g1001_1100.s1053_previous_permutation_with_one_swap.Solution Maven / Gradle / Ivy

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package g1001_1100.s1053_previous_permutation_with_one_swap;

// #Medium #Array #Greedy #2022_02_28_Time_0_ms_(100.00%)_Space_56.1_MB_(27.21%)

/**
 * 1053 - Previous Permutation With One Swap\.
 *
 * Medium
 *
 * 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** (A _swap_ exchanges the positions of two numbers `arr[i]` and `arr[j]`). If it cannot be done, then return the same array.
 *
 * **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
**/
public class Solution {
    public int[] prevPermOpt1(int[] arr) {
        for (int i = arr.length - 1; i >= 0; i--) {
            int diff = Integer.MAX_VALUE;
            int index = i;
            for (int j = i + 1; j < arr.length; j++) {
                if (arr[i] - arr[j] > 0 && diff > arr[i] - arr[j]) {
                    diff = arr[i] - arr[j];
                    index = j;
                }
            }
            if (diff != Integer.MAX_VALUE) {
                int temp = arr[i];
                arr[i] = arr[index];
                arr[index] = temp;
                break;
            }
        }
        return arr;
    }
}