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# Next Permutation LeetCode Solution

## Problem – Next Permutation LeetCode Solution

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 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 C++

``````class Solution {
public:
void nextPermutation(vector<int>& nums) {
int n = nums.size(), k, l;
for (k = n - 2; k >= 0; k--) {
if (nums[k] < nums[k + 1]) {
break;
}
}
if (k < 0) {
reverse(nums.begin(), nums.end());
} else {
for (l = n - 1; l > k; l--) {
if (nums[l] > nums[k]) {
break;
}
}
swap(nums[k], nums[l]);
reverse(nums.begin() + k + 1, nums.end());
}
}
};
``````

### Next Permutation LeetCode Solution in Java

``````public class Solution {
/*0*/ public void nextPermutation(int[] nums) {
// pivot is the element just before the non-increasing (weakly decreasing) suffix
/*2*/   int pivot = indexOfLastPeak(nums) - 1;
// paritions nums into [prefix pivot suffix]
if (pivot != -1) {
int nextPrefix = lastIndexOfGreater(nums, nums[pivot]); // in the worst case it's suffix[0]
// next prefix must exist because pivot < suffix[0], otherwise pivot would be part of suffix
/*4*/       swap(nums, pivot, nextPrefix); // this minimizes the change in prefix
}
/*5*/   reverseSuffix(nums, pivot + 1); // reverses the whole list if there was no pivot
/*6*/ }

/**
* Find the last element which is a peak.
* In case there are multiple equal peaks, return the first of those.
* @return first index of last peak
*/
/*1*/ int indexOfLastPeak(int[] nums) {
for (int i = nums.length - 1; 0 < i; --i) {
if (nums[i - 1] < nums[i]) return i;
}
return 0;
}

/** @return last index where the {@code num > threshold} or -1 if not found */
/*3*/ int lastIndexOfGreater(int[] nums, int threshold) {
for (int i = nums.length - 1; 0 <= i; --i) {
if (threshold < nums[i]) return i;
}
return -1;
}

/** Reverse numbers starting from an index till the end. */
void reverseSuffix(int[] nums, int start) {
int end = nums.length - 1;
while (start < end) {
swap(nums, start++, end--);
}
}

void swap(int[] nums, int i, int j) {
int temp = nums[i];
nums[i] = nums[j];
nums[j] = temp;
}
}
``````

### Next Permutation LeetCode Solution in Python

``````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
``````

### Next Permutation LeetCode Solution in JavaScript

``````var nextPermutation = function(nums) {

for(let i = nums.length-1; i >= 0; i--) {
if(nums[i] < nums[i+1]) {
const large = nextLarge(i);
swap(i, large);
reverse(i+1);
return;
}
}

// If there is no next permutation reverse the arr
nums.reverse()

function swap(i, j) {
[nums[i], nums[j]] = [nums[j], nums[i]];
}

function reverse(idx) {
let start = idx, end = nums.length-1;

while(start < end) {
swap(start, end);
start++;
end--;
}
}

function nextLarge(idx) {
for(let i = nums.length-1; i > idx; i--) {
if(nums[i] > nums[idx]) return i;
}
}
};
``````
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