort and count:\n1, sort all arrays by non-decreasing order;\n2, for each i, shrink from left and right as j and k, then find each triples\n3, handle special case for dups, such as[1, 1], [1, 1, 1]\nTime: O(M*N)\nSpace: O(1)\n\n```\nclass Solution {\n public int numTriplets(int[] a, int[] b) {\n ... | 15 | 4 | [] | 4 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | c++ solution O(n*m) using hash_map | c-solution-onm-using-hash_map-by-dilipsu-8x3m | \n#define ll long long int\nclass Solution {\npublic:\n int find(vector<int>nums1,vector<int>nums2)\n {\n unordered_map<ll,ll>mp;\n for(int | dilipsuthar17 | NORMAL | 2021-03-01T11:23:02.867305+00:00 | 2021-03-01T11:23:02.867346+00:00 | 656 | false | ```\n#define ll long long int\nclass Solution {\npublic:\n int find(vector<int>nums1,vector<int>nums2)\n {\n unordered_map<ll,ll>mp;\n for(int i=0;i<nums2.size();i++)\n {\n for(int j=i+1;j<nums2.size();j++)\n {\n ll p=(long long)nums2[i]*nums2[j];\n ... | 11 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python | Instructions to code | Intuitive | python-instructions-to-code-intuitive-by-bsuj | \nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def is_perf_sqrt(n): \n """\n returns bo | prodcode | NORMAL | 2020-09-06T04:05:32.412492+00:00 | 2020-09-06T04:18:47.924374+00:00 | 886 | false | ```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def is_perf_sqrt(n): \n """\n returns bool if a number is a perfect square\n like 4, 16, 25 --> True\n """\n return int(math.sqrt(n) + 0.5) ** 2 == n\n\n num... | 11 | 0 | [] | 5 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java solution using sorting. With explanation and comments. 4 ms. Faster than 100%. | java-solution-using-sorting-with-explana-aed1 | Sort both arrays.\n2. Similar to the Two sum we have to find all the pairs that matches to the target. Target is the Square number. \n3. Start with left pointer | shaishavjogani | NORMAL | 2020-09-06T05:05:43.505364+00:00 | 2020-09-08T06:20:11.271116+00:00 | 662 | false | 1. Sort both arrays.\n2. Similar to the Two sum we have to find all the pairs that matches to the target. Target is the Square number. \n3. Start with left pointer set to 0, and right pointer set to length-1.\n4. Must handle duplicates. \n\t* \te.g. Target: 4, nums = [1,1,4,4]\n\t* \tSo, total 4 pairs with indexes (0,2... | 7 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python3 | Using Counter | python3-using-counter-by-tbvho-rsoo | For both nums1 and nums2, aggregate all the squares in nums1 and nums2. We will use this to easily track all occurences of nums1[i] * nums1[j] in nums2 and num | tbvho | NORMAL | 2020-09-06T04:14:05.027809+00:00 | 2020-09-06T04:14:05.027871+00:00 | 429 | false | For both nums1 and nums2, aggregate all the squares in nums1 and nums2. We will use this to easily track all occurences of nums1[i] * nums1[j] in nums2 and nums2[i] * nums2[j] in nums1. Then just use nested loops to generate possible combinations that could be in either frequency table.\n\n\n```\nimport collections\n\... | 7 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ | Two Pointer | 100 % faster | c-two-pointer-100-faster-by-csachdeva-50t0 | \nclass Solution {\npublic:\n int countTriplet(vector<int>& nums1, vector<int>& nums2) {\n int count = 0, left, right;\n long num1Val, num2Val; | csachdeva | NORMAL | 2020-09-06T04:37:44.518554+00:00 | 2020-09-06T09:03:42.497546+00:00 | 969 | false | ```\nclass Solution {\npublic:\n int countTriplet(vector<int>& nums1, vector<int>& nums2) {\n int count = 0, left, right;\n long num1Val, num2Val;\n unordered_map<int, int> freq;\n \n for(int i = 0; i < nums2.size(); i++) freq[nums2[i]]++;\n \n for(int i = 0; i < nums... | 6 | 0 | [] | 2 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Straightforward solution - brute force with optimized search using set | straightforward-solution-brute-force-wit-5pcz | \nclass Solution {\npublic:\n bool perfectSquare(long long int n) {\n long long int ans = sqrt(n);\n return n == ans*ans;\n }\n \n int | interviewrecipes | NORMAL | 2020-09-06T04:04:04.484495+00:00 | 2020-09-06T04:04:04.484538+00:00 | 631 | false | ```\nclass Solution {\npublic:\n bool perfectSquare(long long int n) {\n long long int ans = sqrt(n);\n return n == ans*ans;\n }\n \n int squareRoot(long long int n) {\n int ans = sqrt(n);\n return ans;\n }\n \n int findTriplets(vector<int> a, vector<int> b) {\n i... | 6 | 0 | [] | 2 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | 2 pointer approach || O(1) space complexity || 98% faster | 2-pointer-approach-o1-space-complexity-9-n4s9 | \nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n Arrays.sort(nums1);\n Arrays.sort(nums2);\n return count(nums1 | harshitgoel1110 | NORMAL | 2021-06-26T13:08:12.317898+00:00 | 2021-06-26T13:08:12.317941+00:00 | 469 | false | ```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n Arrays.sort(nums1);\n Arrays.sort(nums2);\n return count(nums1 , nums2) + count(nums2 , nums1);\n }\n \n public int count(int a[] , int b[]){\n int n = a.length;\n int m = b.length;\n int c... | 5 | 1 | ['Two Pointers', 'Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++||Two Pointers||Easy to Understand | ctwo-pointerseasy-to-understand-by-retur-cnxx | ```\nclass Solution {\npublic:\n int countTriplet(vector &nums1,vector &nums2)\n {\n int ans=0,left,right;\n unordered_map freq;\n fo | return_7 | NORMAL | 2022-09-04T20:48:27.635595+00:00 | 2022-09-04T20:48:27.635641+00:00 | 663 | false | ```\nclass Solution {\npublic:\n int countTriplet(vector<int> &nums1,vector<int> &nums2)\n {\n int ans=0,left,right;\n unordered_map<int,int> freq;\n for(auto &n:nums2)\n {\n freq[n]++;\n }\n for(int i=0;i<nums1.size();i++)\n {\n long long t=(... | 4 | 0 | ['Two Pointers', 'C'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++, 20ms, 12MB, max(O(n1+n2), O(d1*d2)) (d1, d2 are number of distinct numbers in nums1 and nums2) | c-20ms-12mb-maxon1n2-od1d2-d1-d2-are-num-jv7q | Firstly we count how many times each number appeared in each vector.\n\nFor each distinct number n1 in one set, we try to find in the other set:\n 1. n2 == n1 | alvin-777 | NORMAL | 2020-09-06T05:27:06.510237+00:00 | 2020-09-06T06:32:50.219446+00:00 | 601 | false | Firstly we count how many times each number appeared in each vector.\n\nFor each distinct number n1 in one set, we try to find in the other set:\n 1. `n2 == n1` and n2 appeared at least twice\n 2. `n2 < n1` and exists n2x that `n2*n2x == n1*n1` => `n2x == n1*n1/n2`\n\nFor `1`, if n1 appeared f1 times and n2 appeared ... | 4 | 1 | ['C'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | 5 LINE JAVA SOLUTION USING HASHMAP WITH EXPLANATION | 5-line-java-solution-using-hashmap-with-zh5ux | 1. Store the value of num1*num1 from the first Array in a map.\n2. Using two for loops generate all possible combination of num2 and check if the set contains t | visnunathan | NORMAL | 2020-09-06T04:38:18.290999+00:00 | 2020-09-06T06:36:24.865667+00:00 | 444 | false | **1. Store the value of num1*num1 from the first Array in a map.\n2. Using two for loops generate all possible combination of num2 and check if the set contains those value.\n3. If it contains add it to the sum and return it.**\n```\npublic int calculate(int[] num1, int[] num2) {\n HashMap<Long, Integer> map = n... | 4 | 1 | ['Java'] | 2 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ code using unordered_map | c-code-using-unordered_map-by-shalinipan-j93h | \nclass Solution {\npublic:\n int get(vector<int>& nums1, vector<int>& nums2)\n {\n unordered_map<long long int,long long int>m;int ans=0;\n | shalinipandey1955 | NORMAL | 2020-09-06T06:41:04.916228+00:00 | 2020-10-26T10:29:34.139047+00:00 | 158 | false | ```\nclass Solution {\npublic:\n int get(vector<int>& nums1, vector<int>& nums2)\n {\n unordered_map<long long int,long long int>m;int ans=0;\n for(int i=0;i<nums1.size();i++)\n {\n m[(long long )nums1[i]*nums1[i]]++;\n }\n for(int i=0;i<nums2.size();i++)\n {\n... | 3 | 1 | ['C'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [C++] Explanation, Approach and Code and Time Complexity | c-explanation-approach-and-code-and-time-80n5 | Approach\n 0. Sort the arrays since you need to search\n For type 1:\n\t\n 1. select first number from nums1\n 2. Square it\n 3. Select number fr | rahulanandyadav2000 | NORMAL | 2020-09-06T04:13:10.557620+00:00 | 2020-09-06T04:33:27.322417+00:00 | 474 | false | Approach\n 0. Sort the arrays since you need to search\n For type 1:\n\t\n 1. select first number from nums1\n 2. Square it\n 3. Select number from nums2 and divide it by square of nums1[i]\n 4. If divisible, search for all occurences of the other pair in nums2\n 5. Lets there be x occurences then ... | 3 | 0 | ['C', 'Binary Tree'] | 2 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Rust || Simple solution | rust-simple-solution-by-user7454af-aql2 | Complexity\n- Time complexity: O(n^2)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n^2)\n Add your space complexity here, e.g. O(n) \n\n | user7454af | NORMAL | 2024-06-06T17:09:17.135202+00:00 | 2024-06-06T17:09:17.135238+00:00 | 29 | false | # Complexity\n- Time complexity: $$O(n^2)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n^2)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nuse std::collections::HashMap;\nimpl Solution {\n pub fn num_triplets(nums1: Vec<i32>, nums2: Vec<i32>) -> i32 {... | 2 | 0 | ['Rust'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Python 3] O(MN) short and concise | python-3-omn-short-and-concise-by-gabhay-bogn | \tclass Solution:\n\t\tdef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\t\t\tdef get_res(a,b):\n\t\t\t\tres,n=0,len(b)\n\t\t\t\tprod=Counter( | gabhay | NORMAL | 2022-07-22T05:06:43.091470+00:00 | 2022-07-22T05:06:43.091508+00:00 | 295 | false | \tclass Solution:\n\t\tdef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\t\t\tdef get_res(a,b):\n\t\t\t\tres,n=0,len(b)\n\t\t\t\tprod=Counter([x*x for x in a])\n\t\t\t\tfor i in range(n):\n\t\t\t\t\tfor j in range(i+1,n):\n\t\t\t\t\t\tres+=prod[b[i]*b[j]]\n\t\t\t\treturn res\n\t\t\treturn get_res(nums... | 2 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Brute force approach | C++ | brute-force-approach-c-by-tusharbhart-ifn5 | \nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n \n unordered_map<long, int> m1, m2;\n for(int | TusharBhart | NORMAL | 2022-02-28T08:25:12.432454+00:00 | 2022-02-28T08:25:12.432484+00:00 | 271 | false | ```\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n \n unordered_map<long, int> m1, m2;\n for(int i : nums1) m1[(long long)i*i]++;\n for(int i : nums2) m2[(long long)i*i]++;\n \n int ans = 0;\n \n for(int i=0; i<nums2.si... | 2 | 0 | ['C'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python intuitive hashmap solution, O(n*m) time | python-intuitive-hashmap-solution-onm-ti-cpqi | \nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n sqr1, sqr2 = defaultdict(int), defaultdict(int)\n m, n | byuns9334 | NORMAL | 2021-12-31T09:31:31.364955+00:00 | 2021-12-31T09:31:31.364991+00:00 | 419 | false | ```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n sqr1, sqr2 = defaultdict(int), defaultdict(int)\n m, n = len(nums1), len(nums2)\n for i in range(m):\n sqr1[nums1[i]**2] += 1\n for j in range(n):\n sqr2[nums2[j]**2] += 1\n ... | 2 | 0 | ['Python', 'Python3'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C+ (A very simple Two Sum like approach) | c-a-very-simple-two-sum-like-approach-by-5psp | \n//Approach-1 (Simple solution)\nclass Solution {\npublic:\n void twoProduct(ull target, vector<int> & nums, int &count) {\n unordered_map<int, int> | mazhar_mik | NORMAL | 2021-06-30T02:05:48.952274+00:00 | 2021-06-30T02:22:11.825408+00:00 | 276 | false | ```\n//Approach-1 (Simple solution)\nclass Solution {\npublic:\n void twoProduct(ull target, vector<int> & nums, int &count) {\n unordered_map<int, int> mp;\n mp[1] = 0;\n for(int i = 0; i<nums.size(); i++) {\n if(target%nums[i] == 0) {\n int remain= target/nums[i];\n ... | 2 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | JavaScript Solution | javascript-solution-by-deadication-tzn9 | \nvar numTriplets = function(nums1, nums2) {\n const squaredFreq1 = countSquareFreq(nums1);\n const squaredFreq2 = countSquareFreq(nums2);\n \n retu | Deadication | NORMAL | 2021-04-26T16:48:27.417270+00:00 | 2021-04-26T16:55:12.777276+00:00 | 140 | false | ```\nvar numTriplets = function(nums1, nums2) {\n const squaredFreq1 = countSquareFreq(nums1);\n const squaredFreq2 = countSquareFreq(nums2);\n \n return countProdFreq(nums1, squaredFreq2) + countProdFreq(nums2, squaredFreq1);\n\t\n \n function countSquareFreq(nums) {\n const freq = new Map();\... | 2 | 0 | ['Hash Table', 'JavaScript'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java O(n * sqrt(n)) solution using prime factorisation | java-on-sqrtn-solution-using-prime-facto-khkc | Take the example \nnums1 -> [4]\nnums2 -> [2,2,2,4,4,8,8]\nfor the SqrNo = 16, possible 2 pairs of factors are (1,16) (2,8) (4,4)\nHave a count hashmap of all c | tywin_lannister97 | NORMAL | 2021-03-14T16:54:01.930850+00:00 | 2021-03-14T16:54:01.930887+00:00 | 188 | false | Take the example \nnums1 -> [4]\nnums2 -> [2,2,2,4,4,8,8]\nfor the SqrNo = 16, possible 2 pairs of factors are (1,16) (2,8) (4,4)\nHave a count hashmap of all count nos and simply check for all these cases. Repeat the same for each of nums2 combination against num1. One edge case is int overflow if n= 10^5 n ^ 2 will c... | 2 | 2 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | 93%/91% Python3, Hash table | 9391-python3-hash-table-by-awong05-b49z | \nfrom collections import Counter\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def helper(A, B):\n | awong05 | NORMAL | 2020-09-17T13:17:44.056126+00:00 | 2020-09-17T13:17:44.056162+00:00 | 160 | false | ```\nfrom collections import Counter\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n def helper(A, B):\n ans = 0\n C = Counter([a*a for a in A])\n D = Counter()\n for b in B:\n for k, v in D.items():\n ... | 2 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Python] Using counter O(n*m) | python-using-counter-onm-by-carloscerlir-0xay | \n def numTriplets(self, A: List[int], B: List[int]) -> int:\n def getTriplets(A, B):\n m, n = len(A), len(B)\n ans = 0\n | carloscerlira | NORMAL | 2020-09-06T04:07:13.144072+00:00 | 2020-09-06T04:20:32.859520+00:00 | 151 | false | ```\n def numTriplets(self, A: List[int], B: List[int]) -> int:\n def getTriplets(A, B):\n m, n = len(A), len(B)\n ans = 0\n counter = Counter()\n for j in range(n):\n for k in range(j+1,n):\n prod = B[j]*B[k]\n c... | 2 | 0 | ['Python'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Java] Use Hashmap to Maintain Frequencies O(N^2 + M^2) | java-use-hashmap-to-maintain-frequencies-zgwa | \n public int numTriplets(int[] nums1, int[] nums2) {\n HashMap<Long, Integer> s1 = new HashMap();\n HashMap<Long, Integer> s2 = new HashMap(); | yuhwu | NORMAL | 2020-09-06T04:06:37.134275+00:00 | 2020-09-06T04:06:37.134341+00:00 | 141 | false | ```\n public int numTriplets(int[] nums1, int[] nums2) {\n HashMap<Long, Integer> s1 = new HashMap();\n HashMap<Long, Integer> s2 = new HashMap();\n int n1 = nums1.length;\n int n2 = nums2.length;\n for(int i=0; i<n1; i++){\n for(int j=i+1; j<n1; j++){\n l... | 2 | 1 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ Easy 3 Sum, 2 pointer (with comments) | O(N*M) time O(1) space | c-easy-3-sum-2-pointer-with-comments-onm-0y1u | \n\nclass Solution {\npublic:\n #define ll long long\n \n\t// returns count of valids triplets (i, j, K) of type 1\n\t// note that to find of type 2, just | chwanshu27 | NORMAL | 2020-09-06T04:04:49.667232+00:00 | 2020-09-06T04:24:55.401567+00:00 | 172 | false | ```\n\nclass Solution {\npublic:\n #define ll long long\n \n\t// returns count of valids triplets (i, j, K) of type 1\n\t// note that to find of type 2, just swap nums1 and nums2 while calling \n int returnCount(vector<int>& nums1, vector<int>& nums2) {\n int n = nums1.size(), m = nums2.size();\n ... | 2 | 0 | ['Two Pointers', 'C'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python3 Solution | python3-solution-by-deleted_user-k82m | python\ndef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\trev, n1, n2 = 0, Counter(nums1), Counter(nums2)\n\n\tfor i in range(len(nums1)-1):\ | deleted_user | NORMAL | 2020-09-06T04:02:50.377735+00:00 | 2020-09-06T04:02:50.377837+00:00 | 376 | false | ```python\ndef numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n\trev, n1, n2 = 0, Counter(nums1), Counter(nums2)\n\n\tfor i in range(len(nums1)-1):\n\t\tfor j in range(i+1, len(nums1)):\n\t\t\tt = (nums1[i] * nums1[j])**(1/2)\n\n\t\t\tif t == int(t) and t in n2:\n\t\t\t\trev += n2[t]\n\n\tfor i in range(... | 2 | 0 | ['Python', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy Java Solution Using HashMap | easy-java-solution-using-hashmap-by-ravi-0faf | 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 | ravikumar50 | NORMAL | 2023-08-05T10:55:36.027232+00:00 | 2023-08-05T10:55:36.027260+00:00 | 34 | 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 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ || Easy Explanation ||O(N^2) | c-easy-explanation-on2-by-avi1273619-i2r7 | // long long temp,t,help are used to keep the product within limits (prevent overflow)\n \n// The idea is to store all the squares in a map(not se | avi1273619 | NORMAL | 2022-08-06T13:39:36.526555+00:00 | 2023-01-15T12:45:32.302121+00:00 | 379 | false | // long long temp,t,help are used to keep the product within limits (prevent overflow)\n \n// The idea is to store all the squares in a map(not set because duplicates may be present an dnot even multiset because we won\'t get the number of times the duplicates appears)\n \n// Use nested loo... | 1 | 0 | ['C'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Javascript] Two Approaches - Hashmap & Two Pointers | javascript-two-approaches-hashmap-two-po-3yp6 | Solution 1: Hashmap\n\nThis approach is similar to the two sum problem.\nWe use a hashmap to keep track of the frequencies of past numbers.\nCount the number of | anna-hcj | NORMAL | 2022-06-27T10:59:14.335469+00:00 | 2022-06-27T10:59:14.335512+00:00 | 142 | false | **Solution 1: Hashmap**\n\nThis approach is similar to the two sum problem.\nWe use a hashmap to keep track of the frequencies of past numbers.\nCount the number of `target / nums2[j]` in the hashmap.\n\n`n = length of nums1`, `m = length of nums2`\nTime Complexity: `O(nm)` 498ms\nSpace Complexity: `O(n + m)` 48.3MB\n`... | 1 | 0 | ['JavaScript'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | python soln | python-soln-by-kumarambuj-6fg5 | \nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n hash1=defaultdict(int)\n hash2=defaultdict(int)\n | kumarambuj | NORMAL | 2022-01-02T08:57:32.870500+00:00 | 2022-01-02T08:57:32.870530+00:00 | 114 | false | ```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n hash1=defaultdict(int)\n hash2=defaultdict(int)\n \n for x in nums1:\n hash1[x*x]+=1\n \n for x in nums2:\n hash2[x*x]+=1\n \n res=0\n for i ... | 1 | 0 | ['Python'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ Brute force | Using Hash_map | c-brute-force-using-hash_map-by-nitink_3-34c4 | \n#define ll long long int\nclass Solution {\npublic:\n \n int find(vector<int>nums1,vector<int>nums2)\n {\n unordered_map<ll,ll>mp;\n // | HustlerNitin | NORMAL | 2022-01-01T18:03:06.428556+00:00 | 2022-01-01T18:03:06.428602+00:00 | 392 | false | ```\n#define ll long long int\nclass Solution {\npublic:\n \n int find(vector<int>nums1,vector<int>nums2)\n {\n unordered_map<ll,ll>mp;\n // O(n^2)\n for(int i=0;i<nums2.size();i++)\n {\n for(int j=i+1;j<nums2.size();j++)\n {\n ll p=(long long)nu... | 1 | 0 | ['C', 'C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ BRUTEFORCE | c-bruteforce-by-the_expandable-p63f | ```\nclass Solution {\npublic:\n #define ll long long int\n int numTriplets(vector& a, vector& b) {\n mapmp1;\n mapmp2;\n sort(a.begi | the_expandable | NORMAL | 2021-09-23T04:57:11.768427+00:00 | 2021-09-23T04:57:11.768472+00:00 | 74 | false | ```\nclass Solution {\npublic:\n #define ll long long int\n int numTriplets(vector<int>& a, vector<int>& b) {\n map<ll,ll>mp1;\n map<ll,ll>mp2;\n sort(a.begin(),a.end());\n sort(b.begin(),b.end());\n for(int i = 0 ; i < a.size();i++)\n {\n ll p = (ll)a[i]*a[i];... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java solution | java-solution-by-keerthy0212-v0md | \nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n return res(nums1,nums2)+res(nums2,nums1);\n }\n public static int res(i | keerthy0212 | NORMAL | 2021-05-09T17:08:20.012478+00:00 | 2021-05-09T17:08:20.012519+00:00 | 255 | false | ```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n return res(nums1,nums2)+res(nums2,nums1);\n }\n public static int res(int[] nums1,int[] nums2)\n {\n int count=0;\n\t\tHashMap<Long,Integer> map=new HashMap<Long,Integer>();\n\t\tfor(long i:nums1)\n\t\t\tmap.put((i*i),... | 1 | 0 | ['Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python O(n^2) function | python-on2-function-by-dev-josh-b9jy | How do we approach such a question?\n You can try the brute force, but it will TLE\n I always try to fall back to the Two Sum problem\n\t Can we build dictionar | dev-josh | NORMAL | 2021-02-16T19:02:26.161304+00:00 | 2021-02-16T19:11:19.360665+00:00 | 249 | false | * How do we approach such a question?\n* You can try the brute force, but it will TLE\n* I always try to fall back to the **Two Sum problem**\n\t* Can we build dictionaries in linear or n^2 time to avoid n^3 time?\n\t\t* We can count the number of `nums1[i]^2` for example\n\t\t* We can also count the number of `nums2[j... | 1 | 0 | ['Python', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Python] Count, beats 100%, O(n * m) | python-count-beats-100-on-m-by-shoomoon-2jot | First count the elements in each array, then find the number of triplets for each element.\nIt is faster if there are lots of duplicates.\n\nclass Solution(obje | shoomoon | NORMAL | 2021-01-28T18:22:59.641677+00:00 | 2021-01-28T18:22:59.641714+00:00 | 148 | false | First count the elements in each array, then find the number of triplets for each element.\nIt is faster if there are lots of duplicates.\n```\nclass Solution(object):\n def numTriplets(self, nums1, nums2):\n """\n :type nums1: List[int]\n :type nums2: List[int]\n :rtype: int\n """... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python Solution Based on Binary Search And Memorization | python-solution-based-on-binary-search-a-ku2z | Time Complexity = O ( N * N * Log N )\n\n\n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int: \n nums1.sort()\ | pochy | NORMAL | 2020-09-08T04:53:10.281496+00:00 | 2020-09-08T04:57:41.853257+00:00 | 184 | false | Time Complexity = O ( N * N * Log N )\n\n\n```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int: \n nums1.sort()\n nums2.sort()\n \n def lowerbound(target, left, right, nums):\n while left < right:\n mid = left + (right - l... | 1 | 0 | ['Memoization', 'Binary Tree', 'Python', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python3 3 liner - Number of Ways Where Square of Number Is Equal to Product of Two Numbers | python3-3-liner-number-of-ways-where-squ-67gh | \nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n sq1 = Counter(map(mul, nums1, nums1))\n sq2 = Counter(m | r0bertz | NORMAL | 2020-09-08T03:20:50.283602+00:00 | 2020-09-08T03:20:50.283656+00:00 | 166 | false | ```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n sq1 = Counter(map(mul, nums1, nums1))\n sq2 = Counter(map(mul, nums2, nums2))\n return sum(sq2[a * b] for a, b in combinations(nums1, 2)) + sum(sq1[a * b] for a, b in combinations(nums2, 2))\n``` | 1 | 0 | ['Python', 'Python3'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java factorization solution O(n + m) | java-factorization-solution-on-m-by-igor-u92m | Algorithm:\n1. Create a map key: (num * num), value: count basing on first array\n2. Go throug the second array, for every number calculate a key containing fa | igor84 | NORMAL | 2020-09-07T14:24:28.135348+00:00 | 2020-09-07T15:58:39.652374+00:00 | 180 | false | Algorithm:\n1. Create a map key: (num * num), value: count basing on first array\n2. Go throug the second array, for every number calculate a key containing factors that are represented odd number of times in number. For example, 60 is 2 * 2 * 3 * 5, the key is "3,5"\n3. Search for the values with the same key process... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++|| O(MN)||Easiest solution with self explanatory code | c-omneasiest-solution-with-self-explanat-fe11 | Please Note that time complexity of find function in unordered_map is O(1)\n\nclass Solution {\npublic:\n int numTriplets(vector<int>& n1, vector<int>& n2) { | tejpratapp468 | NORMAL | 2020-09-07T05:56:06.468623+00:00 | 2020-09-07T05:58:59.159634+00:00 | 79 | false | Please Note that time complexity of find function in unordered_map is O(1)\n```\nclass Solution {\npublic:\n int numTriplets(vector<int>& n1, vector<int>& n2) {\n int n=n1.size();int m=n2.size();\n long ans=0;\n //type 1\n for(int i=0;i<n;i++)\n {\n double a=n1[i];\n ... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C# solution using Dictionary | c-solution-using-dictionary-by-pykan-g1du | \npublic int NumTriplets(int[] nums1, int[] nums2) {\n int w1 = GetWays(nums1, nums2);\n int w2 = GetWays(nums2, nums1);\n return w1 + w2;\ | pykan | NORMAL | 2020-09-06T09:13:30.698743+00:00 | 2020-09-11T13:48:20.834182+00:00 | 72 | false | ```\npublic int NumTriplets(int[] nums1, int[] nums2) {\n int w1 = GetWays(nums1, nums2);\n int w2 = GetWays(nums2, nums1);\n return w1 + w2;\n }\n \n public int GetWays(int[] nums1, int[] nums2) {\n int numWays = 0;\n long product = 0;\n \n Dictionary<long, int... | 1 | 0 | [] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Two Diff Solutions O(m^2+n^2) and O(m*n) | java-two-diff-solutions-om2n2-and-omn-by-gwdl | Method1: \n1. Use two different hashmaps\n2. Add entried to map for diff combinations of product\n3. Iterate over arrays and check if square of num exists in ot | meg20 | NORMAL | 2020-09-06T09:04:43.618580+00:00 | 2020-09-06T11:29:25.121585+00:00 | 116 | false | Method1: \n1. Use two different hashmaps\n2. Add entried to map for diff combinations of product\n3. Iterate over arrays and check if square of num exists in other map\n\nTime: O(m^2 + n^2)\n```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n Map<Long, Integer> productSet1 = new HashM... | 1 | 1 | ['Sorting', 'Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Bruteforce O(N^2) | java-bruteforce-on2-by-janap-fxi9 | ```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n return getCount(nums1, getMap(nums2)) + getCount(nums2, getMap(nums1));\n | janap | NORMAL | 2020-09-06T07:10:18.468508+00:00 | 2020-09-06T07:10:18.468564+00:00 | 43 | false | ```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n return getCount(nums1, getMap(nums2)) + getCount(nums2, getMap(nums1));\n }\n \n Map<Long, Integer> getMap(int[] nums){\n Map<Long, Integer> map = new HashMap<>();\n for(long num: nums)\n map.put(num... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | c++ code using hashmap | c-code-using-hashmap-by-srjohn-312s | \nclass Solution {\npublic:\n unordered_map<double,int> hash1;\n unordered_map<double,int> hash2;\n int numTriplets(vector<int>& nums1, vector<int>& nu | srjohn | NORMAL | 2020-09-06T06:56:50.249655+00:00 | 2020-09-06T06:56:50.249692+00:00 | 68 | false | ```\nclass Solution {\npublic:\n unordered_map<double,int> hash1;\n unordered_map<double,int> hash2;\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n for(int i=0;i<nums1.size();i++)\n {if(hash1.find(nums1[i])==hash1.end()) hash1[nums1[i]]=1;\n else hash1[nums1[i]]++; \n ... | 1 | 1 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ using 2 unordered_map | c-using-2-unordered_map-by-tejasrai-qvjx | \nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n int n=nums1.size();\n int m=nums2.size();\n uno | tejasrai | NORMAL | 2020-09-06T06:18:00.064114+00:00 | 2020-09-06T06:19:30.796917+00:00 | 39 | false | ```\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n int n=nums1.size();\n int m=nums2.size();\n unordered_map<long long int,long long int>s1;\n unordered_map<long long int,long long int>s2;\n int count=0;\n for(int i=0;i<n;i++)\n ... | 1 | 1 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Why 91/92 TCs passed? Please suggest/help. | why-9192-tcs-passed-please-suggesthelp-b-n7nh | \nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n Arrays.sort(nums1);\n Arrays.sort(nums2);\n \n int ans | vvbhandare | NORMAL | 2020-09-06T04:19:38.592012+00:00 | 2020-09-06T04:30:04.838119+00:00 | 155 | false | ```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n Arrays.sort(nums1);\n Arrays.sort(nums2);\n \n int ans = 0;\n int n1 = nums1.length, n2 = nums2.length;\n Map<Long, Integer> S1 = new HashMap<>();\n Map<Long, Integer> S2 = new HashMap<>();\n... | 1 | 0 | ['Sorting'] | 2 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy Python Solution with proper and easy explanation | easy-python-solution-with-proper-and-eas-wlev | Okay so I followed a really simple stragegy:\nI created 2 dictionaries and put up the squared values of all the numbers in both the nums array in seperate dicti | shubhitt | NORMAL | 2020-09-06T04:10:51.851297+00:00 | 2020-09-06T04:10:51.851332+00:00 | 59 | false | Okay so I followed a really simple stragegy:\nI created 2 dictionaries and put up the squared values of all the numbers in both the nums array in seperate dictionaries, with their count\nNow I traverse both the arrays seperately and find out pairs such that, if the pair product is in the opposite dictionary, i.e, if I ... | 1 | 1 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Solution Using HashMap | java-solution-using-hashmap-by-jjy_yang-lrtr | Attention: Need to change type to Long\n\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n int re=0;\n Map<Long,Integer> | jjy_yang | NORMAL | 2020-09-06T04:07:08.933471+00:00 | 2020-09-06T04:07:08.933515+00:00 | 65 | false | Attention: Need to change type to Long\n```\nclass Solution {\n public int numTriplets(int[] nums1, int[] nums2) {\n int re=0;\n Map<Long,Integer> map=new HashMap<>();\n for(int i:nums1){\n long l=(long)i*(long)i;\n int count=map.getOrDefault(l,0);\n map.put(l,co... | 1 | 1 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [C++] 30s. Faster than 100% | c-30s-faster-than-100-by-subbuffer-2mkh | Sort first, and then check whether the 3 numbers can exist accoriding to the condition. For this purpose, we can use a HashMap to expedite the search. Thus, in | SubBuffer | NORMAL | 2020-09-06T04:05:13.729730+00:00 | 2020-09-06T06:52:19.933729+00:00 | 168 | false | Sort first, and then check whether the 3 numbers can exist accoriding to the condition. For this purpose, we can use a HashMap to expedite the search. Thus, in the beginning, we insert the elements into the HashMap (in C++ ```unordered_map```) and then check for the obvious relationship of ```nums1[i]^2 = nums2[j] * nu... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python, Hashmap | python-hashmap-by-yseeker-6plc | \nclass Solution(object):\n def numTriplets(self, nums1, nums2):\n\n cnt = 0\n a = collections.defaultdict(int)\n b = collections.defaul | yseeker | NORMAL | 2020-09-06T04:03:58.069536+00:00 | 2020-09-06T04:03:58.069583+00:00 | 230 | false | ```\nclass Solution(object):\n def numTriplets(self, nums1, nums2):\n\n cnt = 0\n a = collections.defaultdict(int)\n b = collections.defaultdict(int)\n \n for num in nums1:\n a[num*num] += 1\n \n for num in nums2:\n b[num*num] += 1\n \... | 1 | 0 | ['Python'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | [Python3] frequency table | python3-frequency-table-by-ye15-lsvk | \n\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n cnt1, cnt2 = Counter(nums1), Counter(nums2)\n \n | ye15 | NORMAL | 2020-09-06T04:02:49.554518+00:00 | 2020-09-06T04:02:49.554579+00:00 | 205 | false | \n```\nclass Solution:\n def numTriplets(self, nums1: List[int], nums2: List[int]) -> int:\n cnt1, cnt2 = Counter(nums1), Counter(nums2)\n \n def fn(x, y, freq): \n """Return count of triplet of nums[i]**2 = nums[j]*nums[k]."""\n ans = 0\n for xx in x: \n ... | 1 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Straightforward O(N^2) Java Solution with 2 HashMaps | straightforward-on2-java-solution-with-2-h595 | \n public int numTriplets(int[] nums1, int[] nums2) {\n //hashMap: product -> freq\n HashMap<Long, Integer> map1 = new HashMap<>();\n Ha | billtang123 | NORMAL | 2020-09-06T04:02:11.789400+00:00 | 2020-09-06T04:02:11.789450+00:00 | 162 | false | ```\n public int numTriplets(int[] nums1, int[] nums2) {\n //hashMap: product -> freq\n HashMap<Long, Integer> map1 = new HashMap<>();\n HashMap<Long, Integer> map2 = new HashMap<>();\n int n1 = nums1.length;\n int n2 = nums2.length;\n for(int i=0;i<n1; i++){\n fo... | 1 | 0 | [] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Hard Way to solve, but constraints are small | hard-way-to-solve-but-constraints-are-sm-sdoc | IntuitionНужно посчитать тройки, где квадрат одного элемента из одного массива равен произведению двух элементов из другого. Это условие проверяется по всем воз | bezzerli | NORMAL | 2025-04-08T00:47:10.023237+00:00 | 2025-04-08T00:47:10.023237+00:00 | 2 | false | ### Intuition
Нужно посчитать тройки, где квадрат одного элемента из одного массива равен произведению двух элементов из другого. Это условие проверяется по всем возможным парам.
### Approach
1. Посчитаем частоты квадратов всех элементов `nums1` и `nums2` (храним в словарях).
2. Для всех пар в `nums2`, проверим, есть ... | 0 | 0 | ['Array', 'Hash Table', 'Math', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | easy solution | easy-solution-by-owenwu4-q6wd | Intuitionprecompute squared freuencies using a hashmapApproachComplexity
Time complexity:
Space complexity:
Code | owenwu4 | NORMAL | 2025-04-07T21:27:46.922644+00:00 | 2025-04-07T21:27:46.922644+00:00 | 2 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
precompute squared freuencies using a hashmap
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:
<!-- Add your spac... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python, Hashing, Brute Force | python-hashing-brute-force-by-mickeyhao-n5x6 | Intuition & ApproachPrecalculate the square of numbers in both lists, and store the frequency. Then iterate through both lists to see if the product of nums2[j] | MickeyHao | NORMAL | 2025-01-13T19:27:38.258977+00:00 | 2025-01-13T19:27:38.258977+00:00 | 12 | false | # Intuition & Approach
Precalculate the square of numbers in both lists, and store the frequency. Then iterate through both lists to see if the product of `nums2[j]*nums2[k]` is in the frequency table of `nums1`. Do the same to the other list.
# Complexity
- Time complexity: $O(N^2)$
- Space complexity: $O(N)$
# Co... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | 1577. Number of Ways Where Square of Number Is Equal to Product of Two Numbers | 1577-number-of-ways-where-square-of-numb-cai4 | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | G8xd0QPqTy | NORMAL | 2025-01-13T16:52:31.716673+00:00 | 2025-01-13T16:52:31.716673+00:00 | 12 | 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 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Simple | Intuitive | simple-intuitive-by-richardleee-147y | IntuitionApproachComplexity
Time complexity:
Space complexity:
Code | RichardLeee | NORMAL | 2025-01-12T16:10:50.356357+00:00 | 2025-01-12T16:10:50.356357+00:00 | 11 | 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 | ['Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ O(NM) Time, O(1) Auxiliary space, beat 95% runtime solution w/ Detailed explanation | c-onm-time-o1-auxiliary-space-beat-95-ru-hhz7 | Intuition\nJust considering a single type (where nums1 is the "target" and nums2 is the "combinations" vector). We may make a few observations:\n\nObservations: | nicklai | NORMAL | 2024-11-04T08:20:57.500917+00:00 | 2024-11-04T08:20:57.500949+00:00 | 16 | false | # Intuition\nJust considering a single type (where nums1 is the "target" and nums2 is the "combinations" vector). We may make a few observations:\n\nObservations:\n- The values of the numbers are **non-negative**, limited to the range 1~100,000\n - If we deal with **sorted** nums1 and nums2, we can we can do some le... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Very intuitive solution using dictionary | very-intuitive-solution-using-dictionary-xpd2 | 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 | belka | NORMAL | 2024-10-28T19:39:32.897017+00:00 | 2024-10-28T19:39:32.897041+00:00 | 10 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy simple code || Beats 90|| O(M*N)|| O(M+N) | easy-simple-code-beats-90-omn-omn-by-sum-s1t6 | 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 | sumanth2328 | NORMAL | 2024-10-15T04:16:15.661749+00:00 | 2024-10-15T04:16:15.661780+00:00 | 24 | 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 | ['Array', 'Hash Table', 'Math', 'Counting', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ Solution || Using Maps | c-solution-using-maps-by-vaibhav_arya007-8ba1 | Code\ncpp []\nclass Solution {\n // Function to count the number of valid triplets where square of a number from `A`\n // is equal to the product of two n | Vaibhav_Arya007 | NORMAL | 2024-10-03T22:04:54.421323+00:00 | 2024-10-03T22:04:54.421347+00:00 | 24 | false | # Code\n```cpp []\nclass Solution {\n // Function to count the number of valid triplets where square of a number from `A`\n // is equal to the product of two numbers from `B`\n int count(vector<int> &A, vector<int> &B) {\n int ans = 0;\n unordered_map<int, int> freqMap;\n\n // Build a freq... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ Solution || Brute Force | c-solution-brute-force-by-vaibhav_arya00-usht | Code\ncpp []\nclass Solution {\npublic:\n // Update the function to use long long for calculations\n int solve(vector<int>& nums1, vector<int>& nums2) {\n | Vaibhav_Arya007 | NORMAL | 2024-10-03T21:56:06.753612+00:00 | 2024-10-03T21:56:06.753655+00:00 | 4 | false | # Code\n```cpp []\nclass Solution {\npublic:\n // Update the function to use long long for calculations\n int solve(vector<int>& nums1, vector<int>& nums2) {\n int count = 0;\n\n for (int i = 0; i < nums1.size(); i++) {\n long long target = (long long)nums1[i] * nums1[i]; // Square of num... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Hashing || Super Simple || C++ | hashing-super-simple-c-by-lotus18-ppmx | Code\n\nclass Solution \n{\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) \n {\n map<long long,int> m1, m2;\n for(auto n | lotus18 | NORMAL | 2024-08-18T17:41:20.534244+00:00 | 2024-08-18T17:41:20.534296+00:00 | 20 | false | # Code\n```\nclass Solution \n{\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) \n {\n map<long long,int> m1, m2;\n for(auto n: nums1) m1[(long long)n*n]++;\n for(auto n: nums2) m2[(long long)n*n]++;\n int ans=0;\n for(int x=0; x<nums2.size(); x++)\n {\... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Simple and Easy JAVA Solution using HashMap | simple-and-easy-java-solution-using-hash-jwoj | 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 | Triyaambak | NORMAL | 2024-07-29T18:02:41.906735+00:00 | 2024-07-29T18:02:41.906788+00:00 | 26 | 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 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | ✅ C++ Solution in O(mn) complexity clearly explained with comments | c-solution-in-omn-complexity-clearly-exp-1303 | 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 | rohith_nair_2021 | NORMAL | 2024-07-15T19:12:19.482418+00:00 | 2024-07-15T19:12:19.482439+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: ```O(mn)```\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: ```O(n) or O(m) depending on which is larg... | 0 | 0 | ['Array', 'Hash Table', 'C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Clean Python Solution, TC : O(n*m) , SC : O(n+m) | clean-python-solution-tc-onm-sc-onm-by-a-qu8f | Approach\n Describe your approach to solving the problem. \n1. We need to take every squares of nums1 and find product pairs in nums2 and viceversa.\n\n2. So we | AGHIL_P | NORMAL | 2024-07-03T17:44:47.259616+00:00 | 2024-07-03T17:44:47.259659+00:00 | 48 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. We need to take every squares of nums1 and find product pairs in nums2 and viceversa.\n\n2. So we can take the count of squares in both array to minimise repititive work, like if there are same numbers mutiple times. And save the squares and their ... | 0 | 0 | ['Hash Table', 'Counting', 'Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | O(N2) complexity | on2-complexity-by-sirispandana77-hppb | Intuition\n Describe your first thoughts on how to solve this problem. \nIf the target value is a multiple of the element in the list, then the % will be 0. To | sirispandana77 | NORMAL | 2024-07-01T18:37:13.303930+00:00 | 2024-07-01T18:37:13.303956+00:00 | 10 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nIf the target value is a multiple of the element in the list, then the % will be 0. To avaoid mulitple calculation a dict can be used to store the values.\n# Approach\n<!-- Describe your approach to solving the problem. -->\nTo calculate ... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy Two Maps Approach | easy-two-maps-approach-by-aman_91-d6ny | 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 | aman_91 | NORMAL | 2024-05-20T10:03:19.686405+00:00 | 2024-05-20T10:03:19.686432+00:00 | 31 | 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 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Solution Number of Ways Where Square of Number Is Equal to Product of Two Numbers | solution-number-of-ways-where-square-of-5c5yu | 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 | 2024-05-18T09:42:03.244072+00:00 | 2024-05-18T09:42:03.244101+00:00 | 0 | 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 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Short & Simple. O(n^2 + m^2) | short-simple-on2-m2-by-3s_akb-dzzg | \npublic class Solution {\n\n int c(int[] a, int[] b)\n {\n int res = 0;\n var aa = a.Select(x => (long)x * (long)x).GroupBy(x => x).ToDicti | 3S_AKB | NORMAL | 2024-05-07T15:29:09.710580+00:00 | 2024-05-07T15:29:09.710600+00:00 | 3 | false | \npublic class Solution {\n\n int c(int[] a, int[] b)\n {\n int res = 0;\n var aa = a.Select(x => (long)x * (long)x).GroupBy(x => x).ToDictionary(x => x.Key, x => x.Count());\n for (int i = 0; i < b.Length; i++)\n for (int k = i + 1; k < b.Length; k++)\n if (aa.Conta... | 0 | 0 | ['C#'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java Solution with HashMap and binary search | java-solution-with-hashmap-and-binary-se-h2iu | Intuition\nCounting, HashMap and binary search\n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity:\n Add your ti | jhapranay | NORMAL | 2024-04-30T19:24:29.044549+00:00 | 2024-04-30T19:24:29.044579+00:00 | 20 | false | # Intuition\nCounting, HashMap and binary search\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. $$O(n)$$ -->\n\n# Code\n```\nclass Solution... | 0 | 0 | ['Hash Table', 'Binary Search', 'Counting', 'Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python 3: FT 100%, LMT 90%: TC O(N1 log(N1) + N2 log(N2) + S1 + S2): Count and Sort Uniques | python-3-ft-100-lmt-90-tc-on1-logn1-n2-l-oy0z | Intuition\n\nWelcome to my ~6th FT 100% algorithm!\n\nAnd asymptotically it could be even faster on large inputs, see below about iterating over divisors and th | biggestchungus | NORMAL | 2024-03-23T21:33:25.248135+00:00 | 2024-03-23T21:33:25.248151+00:00 | 30 | false | # Intuition\n\nWelcome to my ~6th FT 100% algorithm!\n\nAnd asymptotically it could be even faster on large inputs, see below about iterating over divisors and the excellent average case complexity.\n\nSuper brute force is to do a three-pointer algorithm, checking all triplets and counting up the number of matches. But... | 0 | 0 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python (Simple Hashmap) | python-simple-hashmap-by-rnotappl-zm56 | 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 | 2024-03-15T10:49:44.100186+00:00 | 2024-03-15T10:49:44.100209+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 | ['Python3'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | C++ || Similar to two sum problem with hash map | c-similar-to-two-sum-problem-with-hash-m-97fq | \n\n# Code\n\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n std::unordered_map<int, int> m1, m2;\n\n | Gismet | NORMAL | 2024-03-04T08:41:56.252004+00:00 | 2024-03-04T08:41:56.252037+00:00 | 28 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int numTriplets(vector<int>& nums1, vector<int>& nums2) {\n std::unordered_map<int, int> m1, m2;\n\n for(int i : nums1)\n m1[i]++;\n \n for(int i : nums2)\n m2[i]++;\n\n int ans = 0;\n for(const auto& [k1, v1... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Easy solution Using two Hashmaps. | easy-solution-using-two-hashmaps-by-abhi-8uex | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n1. Create two hashmaps | abhishek_biyani08 | NORMAL | 2024-02-26T14:59:40.620685+00:00 | 2024-02-26T14:59:40.620721+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. -->\n1. Create two hashmaps to store the products of each pair of numbers in each array.\n2. Add the products in the hashmaps respectively.\n3. Using for loop check if the ... | 0 | 0 | ['C++'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | JS || Solution by Bharadwaj | js-solution-by-bharadwaj-by-manu-bharadw-gynl | Code\n\nvar numTriplets = function (nums1, nums2) {\n return Triplets(nums1, nums2) + Triplets(nums2, nums1);\n\n function Triplets(nums1, nums2) {\n | Manu-Bharadwaj-BN | NORMAL | 2024-02-01T07:24:15.244179+00:00 | 2024-02-01T07:24:15.244210+00:00 | 16 | false | # Code\n```\nvar numTriplets = function (nums1, nums2) {\n return Triplets(nums1, nums2) + Triplets(nums2, nums1);\n\n function Triplets(nums1, nums2) {\n let res = 0;\n for (let i = 0; i < nums1.length; i++) {\n let target = nums1[i] * nums1[i], map = new Map();\n for (let j =... | 0 | 0 | ['JavaScript'] | 1 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Java simple hashmap solution | java-simple-hashmap-solution-by-dsa_geek-4alj | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \ncreated a check functio | dsa_geek | NORMAL | 2024-01-01T05:52:58.853158+00:00 | 2024-01-01T05:52:58.853206+00:00 | 34 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\ncreated a check function which checks the condition given in the question. and this check function is called twice as asked in question.\n\n# Code\n```\nclass Solution... | 0 | 0 | ['Java'] | 0 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | java simple HashMap | java-simple-hashmap-by-trivedi_cs1-pvio | 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 | trivedi_cs1 | NORMAL | 2023-12-29T09:35:20.014554+00:00 | 2023-12-29T09:35:20.014576+00:00 | 9 | 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 |
number-of-ways-where-square-of-number-is-equal-to-product-of-two-numbers | Python Simple Approach | python-simple-approach-by-raj_zinzuvadiy-64ax | 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 | Raj_Zinzuvadiya | NORMAL | 2023-12-27T19:23:28.978132+00:00 | 2023-12-27T19:23:28.978159+00:00 | 31 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: $$O(n^2)$$ \n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$ \n<!-- Add your space complexity ... | 0 | 0 | ['Python3'] | 0 |
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