# Next Permutation Leetcode Solution 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.

One 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.

## Problem

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 <= 100`
• `0 <= nums[i] <= 100`

### Next Permutation Leetcode Solution in Python

```class 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 CPP

```class 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 Java

```class 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.