Next Permutation Python Solution with Tests

LeetCode 31, Medium. Topics: Array, Two Pointers. View on LeetCode. Generate this problem as a practice environment: tested reference solution, 15 parametrized pytest cases, and a playground notebook:

lcpy gen -n 31   # by problem number
lcpy gen -s next_permutation   # by problem name

Problem

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.

Examples

Input: nums = [1,2,3]
Output: [1,3,2]

Input: nums = [3,2,1]
Output: [1,2,3]

Input: nums = [1,1,5]
Output: [1,5,1]

Constraints

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

Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(n)
    # Space: O(1)
    def next_permutation(self, nums: list[int]) -> None:
        # Find pivot (rightmost ascending pair)
        i = len(nums) - 2
        while i >= 0 and nums[i] >= nums[i + 1]:
            i -= 1

        if i >= 0:
            # Find successor (rightmost element > pivot)
            j = len(nums) - 1
            while nums[j] <= nums[i]:
                j -= 1
            nums[i], nums[j] = nums[j], nums[i]

        # Reverse suffix
        left, right = i + 1, len(nums) - 1
        while left < right:
            nums[left], nums[right] = nums[right], nums[left]
            left += 1
            right -= 1

Complexity

Time	Space
O(n)	O(1)

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Next Permutation Python Solution with Tests

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