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find-n-unique-integers-sum-up-to-zero | Easy and Simple Solution in Java, C++ and Python ||0ms Runtime in Java and C++|| | easy-and-simple-solution-in-java-c-and-p-c0jp | Java []\nclass Solution {\n public int[] sumZero(int n) {\n int ans[] = new int[n];\n for(int i=0;i<n;i++)\n {\n ans[i] = i*2 | _veer_singh04_ | NORMAL | 2023-07-05T17:44:59.266660+00:00 | 2023-07-05T17:57:20.961140+00:00 | 670 | false | ```Java []\nclass Solution {\n public int[] sumZero(int n) {\n int ans[] = new int[n];\n for(int i=0;i<n;i++)\n {\n ans[i] = i*2 -n+1;\n }\n return ans;\n }\n}\n```\n```python []\nclass Solution(object):\n def sumZero(self, n):\n return [i*2-n+1 for i in ran... | 4 | 0 | ['Array', 'Math', 'Python', 'C++', 'Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | Python Solution, simple and easy to follow | python-solution-simple-and-easy-to-follo-006f | \ndef uniqueSum(n):\n result = []\n\n for i in range(n):\n result.append(i*2 - n + 1)\n return result\n | vikingdev | NORMAL | 2022-09-13T15:18:32.341774+00:00 | 2022-09-13T15:18:32.349839+00:00 | 551 | false | ```\ndef uniqueSum(n):\n result = []\n\n for i in range(n):\n result.append(i*2 - n + 1)\n return result\n``` | 4 | 0 | ['Array', 'Python', 'Python3'] | 1 |
find-n-unique-integers-sum-up-to-zero | Beginner Friendly JAVA easy solution with comments || 0 ms 100% fast | beginner-friendly-java-easy-solution-wit-j2w2 | 100% Fastest JAVA SOLUTION\nREAD the comments along the steps...........feel free to comment any query\n\nDon\'T forget to UPVOTE....it means a lot and encourag | satish2203 | NORMAL | 2022-01-22T17:46:15.825811+00:00 | 2022-01-22T17:46:15.825839+00:00 | 222 | false | **100% Fastest JAVA SOLUTION**\n**READ the comments along the steps...........feel free to comment any query**\n\n**Don\'T forget to UPVOTE....it means a lot and encourage to post more**\n\n```\n public int[] sumZero(int n) {\n \n int arr[]=new int[n]; // create a array of given size\n int pos=1; ... | 4 | 0 | ['Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | Python 3 simple solution | python-3-simple-solution-by-derek-y-qqt6 | ```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n res = [0] if n % 2 else []\n for i in range(1, n // 2 + 1):\n res.a | derek-y | NORMAL | 2021-11-26T16:46:35.345531+00:00 | 2021-11-26T16:46:35.345562+00:00 | 420 | false | ```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n res = [0] if n % 2 else []\n for i in range(1, n // 2 + 1):\n res.append(i)\n res.append(-i)\n return res | 4 | 1 | ['Python', 'Python3'] | 0 |
find-n-unique-integers-sum-up-to-zero | Very easy JAVA Solution. Runtime: 0 ms, faster than 100.00% | very-easy-java-solution-runtime-0-ms-fas-4j46 | \nclass Solution\n{\n public int[] sumZero(int n)\n {\n int arr[] = new int[n];\n if(n%2==0)\n {\n for(int i=0, val=1; i<n | ghosharindam195 | NORMAL | 2021-02-17T06:26:31.515784+00:00 | 2021-02-17T06:26:31.515811+00:00 | 364 | false | ```\nclass Solution\n{\n public int[] sumZero(int n)\n {\n int arr[] = new int[n];\n if(n%2==0)\n {\n for(int i=0, val=1; i<n; i+=2,val++)\n {\n arr[i]=val;\n arr[i+1]=-val;\n }\n }\n else\n {\n arr... | 4 | 1 | ['Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | C++ Super Easy Solution, 0ms faster than 100% | c-super-easy-solution-0ms-faster-than-10-oa8r | \nclass Solution {\npublic:\n vector<int> sumZero(int n) {\n vector<int> res(n);\n int a = 1;\n for (int i = 0; i < n; i+=2) {\n | yehudisk | NORMAL | 2020-12-17T15:38:26.276634+00:00 | 2020-12-17T15:38:26.276675+00:00 | 267 | false | ```\nclass Solution {\npublic:\n vector<int> sumZero(int n) {\n vector<int> res(n);\n int a = 1;\n for (int i = 0; i < n; i+=2) {\n if (i == n-1) {\n res[i] = 0;\n return res;\n }\n \n res[i] = a;\n res[i+1]... | 4 | 1 | ['C'] | 0 |
find-n-unique-integers-sum-up-to-zero | Python Solution | python-solution-by-rachitsxn292-zac0 | \nclass Solution(object):\n def sumZero(self, n):\n return range(1-n, n, 2)\n | rachitsxn292 | NORMAL | 2020-04-01T21:24:21.272793+00:00 | 2020-04-01T21:24:21.272845+00:00 | 360 | false | ```\nclass Solution(object):\n def sumZero(self, n):\n return range(1-n, n, 2)\n``` | 4 | 0 | ['Python'] | 1 |
find-n-unique-integers-sum-up-to-zero | Easy Python Solution | easy-python-solution-by-ignasialemany-vltj | \timport random\n\tclass Solution:\n\t\tdef sumZero(self, n: int) -> List[int]:\n\t\t\tanswer = []\n\t\t\tanswer = random.sample(range(1,1000),n-1)\n\t\t\tanswe | ignasialemany | NORMAL | 2020-01-26T08:18:08.786758+00:00 | 2020-01-26T08:18:08.786795+00:00 | 355 | false | \timport random\n\tclass Solution:\n\t\tdef sumZero(self, n: int) -> List[int]:\n\t\t\tanswer = []\n\t\t\tanswer = random.sample(range(1,1000),n-1)\n\t\t\tanswer.append(-sum(answer))\n\t\t\treturn answer | 4 | 0 | [] | 2 |
find-n-unique-integers-sum-up-to-zero | [JAVA] simple solution | java-simple-solution-by-nandathantsin-ye0k | \nclass Solution {\n public int[] sumZero(int n) {\n int[] res = new int[n];\n for(int i=0;i<n/2;i++){\n res[i]=-1*(i+1);\n | nandathantsin | NORMAL | 2019-12-29T04:04:34.777917+00:00 | 2019-12-29T04:04:34.777954+00:00 | 806 | false | ```\nclass Solution {\n public int[] sumZero(int n) {\n int[] res = new int[n];\n for(int i=0;i<n/2;i++){\n res[i]=-1*(i+1);\n res[n-i-1]=(i+1);\n }\n return res;\n }\n}\n``` | 4 | 1 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Beats 100% || JAVA|| | beats-100-java-by-novasynapse17-3ifr | Intuition\nOnly thought was to add positive and negative numbers to sum up zero but wait there are two cases (odd and even)\n\n# Approach\nMade two seperate cas | theDummy | NORMAL | 2024-09-05T08:53:31.793147+00:00 | 2024-09-05T08:53:31.793180+00:00 | 396 | false | # Intuition\nOnly thought was to add positive and negative numbers to sum up zero but wait there are two cases (odd and even)\n\n# Approach\nMade two seperate cases for odd and even \n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space co... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Brute force approach || Easy Solution || 100% runtime | brute-force-approach-easy-solution-100-r-fsi1 | Intuition\nIf n is odd then only add zero at last and if n is even then don\'t add zero in the array and return it and first add 1 and -1 pair, 2 and -2 pair, 3 | Sushant_3999 | NORMAL | 2024-08-13T17:51:55.773662+00:00 | 2024-08-13T17:51:55.773679+00:00 | 394 | false | # Intuition\nIf n is odd then only add zero at last and if n is even then don\'t add zero in the array and return it and first add 1 and -1 pair, 2 and -2 pair, 3 and -3 pair and so on until n-1 times.\n\n# Approach\n1. Create a array sumArr with n size.\n2. create a for() loop with (i=0; i<n ; i++):\ni.check if(n%2==1... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Easy | O(n) | Java | | easy-on-java-by-suraj_ph116-amd4 | Intuition\nIf n is odd add 0 to the resultant array.\n\n# Approach\nMaintain the symmetry of integers i.e. resultant array must contain +ve as well as -ve value | suraj_ph116 | NORMAL | 2024-03-19T16:58:30.812416+00:00 | 2024-03-19T16:58:30.812452+00:00 | 261 | false | # Intuition\nIf n is odd add 0 to the resultant array.\n\n# Approach\nMaintain the symmetry of integers i.e. resultant array must contain +ve as well as -ve value of the number that is being inserted.\n\n# Complexity\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n# Code\n```\nclass Solution {\n public int... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | very simple solution 💯|| interesting | very-simple-solution-interesting-by-moha-m1kr | Intuition\n- simple solution\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n- unique elements \uD83D\uDE02\n Describe your approac | Mohanraj04 | NORMAL | 2023-12-06T10:06:30.666336+00:00 | 2023-12-06T10:06:30.666369+00:00 | 457 | false | # Intuition\n- simple solution\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n- unique elements \uD83D\uDE02\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<!... | 3 | 0 | ['Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | Beats 100.00%of users with C++ || Easy To Understand | beats-10000of-users-with-c-easy-to-under-vhtk | \n\n# Approach : \n\n1. Check if n is equal to 1.\n- If yes, return a vector containing [0].\n- This handles the special case where the sequence length is 1.\n | VenkateshMundra | NORMAL | 2023-10-21T13:30:03.690980+00:00 | 2023-10-21T13:31:42.338664+00:00 | 42 | false | \n\n# Approach : \n\n1. Check if **n** is equal to 1.\n- If yes, return a vector containing [0].\n- This handles the special case where the sequence length is 1.\n\n2. Check if **n** is even (e.g., 2, 4, 6, etc.).\n- If **n** is even, we want to generate a sequence with both negative and positive numbers.\n- Initializ... | 3 | 0 | ['C++'] | 0 |
find-n-unique-integers-sum-up-to-zero | Beats 100.00% runtime and 99.31% memory | beats-10000-runtime-and-9931-memory-by-c-7x2z | 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 | chaitanyanitin59 | NORMAL | 2023-09-15T06:04:27.524476+00:00 | 2023-09-15T06:04:27.524508+00:00 | 238 | 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)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(n)\n<!-- Add your space complexity here, e.g. $... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Python || C++ || Java || Easy and Neat Solution (0 ms runtime) | python-c-java-easy-and-neat-solution-0-m-48gi | Code\nPython []\nclass Solution(object):\n def sumZero(self, n):\n return [i*2-n+1 for i in range(0,n)]\n \n\nJava []\nclass Solution {\n pu | phalakbh | NORMAL | 2023-07-05T17:59:58.147685+00:00 | 2023-07-05T17:59:58.147713+00:00 | 635 | false | # Code\n```Python []\nclass Solution(object):\n def sumZero(self, n):\n return [i*2-n+1 for i in range(0,n)]\n \n```\n```Java []\nclass Solution {\n public int[] sumZero(int n) {\n int ans[] = new int[n];\n for(int i=0;i<n;i++)\n {\n ans[i] = i*2 -n+1;\n }\n ... | 3 | 0 | ['Array', 'Math', 'Python', 'C++', 'Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Simplest Java Solution - based on intuition rather then series operations - O(n). | simplest-java-solution-based-on-intuitio-hlu8 | Intuition\nIn order to make an array sum 0, the array will def have complementary pairs that add upto 0: \n- (0), (-1,1), (-2, 2)...(-n,n) etc.\n\nNow if the si | pateltrushit1710 | NORMAL | 2023-06-02T06:20:07.428372+00:00 | 2023-06-07T04:20:59.034715+00:00 | 404 | false | # Intuition\nIn order to make an array sum 0, the array will def have complementary pairs that add upto 0: \n- (0), (-1,1), (-2, 2)...(-n,n) etc.\n\nNow if the size of the array is n, then we start filling the ends of array with the negative pairs starting from n and then decrement by 1:\n- Even n: -n, -(n-1), ..., -1,... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Java | Easy Solution | 0 ms - 100% beats | java-easy-solution-0-ms-100-beats-by-ako-71i9 | Approach\n\n1. The sumZero method takes an integer n as input and returns an array of n unique integers.\n2. Initialize an integer array res of size n to store | akobirswe | NORMAL | 2023-05-30T04:27:16.160925+00:00 | 2023-05-30T04:27:16.160974+00:00 | 252 | false | # Approach\n\n1. The `sumZero` method takes an integer `n` as input and returns an array of `n` unique integers.\n2. Initialize an integer array `res` of size `n` to store the resulting array.\n3. Iterate from `i = 0` to `n/2 - 1`. This iteration covers half of the array because for each positive number generated, its ... | 3 | 0 | ['Array', 'Math', 'Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | PHP and Python solution | php-and-python-solution-by-__bakhtiyorof-tmkq | Intuition\narray[ i ] = i * 2 - n + 1\n\n- Space complexity:\n Add your space complexity here, e.g. O(n) \n\n# Code\nPHP\n\nclass Solution {\n\n /**\n * | __bakhtiyoroff__ | NORMAL | 2023-04-17T21:25:19.456761+00:00 | 2023-04-17T21:26:10.775345+00:00 | 753 | false | # Intuition\narray[ i ] = i * 2 - n + 1\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\nPHP\n```\nclass Solution {\n\n /**\n * @param Integer $n\n * @return Integer[]\n */\n function sumZero($n) {\n $array= array();\n for ($i = 0; $i < $n; $i++)... | 3 | 0 | ['PHP', 'Python3'] | 1 |
find-n-unique-integers-sum-up-to-zero | 5 Lines | Easy Java Solution | Beats 100% | 0 ms | 5-lines-easy-java-solution-beats-100-0-m-jg4b | Code\nAlso beats 86.68% in terms of memory at the time of submission.\n\nclass Solution {\n public int[] sumZero(int n) {\n int[] answer = new int[n]; | KevinKwan | NORMAL | 2023-01-11T01:52:50.692384+00:00 | 2023-01-18T04:33:15.439777+00:00 | 1,133 | false | # Code\nAlso beats 86.68% in terms of memory at the time of submission.\n```\nclass Solution {\n public int[] sumZero(int n) {\n int[] answer = new int[n];\n for (int i = 1; i<n; i+=2) {\n answer[i-1]=i;\n answer[i]=-i;\n }\n return answer;\n }\n}\n``` | 3 | 0 | ['Java'] | 2 |
find-n-unique-integers-sum-up-to-zero | Q1304 Accepted C++ ✅ 100% Fastest Sol | Simple & Easiest | q1304-accepted-c-100-fastest-sol-simple-anq7u | CRUX\n1) We need to return a vector whose sum is equal to 0.\n2) So there are 2 options either odd or even number.\n3) If odd, 5 then index would be 0,1,2,3,4 t | adityasrathore | NORMAL | 2023-01-06T20:01:53.172731+00:00 | 2023-01-06T20:01:53.172764+00:00 | 730 | false | CRUX\n1) We need to return a vector whose sum is equal to 0.\n2) So there are 2 options either odd or even number.\n3) If odd, 5 then index would be 0,1,2,3,4 thus middle would be zero and right side would be positive and left side would be negative of same number.\n4) Similary for even, 4 index would be same 0,1,2,3 a... | 3 | 0 | ['C'] | 0 |
find-n-unique-integers-sum-up-to-zero | c++ | easy understanding | 100% | 0ms | c-easy-understanding-100-0ms-by-nehagupt-zxnr | \nclass Solution {\npublic:\n vector<int> sumZero(int n) {\n vector<int>ans;\n for(int i=1;i<=n/2;i++)\n {\n ans.push_back(i) | NehaGupta_09 | NORMAL | 2022-09-30T05:55:01.128280+00:00 | 2022-09-30T05:55:01.128322+00:00 | 410 | false | ```\nclass Solution {\npublic:\n vector<int> sumZero(int n) {\n vector<int>ans;\n for(int i=1;i<=n/2;i++)\n {\n ans.push_back(i);\n ans.push_back(i*-1);\n }\n if(n%2!=0)\n {\n ans.push_back(0);\n }\n return ans;\n }\n};\n``` | 3 | 0 | ['C', 'C++'] | 2 |
find-n-unique-integers-sum-up-to-zero | Easiest solution in Java | easiest-solution-in-java-by-silvan8124-f3nz | One of the easiest solutions\n\nclass Solution {\n public int[] sumZero(int n) {\n int [] result = new int[n];\n int sum = 0;\n \n | silvan8124 | NORMAL | 2022-09-23T17:36:17.663876+00:00 | 2022-09-23T17:36:17.663904+00:00 | 180 | false | One of the easiest solutions\n\nclass Solution {\n public int[] sumZero(int n) {\n int [] result = new int[n];\n int sum = 0;\n \n for(int i = 1 ; i < n ; i++) {\n result[i] = i;\n sum += i;\n }\n \n result[0] = -sum;\n \n return re... | 3 | 0 | ['Array'] | 0 |
find-n-unique-integers-sum-up-to-zero | Python Easy Solution in 5 lines | python-easy-solution-in-5-lines-by-pulki-seug | \nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n a=[]\n if n%2!=0:\n a.append(0)\n for i in range(1,n,2):\n | pulkit_uppal | NORMAL | 2022-09-20T04:31:22.357026+00:00 | 2022-09-20T04:33:47.222916+00:00 | 939 | false | ```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n a=[]\n if n%2!=0:\n a.append(0)\n for i in range(1,n,2):\n a.append(i)\n a.append(i*(-1))\n return a\n \n \n \n ``` | 3 | 0 | ['Python'] | 1 |
find-n-unique-integers-sum-up-to-zero | JS | One-liner | faster than 81.31% | js-one-liner-faster-than-8131-by-diff64-12iz | \nvar sumZero = (n) => (arr = new Int32Array(n).map((el, idx) => idx), arr[0] = -n * (n-1) / 2, arr)\n | Diff64 | NORMAL | 2022-08-16T22:01:46.650767+00:00 | 2022-11-08T19:40:31.528826+00:00 | 374 | false | ```\nvar sumZero = (n) => (arr = new Int32Array(n).map((el, idx) => idx), arr[0] = -n * (n-1) / 2, arr)\n``` | 3 | 0 | ['JavaScript'] | 0 |
find-n-unique-integers-sum-up-to-zero | My Javascript Answer | my-javascript-answer-by-parkryan0128-50qz | easy javascript answer\n\n\nvar sumZero = function(n) {\n let arr = []\n let sum = 0\n for (let i=1; i < n; i++) {\n arr.push(i)\n sum = | parkryan0128 | NORMAL | 2022-06-12T00:35:54.409336+00:00 | 2022-06-12T00:35:54.409366+00:00 | 355 | false | easy javascript answer\n\n```\nvar sumZero = function(n) {\n let arr = []\n let sum = 0\n for (let i=1; i < n; i++) {\n arr.push(i)\n sum = sum + i\n }\n arr.push(-sum)\n return arr\n};\n``` | 3 | 0 | ['JavaScript'] | 1 |
find-n-unique-integers-sum-up-to-zero | Simple Java Approach: 0ms | simple-java-approach-0ms-by-bazinga18-1gpm | Pair up the indices, in the first loop populate 0th and 1st index, then 2nd and 3rd index and so on...\nThis way you can uniquely populate even number of indice | bazinga18_ | NORMAL | 2022-04-09T01:01:39.018228+00:00 | 2022-04-09T01:01:39.018265+00:00 | 115 | false | Pair up the indices, in the first loop populate 0th and 1st index, then 2nd and 3rd index and so on...\nThis way you can uniquely populate even number of indices.\nIf n is odd, then we would just populate the (n - 1) even bits using the same approach explained above and leave the last element as 0.\n\nFor example:\nIf ... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Python - Pretty Easy | python-pretty-easy-by-siddheshsagar-ul9u | \ndef sumZero(self, n: int) -> List[int]:\n ans = [] \n i = 1 \n if n % 2 == 0:\n j = 0\n else:\n j = 1\n | SiddheshSagar | NORMAL | 2022-03-24T03:10:32.652419+00:00 | 2022-07-31T19:14:15.255482+00:00 | 413 | false | ```\ndef sumZero(self, n: int) -> List[int]:\n ans = [] \n i = 1 \n if n % 2 == 0:\n j = 0\n else:\n j = 1\n ans.append(0)\n while j < n:\n ans.append(i)\n ans.append(-i)\n i += 1\n j += 2\n return ans... | 3 | 0 | ['Python', 'Python3'] | 0 |
find-n-unique-integers-sum-up-to-zero | [ Java ] Simple solution with explanation beats 100% O(n) | java-simple-solution-with-explanation-be-wgmz | Approach: \n\n1. Case 1 : n is even => return array in range [-n/2,n/2] but exclude 0.\n\tEg : n=8 return {-4,-3,-2,-1, 1, 2, 3, 4}\n2. Case 2 : n is odd => ret | KapProDes | NORMAL | 2021-12-10T10:35:51.623754+00:00 | 2021-12-10T10:35:51.623792+00:00 | 154 | false | __Approach__: \n\n1. __Case 1__ : n is even => return array in range [-n/2,n/2] but exclude 0.\n\tEg : n=8 return {-4,-3,-2,-1, 1, 2, 3, 4}\n2. __Case 2__ : n is odd => return array in range [-n/2,n/2] including 0.\n\tEg : n=7 return {-3,-2,-1, 0, 1, 2, 3}\n\n__Code__ :\n\n```\nclass Solution {\n public int[] sumZer... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Rust one liner | rust-one-liner-by-bigmih-1ijn | \nimpl Solution {\n pub fn sum_zero(n: i32) -> Vec<i32> {\n (1..n).chain(std::iter::once(-n * (n - 1) / 2)).collect()\n }\n}\n | BigMih | NORMAL | 2021-08-06T11:28:39.420264+00:00 | 2021-08-06T11:28:39.420306+00:00 | 71 | false | ```\nimpl Solution {\n pub fn sum_zero(n: i32) -> Vec<i32> {\n (1..n).chain(std::iter::once(-n * (n - 1) / 2)).collect()\n }\n}\n``` | 3 | 0 | ['Rust'] | 0 |
find-n-unique-integers-sum-up-to-zero | [Java] fill up to n-1 and last index put negative of the sum | java-fill-up-to-n-1-and-last-index-put-n-bn0x | \nclass Solution {\n public int[] sumZero(int n) {\n if (n == 1) return new int[]{0};\n \n int[] arr = new int[n];\n int sum = 0; | vinsinin | NORMAL | 2021-03-02T01:38:26.503257+00:00 | 2021-03-02T01:38:26.503299+00:00 | 144 | false | ```\nclass Solution {\n public int[] sumZero(int n) {\n if (n == 1) return new int[]{0};\n \n int[] arr = new int[n];\n int sum = 0;\n for (int i = 0;i<n-1;i++){\n arr[i] = i+1;\n sum += arr[i];\n }\n arr[n-1] = -sum;\n return arr;\n }\... | 3 | 1 | ['Java'] | 3 |
find-n-unique-integers-sum-up-to-zero | Java 0ms, O(n) with comments | java-0ms-on-with-comments-by-denz1994-9k5a | \nclass Solution {\n public int[] sumZero(int n) {\n int[] retArr = new int[n];\n int arrVal = 0;\n \n // Set the leftmost and ri | denz1994 | NORMAL | 2020-10-28T02:38:08.628565+00:00 | 2020-10-28T02:38:20.869441+00:00 | 202 | false | ```\nclass Solution {\n public int[] sumZero(int n) {\n int[] retArr = new int[n];\n int arrVal = 0;\n \n // Set the leftmost and rightmost elements and traverse the array towards the middle \n // Array should be set so... anyRightHalfElement = abs(anyLeftHalfElement)\n for(... | 3 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Python - Faster than 100% | python-faster-than-100-by-juanrodriguez-9r39 | Basically if n is even skip the zero.\n\n\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n mid = n // 2\n if (n & 1): return [num f | juanrodriguez | NORMAL | 2020-10-24T01:32:00.779438+00:00 | 2020-10-24T01:32:00.779470+00:00 | 503 | false | Basically if n is even skip the zero.\n\n```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n mid = n // 2\n if (n & 1): return [num for num in range(-mid, mid + 1)]\n else: return [num for num in range(-mid, mid + 1) if num != 0]\n``` | 3 | 0 | ['Python', 'Python3'] | 0 |
find-n-unique-integers-sum-up-to-zero | Simple Java Solution beats 100% | simple-java-solution-beats-100-by-idiotp-e9nj | Initializes an array matching this pattern\n\nn = 1, [0]\nn = 2, [1, -1]\nn = 3, [1, -1, 0]\nn = 4, [1, -1, 2, -2]\nn = 5, [1, -1, 2, -2, 0]\nn = 6, [1, -1, 2, | idiotplayer | NORMAL | 2020-07-29T06:44:38.371160+00:00 | 2020-07-29T06:44:38.371201+00:00 | 111 | false | Initializes an array matching this pattern\n```\nn = 1, [0]\nn = 2, [1, -1]\nn = 3, [1, -1, 0]\nn = 4, [1, -1, 2, -2]\nn = 5, [1, -1, 2, -2, 0]\nn = 6, [1, -1, 2, -2, 3, -3]\n```\n```\npublic int[] sumZero(int n) {\n int[] result = new int[n];\n \n for (int i=1, j=0; i<=n/2; i++) {\n result[j++] = i... | 3 | 0 | [] | 0 |
find-n-unique-integers-sum-up-to-zero | Python: solution | python-solution-by-onysuke-ydnf | Runtime: 32 ms, faster than 69.06% of Python3 online submissions for Find N Unique Integers Sum up to Zero.\nMemory Usage: 13.9 MB, less than 100.00% of Python3 | onysuke | NORMAL | 2020-05-06T00:18:40.597271+00:00 | 2020-05-06T00:18:40.597303+00:00 | 552 | false | Runtime: 32 ms, faster than 69.06% of Python3 online submissions for Find N Unique Integers Sum up to Zero.\nMemory Usage: 13.9 MB, less than 100.00% of Python3 online submissions for Find N Unique Integers Sum up to Zero.\n\n```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\n ans = []\n a... | 3 | 0 | [] | 0 |
find-n-unique-integers-sum-up-to-zero | javascript o(n) 100%/100% w/ very detailed explanation | javascript-on-100100-w-very-detailed-exp-8mgj | explanation:\nIn brief, we want to fill the result array with a sequence as follows: 0, -1, 1, -2, 2, -3, 3, ...\n\nThis will work for any odd number and is ver | carti | NORMAL | 2020-04-10T23:00:20.576120+00:00 | 2020-04-10T23:06:18.793701+00:00 | 248 | false | **explanation:**\nIn brief, we want to fill the result array with a sequence as follows: `0, -1, 1, -2, 2, -3, 3, ...`\n\nThis will work for any *odd* number and is very easy to generate, as seen in the code below. Example, for `n = 1`, result would be `[0]`. This sums to 0 obviously. For` n = 3`, result would be `[0,-... | 3 | 0 | ['JavaScript'] | 1 |
find-n-unique-integers-sum-up-to-zero | Extremely Simple Logic (beats 100%) | extremely-simple-logic-beats-100-by-nsai-t9d3 | You just need to add the index values in the result array and keep track of the sum of the values.\nInsert the last value of the result array as the negative of | nsai | NORMAL | 2020-03-04T15:18:06.624997+00:00 | 2020-03-11T09:12:44.258330+00:00 | 259 | false | You just need to add the index values in the result array and keep track of the sum of the values.\nInsert the last value of the result array as the negative of the sum so far. This will give unique values with sum of the values equaling zero.\n```\nclass Solution {\n public int[] sumZero(int n) {\n int res[]... | 3 | 0 | ['Java'] | 2 |
find-n-unique-integers-sum-up-to-zero | One Line Solution | one-line-solution-by-fallenranger-20z4 | Logic :\nIf n is odd:\neg: n = 5 \n1, 2, 0, -2, -1\n\nIf n is even:\neg: n = 4\n1, 2, -2, -1\n\nSo if n is odd, array will have 0 in middle and half mirror arra | fallenranger | NORMAL | 2019-12-29T05:31:45.578729+00:00 | 2019-12-29T05:35:27.981151+00:00 | 298 | false | Logic :\nIf n is odd:\neg: n = 5 \n1, 2, 0, -2, -1\n\nIf n is even:\neg: n = 4\n1, 2, -2, -1\n\nSo if n is odd, array will have 0 in middle and half mirror arrays around 0.\nIf n is even, array will just have two mirror half arrays.\n``` list(range(1,n//2+1))``` will create mirror & negative array of ```list(range(-(n/... | 3 | 1 | ['Python', 'Python3'] | 1 |
find-n-unique-integers-sum-up-to-zero | [Java/Python/C++] Simple solution to create symmetric array with sum 0 - Explained | javapythonc-simple-solution-to-create-sy-1k95 | Explaination\n\n - Insert 0 in the middle if odd length is required\n - Keep adding +ve and -ve nos on the right and left of same magnitude to maintain the sum | sankalpdayal5 | NORMAL | 2019-12-29T04:07:10.083289+00:00 | 2019-12-29T13:05:58.355264+00:00 | 451 | false | **Explaination**\n\n - Insert `0` in the middle if `odd length` is required\n - Keep adding `+ve` and `-ve` nos on the `right` and `left` of *same magnitude* to maintain the sum as `0`\n\n\n\n**Java -**\n```\nclass Solution {\n public int[] sumZero(int n) {\n int[] res = new int[n];\n \n // If o... | 3 | 2 | [] | 1 |
find-n-unique-integers-sum-up-to-zero | Python 3 O(N/2) with explanation | python-3-on2-with-explanation-by-strawbe-6gnq | The idea is this. If you have an even number of elements to append, you can just keep adding any arbitrary element and its inverse (e.g., 5 and -5) until you ha | strawberrykiwi | NORMAL | 2019-12-29T04:02:29.874022+00:00 | 2019-12-29T04:02:29.874063+00:00 | 532 | false | The idea is this. If you have an even number of elements to append, you can just keep adding any arbitrary element and its inverse (e.g., 5 and -5) until you have appended enough. If you have an odd, just append 0, then do what you\'d do for an even.\n\n```\nclass Solution:\n def sumZero(self, n: int) -> List[int]:\... | 3 | 0 | [] | 0 |
find-n-unique-integers-sum-up-to-zero | Generate an Array with Sum Zero || Beats 100% | generate-an-array-with-sum-zero-beats-10-59fq | IntuitionTo create an array of size n where the sum of all elements is zero, we can leverage the property of symmetric pairs. If n is even, we can use numbers l | lokeshthakur8954 | NORMAL | 2025-03-23T07:50:26.377582+00:00 | 2025-03-23T07:50:26.377582+00:00 | 70 | false | # Intuition
To create an array of size n where the sum of all elements is zero, we can leverage the property of symmetric pairs. If n is even, we can use numbers like (-x, x). If n is odd, we include an extra 0 in the middle.
# Approach
Initialize an array of size n.
If n is 1, return [0] directly.
Fill the first ha... | 2 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | 0ms solution beats 100 percent easy to understand just try to dry run it then you will find it easy. | 0ms-solution-beats-100-percent-easy-to-u-vbd3 | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
0(n)
Code | anshgupta1234 | NORMAL | 2025-02-05T15:14:38.951250+00:00 | 2025-02-05T15:14:38.951250+00:00 | 344 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(n)
- Space complexity:
0(n)
# Code
```java []
class Solution {
public int[] sumZero(int n) {
int[] arr=new int[n];
... | 2 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Видео-анализ за 5 минут || Video analysis in 5 minutes | video-analiz-za-5-minut-video-analysis-i-d2dd | Разбор задачи на ютуб
Спасибо за подписки!!
ApproachЕсли n не равны 1 или 2 то создаем переменную, которая будет хранить сумму всех добавленных чисел
int sum=0; | Alexandre_gtg_133 | NORMAL | 2025-01-31T19:32:35.520167+00:00 | 2025-01-31T19:32:35.520167+00:00 | 63 | false | # Разбор задачи на ютуб
https://youtu.be/_9qiFSiSvWM
Спасибо за подписки!!
# Approach
Если **n** не равны **1** или **2** то создаем переменную, которая будет хранить сумму всех добавленных чисел
**... | 2 | 0 | ['C++'] | 1 |
find-n-unique-integers-sum-up-to-zero | Easy CPP Solution | Beats 100% of the solutions | easy-cpp-solution-beats-100-of-the-solut-d0fv | IntuitionApproachComplexity
Time complexity:
O(n)
Space complexity:
O(n)
Code | VaibhaviShah25 | NORMAL | 2025-01-16T18:52:18.692267+00:00 | 2025-01-16T18:52:18.692267+00:00 | 197 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
# Approach
<!-- Describe your approach to solving the problem. -->
# Complexity
- Time complexity:
O(n)
- Space complexity:
O(n)
# Code
```cpp []
class Solution {
public:
vector<int> sumZero(int n) {
int sum = 0;
if(... | 2 | 0 | ['Array', 'Math', 'C++'] | 0 |
find-n-unique-integers-sum-up-to-zero | Zero-Sum Game: Crafting a Balanced Array Using Python | zero-sum-game-crafting-a-balanced-array-4pce4 | Intuition\nThe goal is to create an array of integers where the sum of the elements is zero. A simple way to achieve this is to generate a sequence of numbers a | Krishnaa2004 | NORMAL | 2024-09-06T13:19:34.827341+00:00 | 2024-09-06T13:19:34.827366+00:00 | 213 | false | # Intuition\nThe goal is to create an array of integers where the sum of the elements is zero. A simple way to achieve this is to generate a sequence of numbers and adjust the last element to balance the sum to zero.\n\n# Approach\n1. Generate a sequence of numbers from 1 to \\( n-1 \\).\n2. The sum of these numbers ca... | 2 | 0 | ['Python3'] | 0 |
find-n-unique-integers-sum-up-to-zero | Beginner Level Approach Using One Loop | beginner-level-approach-using-one-loop-b-faav | \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 comp | Akhil1489 | NORMAL | 2024-07-11T07:05:07.635867+00:00 | 2024-07-11T07:05:07.635883+00:00 | 139 | false | \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```\n/**\n * @param {number} n\n * @return {number[]}\n */\nvar sum... | 2 | 0 | ['JavaScript'] | 1 |
find-n-unique-integers-sum-up-to-zero | Python3 || Easy and Optimized Approach... ✌ | python3-easy-and-optimized-approach-by-h-w1kd | Explanation:\n1. Initialization: Initialize an empty list res to store the result.\n2. Generate Pairs: Loop from 1 to n//2 (integer division) and add pairs of i | hanishb81 | NORMAL | 2024-05-15T14:36:58.944871+00:00 | 2024-05-15T14:36:58.944902+00:00 | 281 | false | ### Explanation:\n1. **Initialization**: Initialize an empty list `res` to store the result.\n2. **Generate Pairs**: Loop from `1` to `n//2` (integer division) and add pairs of integers `(i, -i)` to the list. This ensures that each pair sums to zero.\n3. **Handle Odd n**: If `n` is odd, append `0` to the list. This is ... | 2 | 0 | ['Python3'] | 0 |
find-n-unique-integers-sum-up-to-zero | Code by a Newbie(beats 100%) | code-by-a-newbiebeats-100-by-harshith_gr-q6ze | \n\n# Code\n\nclass Solution {\n public int[] sumZero(int n) {\n int a[]=new int[n];\n int c=1;\n int sum=0;\n for(int i=0;i<n-1; | harshith_griet | NORMAL | 2024-04-27T11:04:44.310411+00:00 | 2024-04-27T11:04:44.310435+00:00 | 211 | false | \n\n# Code\n```\nclass Solution {\n public int[] sumZero(int n) {\n int a[]=new int[n];\n int c=1;\n int sum=0;\n for(int i=0;i<n-1;i++)\n {\n sum+=c;\n a[i]=c;\n c++;\n }\n a[n-1]=-1*sum;\n return a; \n }\n}\n``` | 2 | 0 | ['Java'] | 3 |
find-n-unique-integers-sum-up-to-zero | Easy Soln || C++ || Beat 100%🔥🔥 | easy-soln-c-beat-100-by-vishalkumar00-v3t1 | 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 | vishalkumar00 | NORMAL | 2024-03-20T07:57:12.837065+00:00 | 2024-03-20T07:57:12.837103+00:00 | 346 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
find-n-unique-integers-sum-up-to-zero | Easy java sol (100% beat): Dont copy - paste please : first understand then write code | easy-java-sol-100-beat-dont-copy-paste-p-4cl1 | \n\n# Code\n\nclass Solution {\n public int[] sumZero(int n) {\n int[] arr = new int[n];\n \n if(n%2!=0){\n arr[0]=0;\n } | rushikesh3010 | NORMAL | 2024-01-11T16:07:28.935374+00:00 | 2024-01-11T16:07:28.935411+00:00 | 73 | false | \n\n# Code\n```\nclass Solution {\n public int[] sumZero(int n) {\n int[] arr = new int[n];\n \n if(n%2!=0){\n arr[0]=0;\n }\n for(int i=0;i<n-1;i+=2){\n arr[i]=1+i;\n arr[i+1]=-(i+1);\n }\n\n return arr;\n \n }\n}\n```\n\n![upv... | 2 | 0 | ['Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | veryyyyyyy easyyyyyyyy lollllllllllllllllll | veryyyyyyy-easyyyyyyyy-lolllllllllllllll-pkmy | 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 | ramuyadav90597 | NORMAL | 2023-12-17T14:03:56.218942+00:00 | 2023-12-17T14:03:56.218970+00:00 | 251 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | JAVA SOLUTION : simple and easy | java-solution-simple-and-easy-by-jyothii-4fb9 | 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 | jyothii_20 | NORMAL | 2023-10-05T17:17:32.486571+00:00 | 2023-10-05T17:17:32.486605+00:00 | 124 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:0 ms\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'] | 1 |
find-n-unique-integers-sum-up-to-zero | JS Very Easy Solution||Easy to Understand | js-very-easy-solutioneasy-to-understand-m7v52 | javascript []\n\n/**\n * @param {number} n\n * @return {number[]}\n */\nvar sumZero = function(n) {\n let mid = Math.floor(n/2);\n let arr=[]\n for(let | unaisek | NORMAL | 2023-09-18T05:18:35.076118+00:00 | 2023-09-18T05:18:35.076138+00:00 | 82 | false | ```javascript []\n\n/**\n * @param {number} n\n * @return {number[]}\n */\nvar sumZero = function(n) {\n let mid = Math.floor(n/2);\n let arr=[]\n for(let i =1;i<=mid;i++){\n arr.push(i,-i);\n }\n if(n%2 == 1){\n arr.push(0)\n }\n return arr\n};\n``` | 2 | 0 | ['JavaScript'] | 1 |
find-n-unique-integers-sum-up-to-zero | Simple & Easy Code || JAVA || Beats 100% || Runtime 0ms | simple-easy-code-java-beats-100-runtime-24k4f | Intuition\nWe have to return an array in output whose elements make the sum of 0. \n Describe your first thoughts on how to solve this problem. \n\n# Approach\n | rajswapnil50 | NORMAL | 2023-09-06T18:50:23.762304+00:00 | 2023-09-06T18:50:23.762327+00:00 | 30 | false | # Intuition\nWe have to return an array in output whose elements make the sum of 0. \n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nFirstly take an array where you can store the numbers. Now, as we have to output an array in which on adding up all the elements in array give the `sum... | 2 | 0 | ['Array', 'Math', 'Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | Two pointers Approach | two-pointers-approach-by-tausif_02-lrkx | 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 | tausif_02 | NORMAL | 2023-07-30T19:36:23.305504+00:00 | 2023-07-30T19:36:23.305525+00:00 | 184 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | java two pointer approach | java-two-pointer-approach-by-rsnishank-yt66 | Intuition \n Describe your first thoughts on\nbasic two pointer approach how to solve this problem. \n\n# Approach\n Describe your approach to solving the probl | rsnishank | NORMAL | 2023-07-29T08:29:34.312494+00:00 | 2023-07-29T08:29:34.312525+00:00 | 247 | false | # Intuition \n<!-- Describe your first thoughts on\nbasic two pointer approach how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: \n<!-- Add your space com... | 2 | 0 | ['Java'] | 1 |
find-n-unique-integers-sum-up-to-zero | Find N Unique Integers Sum up to Zero 👨🏻💻👨🏻💻 || Java solution code🔥🔥🔥... | find-n-unique-integers-sum-up-to-zero-ja-epds | Code\n\nclass Solution {\n public int[] sumZero(int n) {\n int[] arr = new int[n];\n int sum = 0;\n \n for(int i=0; i<n; i++){\n | Jayakumar__S | NORMAL | 2023-06-30T13:28:28.785913+00:00 | 2023-06-30T13:28:28.785935+00:00 | 698 | false | # Code\n```\nclass Solution {\n public int[] sumZero(int n) {\n int[] arr = new int[n];\n int sum = 0;\n \n for(int i=0; i<n; i++){\n if(i == n-1){\n arr[i] = sum * (-1);\n }\n else{\n arr[i] = i+1;\n sum+= i+1;... | 2 | 0 | ['Java'] | 0 |
find-n-unique-integers-sum-up-to-zero | Beginners friendly | beginners-friendly-by-bhaskarkumar07-iz0t | \n\n# Approach\n Describe your approach to solving the problem. \n there are 2 cases when n is odd \n and when n is even \n in odd one element must be | bhaskarkumar07 | NORMAL | 2023-05-21T12:49:20.551109+00:00 | 2023-05-21T12:49:20.551156+00:00 | 908 | false | \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n there are 2 cases when n is odd \n and when n is even \n in odd one element must be 0;\n and in even there should be no zero. it will work as simple \n just observe the example testcase\n\n# Complexity\n- Time complexity: O(N)\n<!-... | 2 | 0 | ['Java'] | 0 |
maximum-size-of-a-set-after-removals | Easy to understand | C++ | Sets | easy-to-understand-c-sets-by-amaan2510-8u1d | Approach\n Describe your approach to solving the problem. \n1. Calculate the number of unique elements in nums1 and nums2 by storing their elements in sets s1 a | amaan2510 | NORMAL | 2024-01-07T04:15:58.288973+00:00 | 2024-01-07T04:15:58.288992+00:00 | 2,121 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\n1. Calculate the number of unique elements in nums1 and nums2 by storing their elements in sets s1 and s2.\n2. Also calculate total unique elements in both arrays in set s.\n3. The answer is minimum of:\n a. size of s\n b. sum of number ... | 25 | 2 | ['C++'] | 2 |
maximum-size-of-a-set-after-removals | Easy Video Explanation with Proofs 🔥 || Set Intersection 🔥 | easy-video-explanation-with-proofs-set-i-pvt1 | Intuition\n Describe your first thoughts on how to solve this problem. \nTry to use set and basic of math is required. Try to dry run few scenarios for better u | ayushnemmaniwar12 | NORMAL | 2024-01-07T06:05:48.774490+00:00 | 2024-01-07T06:05:48.774519+00:00 | 2,107 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nTry to use set and basic of math is required. Try to dry run few scenarios for better understanding\n\n# ***Easy Video Explanation***\n\nhttps://youtu.be/crNIGc4zb8s\n\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -... | 22 | 0 | ['Math', 'Ordered Set', 'C++', 'Java', 'Python3'] | 2 |
maximum-size-of-a-set-after-removals | Java HashSet O(N) | java-hashset-on-by-hobiter-veea | See comments for details\n\n- Time complexity:\n Add your time complexity here, e.g. O(n) \nO(N)\n- Space complexity:\n Add your space complexity here, e.g. O(n | hobiter | NORMAL | 2024-01-07T08:31:26.194391+00:00 | 2024-01-07T08:33:12.519197+00:00 | 708 | false | See comments for details\n\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\nO(N)\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\nO(N)\n# Code\n```\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n Set<Integer> set1 = toSet(nums... | 12 | 0 | ['Java'] | 3 |
maximum-size-of-a-set-after-removals | Simple set operations, symmetric difference | simple-set-operations-symmetric-differen-u57d | \n\n# Code\npython\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n s1 = set(nums1)\n s2 = set(nums2)\ | cpacc1 | NORMAL | 2024-01-07T04:05:27.389163+00:00 | 2024-01-07T04:06:30.536247+00:00 | 1,295 | false | \n\n# Code\n```python\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n s1 = set(nums1)\n s2 = set(nums2)\n n = len(nums1)\n inter = s1.intersection(s2)\n ex1 = min(len(s1) - len(inter) , n//2)\n ex2 = min(len(s2) - len(inter) , n//2)\... | 11 | 1 | ['Python3'] | 7 |
maximum-size-of-a-set-after-removals | Python: 5 lines Venn diagram for two sets. Few set operations | python-5-lines-venn-diagram-for-two-sets-sqcf | Intuition\nIt is clear that duplicate numbers in nums1 and nums2 do not matter. So you can conver them into two sets.\n\nNow some values can overlap so you have | salvadordali | NORMAL | 2024-01-07T04:29:56.596606+00:00 | 2024-01-07T04:29:56.596626+00:00 | 524 | false | # Intuition\nIt is clear that duplicate numbers in nums1 and nums2 do not matter. So you can conver them into two sets.\n\nNow some values can overlap so you have the following Venn diagram.\n\n![clip_image007\\[1\\].gif](https://assets.leetcode.com/users/images/dafda736-394e-4135-9049-82583da52613_1704601596.873948.gi... | 10 | 0 | ['Python3'] | 5 |
maximum-size-of-a-set-after-removals | Video Explanation | video-explanation-by-codingmohan-q9wa | Explanation\n\nClick here for the video\n\n# Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n int n | codingmohan | NORMAL | 2024-01-07T04:07:42.084134+00:00 | 2024-01-07T04:07:42.084153+00:00 | 505 | false | # Explanation\n\n[Click here for the video](https://youtu.be/63dNVenHJ_c)\n\n# Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n int n = nums1.size();\n unordered_map<int, int> m1, m2;\n \n for (auto i : nums1) m1[i] ++;\n for (au... | 7 | 0 | ['C++'] | 2 |
maximum-size-of-a-set-after-removals | C++ | Easy beginner friendly sol | Deatailed explanation with comment | Set | c-easy-beginner-friendly-sol-deatailed-e-nznn | # Intuition \n\n\n\n\n\n# Complexity\n- Time complexity: O(n) [Insert operation in unordered set takes O(1) time on avg.]\n\n\n- Space complexity: O(n) [To s | abhik2003 | NORMAL | 2024-01-07T04:04:20.038713+00:00 | 2024-01-07T04:56:19.946476+00:00 | 514 | 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)$$ [Insert operation in unordered set takes $$O(1)$$ time on avg.]\n<!-- Add your time complexity here, e.g... | 6 | 0 | ['C++'] | 1 |
maximum-size-of-a-set-after-removals | Set || Easy | set-easy-by-aayush0606-sb9c | Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& A, vector<int>& B) {\n int n=A.size();\n unordered_set<int>s1(A.begin(),A.e | Aayush0606 | NORMAL | 2024-01-07T04:00:52.982094+00:00 | 2024-01-07T04:37:15.936213+00:00 | 1,037 | false | # Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& A, vector<int>& B) {\n int n=A.size();\n unordered_set<int>s1(A.begin(),A.end());\n unordered_set<int>s2(B.begin(),B.end());\n int i=0,j=0;\n unordered_set<int>st;\n // Add all unique elements first\n\n... | 6 | 0 | ['Math', 'Greedy', 'Simulation', 'Ordered Set', 'C++'] | 1 |
maximum-size-of-a-set-after-removals | Java solution using HashSet | java-solution-using-hashset-by-hrishi_ch-yetx | Code\n\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n Set<Integer> s1 = new HashSet(); \n Set<Integer> s2 = new Ha | hrishi_chavan | NORMAL | 2024-01-07T04:27:51.165884+00:00 | 2024-01-08T04:22:49.392416+00:00 | 326 | false | # Code\n```\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n Set<Integer> s1 = new HashSet(); \n Set<Integer> s2 = new HashSet();\n Set<Integer> s3 = new HashSet(); \n \n for (int num: nums1){\n s1.add(num);\n s3.add(num);\n }... | 5 | 0 | ['Java'] | 4 |
maximum-size-of-a-set-after-removals | a few solutions | a-few-solutions-by-claytonjwong-xvwb | Synopsis:\n\nIt is more simple to include values than exclude values!\n\nGame Plan: \n1. Let N be the cardinality of the input array A and let threshold T be N | claytonjwong | NORMAL | 2024-01-16T16:37:00.271717+00:00 | 2024-01-17T14:09:49.710054+00:00 | 81 | false | **Synopsis:**\n\nIt is more simple to *include* values than *exclude* values!\n\n**Game Plan:** \n1. Let `N` be the cardinality of the input array `A` and let threshold `T` be `N / 2`\n2. Let `A` and `B` be sets from input arrays `A` and `B` correspondingly\n3. Further reduce `A` and `B` as **mutually exclusive** sets,... | 4 | 0 | [] | 2 |
maximum-size-of-a-set-after-removals | Easy C++ Solution ✅✅ | easy-c-solution-by-abhi242-smd3 | Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_set<int> s;\n int n1=nums1.size(); | Abhi242 | NORMAL | 2024-01-07T05:17:31.750917+00:00 | 2024-01-07T05:17:31.750945+00:00 | 309 | false | # Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_set<int> s;\n int n1=nums1.size();\n int n2=nums2.size();\n for(int i=0;i<n1;i++){\n s.insert(nums1[i]);\n }\n int use1=s.size();\n s.clear();\... | 4 | 0 | ['C++'] | 0 |
maximum-size-of-a-set-after-removals | SET -Union-Intersection | set-union-intersection-by-satyam_9911-0lvg | Intuition\n ans will not exceed n/2 (unique elements from nums1) + n/2(unique elements from nums2) = n. \n If there are more than n/2 unique elements from nums | satyam_9911 | NORMAL | 2024-01-07T04:41:58.286142+00:00 | 2024-01-07T14:13:21.088065+00:00 | 156 | false | # Intuition\n* ans will not exceed n/2 (unique elements from nums1) + n/2(unique elements from nums2) = `n`. \n* If there are more than n/2 unique elements from nums1 , then we can easily delete `n/2` elements and make the uniqueness factor from nums1 to `n/2`. \n* Similarly for `nums2`. \n\n* `ans = min(n , min(n/2 ,... | 4 | 0 | ['Hash Table', 'Math', 'Union Find', 'Ordered Set', 'C++'] | 1 |
maximum-size-of-a-set-after-removals | CONDITION BASED || C++ | condition-based-c-by-this_is_4503-hpa3 | Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_map<int,int> mp1,mp2;\n unordered_ | SatyamJha03 | NORMAL | 2024-01-07T04:16:05.761503+00:00 | 2024-01-07T04:16:05.761525+00:00 | 1,631 | false | # Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_map<int,int> mp1,mp2;\n unordered_set<int> s;\n for(auto &i: nums1){\n mp1[i]++;\n }\n for(auto &i: nums2){\n mp2[i]++;\n }\n int a ... | 4 | 0 | ['C++'] | 1 |
maximum-size-of-a-set-after-removals | ✅✅ Explanation with Proofs 💯 || C++ || ✅Clean Code || | explanation-with-proofs-c-clean-code-by-dkegs | Intuition:\nThe problem seems to involve finding the maximum possible size of a set that can be formed using elements from two given arrays (nums1 and nums2). T | arslanarsal | NORMAL | 2024-01-20T14:26:58.680715+00:00 | 2024-01-20T14:26:58.680753+00:00 | 130 | false | # Intuition:\nThe problem seems to involve finding the maximum possible size of a set that can be formed using elements from two given arrays (`nums1` and `nums2`). The intuition might involve using sets to represent unique elements in each array and finding the common elements between them.\n\n# Approach:\n1. Create t... | 3 | 0 | ['Array', 'Hash Table', 'Greedy', 'C++'] | 2 |
maximum-size-of-a-set-after-removals | [Python3] sets | python3-sets-by-awice-c7rg | Let N be the number of items we have to pick from each array.\n\nUse sets. Among the elements that only occur in $A$ (ie. $A-B$), we pick them first, to a maxi | awice | NORMAL | 2024-01-07T05:41:51.694999+00:00 | 2024-01-07T05:41:51.695023+00:00 | 139 | false | Let N be the number of items we have to pick from each array.\n\nUse sets. Among the elements that only occur in $A$ (ie. $A-B$), we pick them first, to a maximum of $N$ elements. Similarly for $B$.\n\nNow you have $2 N - \\text{onlyA} - \\text{onlyB}$ picks remaining, and you can pick from $A \\cap B$.\n\n# Code\n``... | 3 | 0 | ['Python3'] | 0 |
maximum-size-of-a-set-after-removals | Java solution | java-solution-by-charlie-tej-123-q58a | \nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n int i,j,n=nums1.length;\n Set<Integer> set1=new HashSet<>();\n | charlie-tej-123 | NORMAL | 2024-01-07T04:33:18.294935+00:00 | 2024-01-07T04:33:18.294958+00:00 | 604 | false | ```\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n int i,j,n=nums1.length;\n Set<Integer> set1=new HashSet<>();\n Set<Integer> set2=new HashSet<>();\n Set<Integer> set3=new HashSet<>();\n for(int x:nums1)\n {\n set1.add(x);\n ... | 3 | 0 | ['Java'] | 1 |
maximum-size-of-a-set-after-removals | Simple HashMaps and Set Solution | simple-hashmaps-and-set-solution-by-gupt-r0u1 | \nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n HashMap<Integer,Integer> map1=new HashMap<>();\n int n=nums1.lengt | guptasheenam1510 | NORMAL | 2024-01-07T04:03:23.172343+00:00 | 2024-01-09T14:42:29.666279+00:00 | 483 | false | ```\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n HashMap<Integer,Integer> map1=new HashMap<>();\n int n=nums1.length;\n HashMap<Integer,Integer> map2=new HashMap<>();\n Set set=new HashSet<>();\n int ans=0;\n for(int i=0;i<nums1.length;i++)\n ... | 3 | 0 | ['Java'] | 3 |
maximum-size-of-a-set-after-removals | Python Simple Solution (sets) (accepted) | python-simple-solution-sets-accepted-by-akxn8 | Code\n\nimport collections\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n max_len = len(nums1) // 2\n | nithil | NORMAL | 2024-01-07T04:03:02.757547+00:00 | 2024-01-07T04:03:02.757577+00:00 | 446 | false | # Code\n```\nimport collections\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n max_len = len(nums1) // 2\n nums1_set = set(nums1)\n nums2_set = set(nums2)\n \n if len(nums1_set) <= max_len and len(nums2_set) <= max_len:\n return... | 3 | 0 | ['Python3'] | 1 |
maximum-size-of-a-set-after-removals | ✅ 97% beats || Python3 || Set | 97-beats-python3-set-by-lutfullo_m-p9wr | \n\n\n\n# Intuition\nTo solve this problem, we can use a greedy approach. We aim to maximize the size of the set s after removing elements from nums1 and nums2. | lutfullo_m | NORMAL | 2024-05-26T12:40:13.802447+00:00 | 2024-05-26T12:40:13.802468+00:00 | 90 | false | \n\n\n\n# Intuition\nTo solve this problem, we can use a greedy approach. We aim to maximize the size of the set `s` after removing elements from `nums1` and `nums2`. We can achieve this by removing element... | 2 | 0 | ['Python3'] | 0 |
maximum-size-of-a-set-after-removals | 100% Solution using HashSet O(n+m) | 100-solution-using-hashset-onm-by-thanus-xmtc | Complexity\n- Time complexity: O(n+m)\n Add your time complexity here, e.g. O(n) \n\n- Space complexity: O(n+m)\n Add your space complexity here, e.g. O(n) \n\n | thanushasm25 | NORMAL | 2024-01-19T06:28:44.779513+00:00 | 2024-01-19T06:28:44.779536+00:00 | 306 | false | # Complexity\n- Time complexity: $$O(n+m)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n+m)$$\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\n public int maximumSetSize(int[] nums1, int[] nums2) {\n Set<Integer> set1 = new HashSe... | 2 | 0 | ['Java'] | 0 |
maximum-size-of-a-set-after-removals | 🔥Greedy💯 || ✅Clean Code | greedy-clean-code-by-adish_21-da61 | Complexity\n\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n\nclass Solution {\npublic:\n int | aDish_21 | NORMAL | 2024-01-11T18:30:59.626309+00:00 | 2024-01-11T18:30:59.626372+00:00 | 200 | false | # Complexity\n```\n- Time complexity:\nO(n)\n\n- Space complexity:\nO(n)\n```\n\n# Code\n## Please Upvote if it helps\uD83E\uDD17\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n int n = nums1.size(), ans = 0, count = 0;\n unordered_set<int> st1(nums1.beg... | 2 | 0 | ['Array', 'Greedy', 'Ordered Set', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | Java Most Easy Solution | java-most-easy-solution-by-vedantzope9-qyw1 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | vedantzope9 | NORMAL | 2024-01-08T02:57:17.838909+00:00 | 2024-01-08T02:57:17.838942+00:00 | 20 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Ordered Set', 'Java'] | 0 |
maximum-size-of-a-set-after-removals | 💡Most Optimized and Intuitive Code💡 | most-optimized-and-intuitive-code-by-nis-o94b | Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_set<int>s1,s2,combined;\n int n = | nishant101 | NORMAL | 2024-01-07T05:23:28.209684+00:00 | 2024-01-07T05:23:28.209715+00:00 | 176 | false | # Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n unordered_set<int>s1,s2,combined;\n int n = nums1.size();\n for(auto it : nums1) s1.insert(it);\n for(auto it : nums2) s2.insert(it);\n for(auto it : s1) combined.insert(it);\n ... | 2 | 0 | ['C++'] | 0 |
maximum-size-of-a-set-after-removals | Simple and Easy Casework | simple-and-easy-casework-by-manmeet8287-vvb5 | Intuition\nThere can be 2 types of elements in each set,\n\n1. Which are common in both the arrays\n2. Which are distinct in both the arrays\n\n\nNow, suppose t | Manmeet8287 | NORMAL | 2024-01-07T05:16:39.505732+00:00 | 2024-01-07T05:16:39.505760+00:00 | 60 | false | # Intuition\nThere can be 2 types of elements in each set,\n```\n1. Which are common in both the arrays\n2. Which are distinct in both the arrays\n```\n\nNow, suppose there are **x** elements in set A which are distinct and **y** elements in set B which are distinct and **m** elements arre common in both of them.\nTher... | 2 | 0 | ['Math', 'C++'] | 1 |
maximum-size-of-a-set-after-removals | 3 sets - Java O(N) | O(N) | 3-sets-java-on-on-by-wangcai20-1dm0 | Intuition\n Describe your first thoughts on how to solve this problem. \nPure logic, we first should pick non-overlap unique numbers from nums1 up to n/2 as wel | wangcai20 | NORMAL | 2024-04-29T01:52:05.352869+00:00 | 2024-04-29T01:52:05.352895+00:00 | 32 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPure logic, we first should pick non-overlap unique numbers from `nums1` up to `n/2` as well as non-overlap unique numbers from `nums2` up to `n/2`. We can see these numbers is safe to pick without affecting the rest.\n\nThen we can add t... | 1 | 0 | ['Java'] | 1 |
maximum-size-of-a-set-after-removals | Greedy Approach | Simple Solution | C++ | greedy-approach-simple-solution-c-by-sam-eec2 | Intuition\nThe value can be seperated into unique and common.\n\n# Approach\nGreedy approach of counting unique and common values.\n\n# Complexity\n- Time compl | sameerahmed56 | NORMAL | 2024-03-23T10:52:30.887727+00:00 | 2024-03-23T10:52:30.887750+00:00 | 287 | false | # Intuition\nThe value can be seperated into unique and common.\n\n# Approach\nGreedy approach of counting unique and common values.\n\n# Complexity\n- Time complexity:\n- O(N)\n\n- Space complexity:\n- O(N)\n\n# Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n ... | 1 | 0 | ['Greedy', 'Counting', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | Intuition Based Solution || Diff to understand || Clean Code | intuition-based-solution-diff-to-underst-og73 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nBrute Force approach.\n | Marlboro_Man_10 | NORMAL | 2024-03-10T01:16:03.245297+00:00 | 2024-03-10T01:16:03.245336+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nBrute Force approach.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->O(NlogN)\n\n- Space complexity:\n<!-- Add your space co... | 1 | 0 | ['Sorting', 'Ordered Set', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | Simple with sets | simple-with-sets-by-kiujpl-jodf | Intuition\nGreedy algorithm: first pick unique elements from each list, then pick common elements.\n\n# Approach\n\nLet A be the (unique) elements that are only | kiujpl | NORMAL | 2024-02-01T13:13:32.787753+00:00 | 2024-02-01T13:13:32.787775+00:00 | 6 | false | # Intuition\nGreedy algorithm: first pick unique elements from each list, then pick common elements.\n\n# Approach\n\nLet A be the (unique) elements that are only in nums1, let B be the (unique) elements that are only in nums2, and let C be the (unique) elements that are in both.\n\nFirst pick `a = min(n/2, len(A))` el... | 1 | 0 | ['Dart'] | 0 |
maximum-size-of-a-set-after-removals | C++ map greedy observation | c-map-greedy-observation-by-uchihaxmadar-cka3 | Intuition\n- The frequency of a number in an array is irrelevant; it will contribute at most once to the final count, whether it appears in one array or both.\n | uchihaXmadara | NORMAL | 2024-01-08T21:25:18.243499+00:00 | 2024-01-08T21:30:54.550764+00:00 | 131 | false | # Intuition\n- The frequency of a number in an array is irrelevant; it will contribute at most once to the final count, whether it appears in one array or both.\n\n- It\'s unnecessary to strictly select n/2 numbers from both arrays. For example, consider nums1: [1, 2, 4, 4, 4, 4, 4, 4], nums2: [1, 2, 3, 4, 5, 6, 7, 8].... | 1 | 0 | ['Hash Table', 'Greedy', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | Easy Sorting + Hashmaps | easy-sorting-hashmaps-by-piyuxh_01-1a65 | 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 | Piyuxh_01 | NORMAL | 2024-01-08T15:11:31.782272+00:00 | 2024-01-08T15:11:31.782302+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:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['C++'] | 0 |
maximum-size-of-a-set-after-removals | C++ Most Optimal and Concise Solution with Comments and Explanation! | c-most-optimal-and-concise-solution-with-huix | Please upvote if you found this helpful!\n\n# Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n // se | Ningck | NORMAL | 2024-01-07T18:48:48.379206+00:00 | 2024-01-07T18:53:15.929380+00:00 | 4 | false | Please upvote if you found this helpful!\n\n# Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n // set1 is the number of unique elements that will be remaining after removal in nums1 (max size will be n/2)\n // set2 is the number of unique elements th... | 1 | 0 | ['C++'] | 0 |
maximum-size-of-a-set-after-removals | C ++ solution . TC -- > O(n) | c-solution-tc-on-by-blueberryy-b7l7 | Intuition\n Describe your first thoughts on how to solve this problem. \nPick unique elements from both arrays and then go for common elements if needed .\n\n# | blueberryy | NORMAL | 2024-01-07T13:58:15.275439+00:00 | 2024-01-07T13:58:15.275471+00:00 | 29 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nPick unique elements from both arrays and then go for common elements if needed .\n\n# Complexity\n- Time complexity: $$O(n)$$\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: $$O(n)$$\n<!-- Add your space com... | 1 | 0 | ['C++'] | 1 |
maximum-size-of-a-set-after-removals | C++ Implementation using Set | c-implementation-using-set-by-ayushnauti-o91i | \n\n# Code\n\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n vector<int>v,v1;\n sort(nums1.begin(), | ayushnautiyal1110 | NORMAL | 2024-01-07T09:47:15.174395+00:00 | 2024-01-07T09:47:15.174423+00:00 | 26 | false | \n\n# Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n vector<int>v,v1;\n sort(nums1.begin(),nums1.end());\n \n sort(nums2.begin(),nums2.end());\n for(auto x:nums1){\n if(v.empty()){\n v.push_back(x);\n ... | 1 | 0 | ['Hash Table', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | C++ | Intuitive solution | c-intuitive-solution-by-sahan_18-bxsu | Intuition\nVenn Diagram\n\n# Approach\nCalculate no. of same and different elements\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\n Add your | sahan_18 | NORMAL | 2024-01-07T06:51:14.930005+00:00 | 2024-01-07T06:51:14.930052+00:00 | 12 | false | # Intuition\nVenn Diagram\n\n# Approach\nCalculate no. of same and different elements\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int maximumSetSize(vector<int>& nums1, vector<int>& nums2) {\n ... | 1 | 0 | ['C++'] | 0 |
maximum-size-of-a-set-after-removals | Easy to understand with approach and by using map | easy-to-understand-with-approach-and-by-mdxlu | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1-Determine Target Size count1 and count2 which is nums1.size()/2.\n\n2-C | 84SiddhantGupta | NORMAL | 2024-01-07T06:02:04.107015+00:00 | 2024-01-07T06:02:04.107046+00:00 | 38 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n**1-Determine Target Size count1 and count2 which is nums1.size()/2.**\n\n**2-Count Occurrences of elements in nums1 and nums2**\n\n**3-Prioritize Arrays: which is having more unique elements.**\nCheck if the size of mp1 is ... | 1 | 0 | ['Ordered Set', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | JavaScript set and logic operations | javascript-set-and-logic-operations-by-t-jva2 | Intuition\n Describe your first thoughts on how to solve this problem. \nThis solution was inspired by @blackbrain2009\n\n\n# Approach\n Describe your approach | Tayomide | NORMAL | 2024-01-07T04:38:44.961562+00:00 | 2024-01-07T04:45:20.876509+00:00 | 127 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThis solution was inspired by [@blackbrain2009](https://leetcode.com/problems/maximum-size-of-a-set-after-removals/solutions/4520698/hash-table-math-t-o-n)\n\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nStep 1: ... | 1 | 0 | ['JavaScript'] | 1 |
maximum-size-of-a-set-after-removals | [JavaScript] 10037. Maximum Size of a Set After Removals | javascript-10037-maximum-size-of-a-set-a-pi8t | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | pgmreddy | NORMAL | 2024-01-07T04:28:45.568269+00:00 | 2024-01-07T04:30:28.915757+00:00 | 66 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 1 | 0 | ['JavaScript'] | 1 |
maximum-size-of-a-set-after-removals | Simple solution using set operations | simple-solution-using-set-operations-by-8yz8u | Code\n\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n s1, s2, n = set(nums1), set(nums2), len(nums1)\n | ajmalhassan | NORMAL | 2024-01-07T04:28:40.063769+00:00 | 2024-01-07T04:28:40.063802+00:00 | 16 | false | # Code\n```\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n s1, s2, n = set(nums1), set(nums2), len(nums1)\n return min(\n min(len(s1 - s2), n // 2) # Numbers in s1 but not in s2, limited to n/2 as we have to pick max of n/2\n + min(len(s2 ... | 1 | 0 | ['Python3'] | 0 |
maximum-size-of-a-set-after-removals | Easy C++✅✅ solution using 'unordered_set' | easy-c-solution-using-unordered_set-by-a-efzx | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Abbas_R | NORMAL | 2024-01-07T04:22:46.539740+00:00 | 2024-01-07T04:22:46.539764+00:00 | 35 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:O(N)\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:O(N)\n<!-- Add your space complexity here, e.g. $$O... | 1 | 0 | ['Ordered Set', 'C++'] | 0 |
maximum-size-of-a-set-after-removals | [Python 🐍] Sets Unique in nums1 PLUS Unique in nums2 PLUS Shared nums1 nums2 | python-sets-unique-in-nums1-plus-unique-od9wb | 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 | xmky | NORMAL | 2024-01-07T04:21:34.607502+00:00 | 2024-01-08T02:48:09.817331+00:00 | 179 | 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'] | 1 |
maximum-size-of-a-set-after-removals | Brute Force | brute-force-by-prithvinadagouda-gevs | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\nDONT LOOK AT THE SIZE ITS UNDERSTANTABLE\n\n# Complexity\n- Time complexi | PrithviNadagouda | NORMAL | 2024-01-07T04:11:16.072851+00:00 | 2024-01-07T04:11:16.072891+00:00 | 490 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\nDONT LOOK AT THE SIZE ITS UNDERSTANTABLE\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``... | 1 | 0 | ['Ordered Map', 'C++'] | 2 |
maximum-size-of-a-set-after-removals | Brute Force Python Sol. | brute-force-python-sol-by-r34per-twri | Code\n\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n cnt1=Counter(nums1)\n cnt2=Counter(nums2)\n | R34PER | NORMAL | 2024-01-07T04:08:08.721502+00:00 | 2024-01-07T04:08:08.721591+00:00 | 125 | false | # Code\n```\nclass Solution:\n def maximumSetSize(self, nums1: List[int], nums2: List[int]) -> int:\n cnt1=Counter(nums1)\n cnt2=Counter(nums2)\n \n target=len(nums1)//2\n temp=set()\n \n c1,c2=0,0\n for key,val in cnt1.items():\n if c1>=target:\n ... | 1 | 0 | ['Python3'] | 1 |
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