Previous Permutation With One Swap — LeetCode 1053 Python Solution

The problem

Given an array of positive integers arr (not necessarily distinct), return the lexicographically largest permutation that is smaller than arr, that can be made with exactly one swap. If it cannot be done, then return the same array.

Example

Input

arr = [3,2,1]

Output

[3,1,2]

Explanation

Swapping 2 and 1.

Python solution

class Solution:
    def prevPermOpt1(self, arr: List[int]) -> List[int]:
        n = len(arr)
        for i in range(n - 1, 0, -1):
            if arr[i - 1] > arr[i]:
                for j in range(n - 1, i - 1, -1):
                    if arr[j] < arr[i - 1] and arr[j] != arr[j - 1]:
                        arr[i - 1], arr[j] = arr[j], arr[i - 1]
                        return arr
        return arr

Complexity

Pattern: Greedy

Take the locally best option every time — when you can prove that never costs you later. LeetCode 1053. Previous Permutation With One Swap is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Greedy.

The greedy guide has the Python template for the pattern and the 346 LeetCode problems that use it.

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