1053. Previous Permutation With One Swap

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.

Note that a swap exchanges the positions of two numbers arr[i] and arr[j]

  Example 1:

Input: arr = [3,2,1]
Output: [3,1,2]
Explanation: Swapping 2 and 1.

Example 2:

Input: arr = [1,1,5]
Output: [1,1,5]
Explanation: This is already the smallest permutation.

Example 3:

Input: arr = [1,9,4,6,7]
Output: [1,7,4,6,9]
Explanation: Swapping 9 and 7.

  Constraints:

• 1 <= arr.length <= 104

• 1 <= arr[i] <= 104

Solution

/**
 * @param {number[]} arr
 * @return {number[]}
 */
var prevPermOpt1 = function(arr) {
    for (var i = arr.length - 2; i >= 0; i--) {
        if (arr[i] <= arr[i + 1]) continue;
        var max = 0;
        var maxIndex = -1;
        for (var j = i + 1; j < arr.length; j++) {
            if (arr[j] > max && arr[j] < arr[i]) {
                max = arr[j];
                maxIndex = j;
            }
        }
        swap(arr, i, maxIndex);
        break;
    }
    return arr;
};

var swap = function(arr, i, j) {
    var tmp = arr[i];
    arr[i] = arr[j];
    arr[j] = tmp;
};

Explain:

1. we need a smaller array than the current one, so that we need to swap a smaller number from right to left
2. we need a largest array from all the possible result, so that we are going to find the first possible index to swap, from the right of array
3. from right find a possible index to swap, find biggest number smaller than this one to swap on the right

Complexity:

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

Solution 2

/**
 * @param {number[]} arr
 * @return {number[]}
 */
var prevPermOpt1 = function(arr) {
    for (var i = arr.length - 2; i >= 0; i--) {
        if (arr[i] <= arr[i + 1]) continue;
        for (var j = arr.length; j > i; j--) {
            if (arr[j] < arr[i] && arr[j] !== arr[j - 1]) {
                swap(arr, i, j);
                return arr;
            }
        }
    }
    return arr;
};

var swap = function(arr, i, j) {
    var tmp = arr[i];
    arr[i] = arr[j];
    arr[j] = tmp;
};

Explain:

because we know that numbers from right to left is in order, (from solution 1), we can just find the first one from right.

Complexity:

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