In this post, you will know how to solve the Next Permutation Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.Next Permutation Leetcode SolutionsOne more thing to add, don’t directly look for the solutions, first try to solve the problems of Leetcode by yourself. If you find any difficulty after trying several times, then you can look for solutions.ProblemA permutation of an array of integers is an arrangement of its members into a sequence or linear order.For example, for arr = [1,2,3], the following are all the permutations of arr: [1,2,3], [1,3,2], [2, 1, 3], [2, 3, 1], [3,1,2], [3,2,1].The next permutation of an array of integers is the next lexicographically greater permutation of its integer. More formally, if all the permutations of the array are sorted in one container according to their lexicographical order, then the next permutation of that array is the permutation that follows it in the sorted container. If such arrangement is not possible, the array must be rearranged as the lowest possible order (i.e., sorted in ascending order).For example, the next permutation of arr = [1,2,3] is [1,3,2].Similarly, the next permutation of arr = [2,3,1] is [3,1,2].While the next permutation of arr = [3,2,1] is [1,2,3] because [3,2,1] does not have a lexicographical larger rearrangement.Given an array of integers nums, find the next permutation of nums.The replacement must be in place and use only constant extra memory.Example 1:Input: nums = [1,2,3] Output: [1,3,2] Example 2:Input: nums = [3,2,1] Output: [1,2,3] Example 3:Input: nums = [1,1,5] Output: [1,5,1] Constraints:1 <= nums.length <= 1000 <= nums[i] <= 100Next Permutation Leetcode Solution in Pythonclass Solution: def nextPermutation(self, nums: List[int]) -> None: n = len(nums) # From back to front, find the first num < nums[i + 1] i = n - 2 while i >= 0: if nums[i] < nums[i + 1]: break i -= 1 # From back to front, find the first num > nums[i], swap it with nums[i] if i >= 0: for j in range(n - 1, i, -1): if nums[j] > nums[i]: nums[i], nums[j] = nums[j], nums[i] break def reverse(nums: List[int], l: int, r: int) -> None: while l < r: nums[l], nums[r] = nums[r], nums[l] l += 1 r -= 1 # Reverse nums[i + 1..n - 1] reverse(nums, i + 1, len(nums) - 1) Next Permutation Leetcode Solutions in CPPclass Solution { public: void nextPermutation(vector<int>& nums) { const int n = nums.size(); // From back to front, find the first num < nums[i + 1] int i; for (i = n - 2; i >= 0; --i) if (nums[i] < nums[i + 1]) break; // From back to front, find the first num > nums[i], swap it with nums[i] if (i >= 0) for (int j = n - 1; j > i; --j) if (nums[j] > nums[i]) { swap(nums[i], nums[j]); break; } // Reverse nums[i + 1..n - 1] reverse(nums, i + 1, n - 1); } private: void reverse(vector<int>& nums, int l, int r) { while (l < r) swap(nums[l++], nums[r--]); } }; Next Permutation Leetcode Solutions in Javaclass Solution { public void nextPermutation(int[] nums) { final int n = nums.length; // From back to front, find the first num < nums[i + 1] int i; for (i = n - 2; i >= 0; --i) if (nums[i] < nums[i + 1]) break; // From back to front, find the first num > nums[i], swap it with nums[i] if (i >= 0) for (int j = n - 1; j > i; --j) if (nums[j] > nums[i]) { swap(nums, i, j); break; } // Reverse nums[i + 1..n - 1] reverse(nums, i + 1, n - 1); } private void reverse(int[] nums, int l, int r) { while (l < r) swap(nums, l++, r--); } private void swap(int[] nums, int i, int j) { final int temp = nums[i]; nums[i] = nums[j]; nums[j] = temp; } } Note: This problem Next Permutation is generated by Leetcode but the solution is provided by BrokenProgrammers. This tutorial is only for Educational and Learning purposes.Next: Swap Nodes in Pairs Leetcode Solution Post navigationSubstring with Concatenation of All Words Leetcode Solution Swap Nodes in Pairs Leetcode Solution