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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0701_0800.s0765_couples_holding_hands;

// #Hard #Depth_First_Search #Greedy #Breadth_First_Search #Graph #Union_Find
// #2022_03_26_Time_0_ms_(100.00%)_Space_41.6_MB_(54.28%)

/**
 * 765 - Couples Holding Hands\.
 *
 * Hard
 *
 * There are `n` couples sitting in `2n` seats arranged in a row and want to hold hands.
 *
 * The people and seats are represented by an integer array `row` where `row[i]` is the ID of the person sitting in the i^th seat. The couples are numbered in order, the first couple being `(0, 1)`, the second couple being `(2, 3)`, and so on with the last couple being `(2n - 2, 2n - 1)`.
 *
 * Return _the minimum number of swaps so that every couple is sitting side by side_. A swap consists of choosing any two people, then they stand up and switch seats.
 *
 * **Example 1:**
 *
 * **Input:** row = [0,2,1,3]
 *
 * **Output:** 1
 *
 * **Explanation:** We only need to swap the second (row[1]) and third (row[2]) person.
 *
 * **Example 2:**
 *
 * **Input:** row = [3,2,0,1]
 *
 * **Output:** 0
 *
 * **Explanation:** All couples are already seated side by side.
 *
 * **Constraints:**
 *
 * *   `2n == row.length`
 * *   `2 <= n <= 30`
 * *   `n` is even.
 * *   `0 <= row[i] < 2n`
 * *   All the elements of `row` are **unique**.
**/
public class Solution {
    public int minSwapsCouples(int[] row) {
        int swaps = 0;
        for (int i = 0; i < row.length - 1; i += 2) {
            int coupleValue = row[i] % 2 == 0 ? row[i] + 1 : row[i] - 1;
            if (row[i + 1] != coupleValue) {
                swaps++;
                int coupleIndex = findIndex(row, coupleValue);
                swap(row, coupleIndex, i + 1);
            }
        }
        return swaps;
    }

    private void swap(int[] row, int i, int j) {
        int tmp = row[i];
        row[i] = row[j];
        row[j] = tmp;
    }

    private int findIndex(int[] row, int value) {
        for (int i = 0; i < row.length; i++) {
            if (row[i] == value) {
                return i;
            }
        }
        return -1;
    }
}