# Next Permutation Program Solution

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Problem Statement

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`

• Time: O(n)O(n)
• Space: O(1)O(1)

### Next Permutation Program Solution In C++

``````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 Program Solution 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;
}
}
``````

### Next Permutation Program Solution 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)
``````
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