Next Permutation

A permutation of an array of integers is an arrangement of its members into a sequence or linear order. 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). Given an array of integers nums, find the next permutation of nums. The replacement must be in place and use only constant extra memory.

(from leetcode 31)

Explanation

To find the next permutation of a given array, we need to increase the order of the array as little as we can. Notice that the maximum order form of an array is when it is sorted in a non-ascending order. For example, [4,3,1] is the maximum order form of the array [3,1,4]. According to the question definition, the next permutation of a maximum order form is the lowest possible order (i.e., the next permutation of [4,3,1] is [1,3,4]).

The overall steps for finding the next permutation of an array are:

1. Find the non-descending index in the array reversely. The subarray nums[non_desc_idx+1:] is a subarray in its max order form. If input array is already in its max order form (i.e., non_desc_idx==-1), go to step 3.
2. Swap nums[non_desc_idx] with nums[swap_idx] where nums[swap_idx] is the smallest element larger than nums[non_desc_idx]. Since we want to increment the order as little as possible, we replace the element before subarray nums[non_desc_idx+1:] with a slightly larger one we can find in nums[non_desc_idx+1:].
3. Reverse nums[non_desc_idx+1:] by swapping. The will change nums[non_desc_idx+1:] to its min order form.

The time complexity of this solution is O(n) and space complexity is O(1).

Example

input:  nums = [3,4,6,5,2,1]
output: nums = [3,5,1,2,4,6]

Python3 Implementation

def nextPermutation(self, nums: List[int]) -> None:

    def reverse_by_swap(l, r):
        # l, r are idx in array inclusive
        for i in range((r-l+1)//2):
            nums[l+i], nums[r-i] = nums[r-i], nums[l+i]
        return 

    n = len(nums)
    non_desc_idx = n-2
    while non_desc_idx >= 0:
        if nums[non_desc_idx] >= nums[non_desc_idx+1]:
            non_desc_idx -= 1
        else:
            break
    
    # if input array is in its max order form
    if non_desc_idx==-1:
        reverse_by_swap(0,n-1)
        return
    
    for swap_idx in range(n-1,non_desc_idx,-1):
        if nums[swap_idx] > nums[non_desc_idx]:
            nums[non_desc_idx], nums[swap_idx] = nums[swap_idx], nums[non_desc_idx]
            break
    
    reverse_by_swap(non_desc_idx+1, n-1)
    return