Day 31 - Next Permutation

ruarfff

Ruairí O'Brien

Posted on January 31, 2021

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]
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Example 2:

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

Input: nums = [1,1,5]
Output: [1,5,1]
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Example 4:

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

Alt Text

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ruarfff
Ruairí O'Brien

Posted on January 31, 2021

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