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# 31. Next Permutation

## Problem Statement

<br>

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 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**](http://en.wikipedia.org/wiki/In-place_algorithm) and use only constant extra memory.

&#x20;

**Example 1:**

<pre><code><strong>Input: nums = [1,2,3]
</strong><strong>Output: [1,3,2]
</strong></code></pre>

**Example 2:**

<pre><code><strong>Input: nums = [3,2,1]
</strong><strong>Output: [1,2,3]
</strong></code></pre>

**Example 3:**

<pre><code><strong>Input: nums = [1,1,5]
</strong><strong>Output: [1,5,1]
</strong></code></pre>

&#x20;

**Constraints:**

* `1 <= nums.length <= 100`
* `0 <= nums[i] <= 100`

## Intuition

```
From the end find the break point and then swap
sorting the array after the index

eg :

5 1 4 3 2
Next permutation is 5 2 1 3 4
So Finding Increasing from end and thus swap and sort the end array


For edge case 
5 4 3 2 1

Reverse and return the array
```

### Links

<https://leetcode.com/problems/next-permutation/>

### Video Links

<https://www.youtube.com/watch?v=JDOXKqF60RQ&ab_channel=takeUforward>

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    void nextPermutation(vector<int>& nums) {
        int n = nums.size();
        int index = -1;

        for(int i=n-2; i>=0; i--){
            if(nums[i] < nums[i+1]){
                index = i;
                break;
            }
        }

        if(index == -1){
            reverse(nums.begin(), nums.end());
            return ;
        }

        int just_large;
        for(int i=n-1; i>index; i--){
            if(nums[index] < nums[i]){
                just_large = i;
                break;
            }
        }

        swap(nums[just_large], nums[index]);
        sort(nums.begin() + index+1, nums.end());


    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###