# 31. Next Permutation

A **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 considered permutations of `arr`: `[1,2,3]`, `[1,3,2]`, `[3,1,2]`, `[2,3,1]`.

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**](http://en.wikipedia.org/wiki/In-place_algorithm) 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`

```python
class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        # Approach
        # 1. From end find first index where index-1 < index. It means all elements after index are decreasing.
        # 2. You can think of above sequence that we try to find out is [3,2,1]
        # 3. If index == 0 it means it is something like [3,2,1], so only permumation is to reverse org. array
        # 4. If not equal to 0 , then find first number greater than (index-1)
        # 5. Swap the two, then reverse whole sub-array arr[index:]
        i = j = len(nums)-1
        while i > 0 and nums[i-1] >= nums[i]:
            i -= 1
        if i == 0:
            nums.reverse()
            return
        while nums[j] <= nums[i-1]:
            j -= 1
        nums[i-1] , nums[j] = nums[j], nums[i-1]
        left = i
        right = len(nums)-1
        while left < right:
            nums[left], nums[right] = nums[right] , nums[left]
            left += 1
            right -= 1
            
        
```