31. Next Permutation - Explanation

Problem Link

Description

A permutation of an array of integers is an arrangement of its elements 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.

You are 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]

Constraints:

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

Company Tags

Please upgrade to NeetCode Pro to view company tags.



1. Brute Force

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        permutations = self.permute(nums[:])
        permutations.sort()
        for i, p in enumerate(permutations):
            if p == nums:
                nextP = permutations[(i + 1) % len(permutations)]
                for j in range(len(nums)):
                    nums[j] = nextP[j]
                break


    def permute(self, nums: List[int]) -> List[List[int]]:
        res = []

        def dfs(i):
            if i == len(nums):
                res.append(nums.copy())
                return

            for j in range(i, len(nums)):
                if j > i and nums[i] == nums[j]:
                    continue

                nums[i], nums[j] = nums[j], nums[i]
                dfs(i + 1)

            for j in range(len(nums) - 1, i, -1):
                nums[j], nums[i] = nums[i], nums[j]

        nums.sort()
        dfs(0)
        return res

Time & Space Complexity

  • Time complexity: O(n!∗n)O(n! * n)
  • Space complexity: O(n!∗n)O(n! * n)

2. Greedy

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

        if i >= 0:
            j = n - 1
            while nums[j] <= nums[i]:
                j -= 1
            nums[i], nums[j] = nums[j], nums[i]

        l, r = i + 1, n - 1
        while l < r:
            nums[l], nums[r] = nums[r], nums[l]
            l += 1
            r -= 1

Time & Space Complexity

  • Time complexity: O(n)O(n)
  • Space complexity: O(1)O(1)