31.Next_Permutation(Medium).md

31. Next Permutation (Medium)

Date and Time: Dec 18, 2024, 10:12 (EST)

Link: https://leetcode.com/problems/next-permutation

Question:

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.

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]

Edge Case:

Input: nums = [1,5,8,4,7,6,5,3,1]

Output: [1,5,8,5,1,3,4,6,7]

Constraints:

• 1 <= nums.length <= 100

• 0 <= nums[i] <= 100

Walk-through:

Notice the pattern that if nums is decending, then we don't have next permutation, we have to traverse the whole nums.

From the right to left of nums, the elements are decending, so if we find an element such that nums[i] < nums[i+1], then we can find another element such that nums[j] > nums[i] (if i >= 0), because these two elements will be the smallest digits to swap, so we swap them to form the next permutation.

Finally, we need to reverse nums[i+1:] to make them to be the smallest, since the right part of nums[i:] is decreasing. And also, we check if i >= 0 so even when i = -1, we can swap the whole nums to be ascending.

Python Solution:

class Solution:
    def nextPermutation(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        # Find an element such that `nums[i] < nums[i+1]`, find another element such that `nums[j] > nums[i]`, swap them, then reverse `nums[i+1:]`

        # TC: O(n), n=len(nums), SC: O(1)
        i, j = len(nums)-2, len(nums)-1
        # Find elememt that nums[i] < nums[i+1]
        while i >= 0 and nums[i] >= nums[i+1]:
            i -= 1
        # Check if nums has next permutation
        if i >= 0:
            while j >= 0 and nums[j] <= nums[i]:
                j -= 1
            # Swap two elements
            nums[i], nums[j] = nums[j], nums[i]
        # Reverse
        nums[i+1:] = nums[i+1:][::-1]

Time Complexity: $O(n)$
Space Complexity: $O(1)$