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Ruairí O'Brien

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# Day 31 - Next Permutation

## The Problem

Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers.

If such an arrangement is not possible, it must rearrange it as the lowest possible order (i.e., sorted in ascending order).

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]
``````

Example 4:

``````Input: nums = [1]
Output: [1]
``````

Constraints:

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

## Tests

``````import pytest
from .Day31_NextPermutation import Solution

s = Solution()

@pytest.mark.parametrize(
"nums,expected",
[
([1, 2, 3], [1, 3, 2]),
([3, 2, 1], [1, 2, 3]),
([1, 1, 5], [1, 5, 1]),
([1], [1]),
],
)
def test_next_permutation(nums, expected):
s.nextPermutation(nums)

assert nums == expected
``````

## Solution

``````from typing import List

class Solution:
def nextPermutation(self, nums: List[int]) -> None:
"""
Do not return anything, modify nums in-place instead.
"""
i = len(nums) - 2
while i >= 0 and nums[i + 1] <= nums[i]:
i -= 1

if i >= 0:
j = len(nums) - 1
while j >= 0 and nums[j] <= nums[i]:
j -= 1
nums[i], nums[j] = nums[j], nums[i]

k = len(nums) - 1
while i < k:
i += 1
nums[i], nums[k] = nums[k], nums[i]
k -= 1
``````