# Next Permutation – LeetCode Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

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** 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`

# C++ Next Permutation LeetCode Solution

```
``````
void nextPermutation(vector<int>& nums) {
int i = nums.size() - 1, k = i;
while (i > 0 && nums[i-1] >= nums[i])
i--;
for (int j=i; j<k; j++, k--)
swap(nums[j], nums[k]);
if (i > 0) {
k = i--;
while (nums[k] <= nums[i])
k++;
swap(nums[i], nums[k]);
}
}
```

# Java Next Permutation LeetCode Solution

```
``````
public void nextPermutation(int[] A) {
if(A == null || A.length <= 1) return;
int i = A.length - 2;
while(i >= 0 && A[i] >= A[i + 1]) i--; // Find 1st id i that breaks descending order
if(i >= 0) { // If not entirely descending
int j = A.length - 1; // Start from the end
while(A[j] <= A[i]) j--; // Find rightmost first larger id j
swap(A, i, j); // Switch i and j
}
reverse(A, i + 1, A.length - 1); // Reverse the descending sequence
}
public void swap(int[] A, int i, int j) {
int tmp = A[i];
A[i] = A[j];
A[j] = tmp;
}
public void reverse(int[] A, int i, int j) {
while(i < j) swap(A, i++, j--);
}
```

# Python 3 Next Permutation LeetCode Solution

```
``````
def nextPermutation(self, nums):
i = j = len(nums)-1
while i > 0 and nums[i-1] >= nums[i]:
i -= 1
if i == 0: # nums are in descending order
nums.reverse()
return
k = i - 1 # find the last "ascending" position
while nums[j] <= nums[k]:
j -= 1
nums[k], nums[j] = nums[j], nums[k]
l, r = k+1, len(nums)-1 # reverse the second part
while l < r:
nums[l], nums[r] = nums[r], nums[l]
l +=1 ; r -= 1
```

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