Description

N couples sit in 2N seats arranged in a row and want to hold hands. We want to know 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.

The people and seats are represented by an integer from 0 to 2N-1, 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).

The couples' initial seating is given by row[i] being the value of the person who is initially sitting in the i-th seat.

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.

Note:

1. len(row) is even and in the range of [4, 60].
2. row is guaranteed to be a permutation of 0...len(row)-1.

Solution

Greedy, O(N ^ 2), S(1)

class Solution {
    public int minSwapsCouples(int[] row) {
        int n = row.length;
        int count = 0;
        
        for (int i = 0; i < n; i += 2) {
            if (isCouple(row[i], row[i + 1])) {
                continue;
            }
            
            int target = search(row, i + 2, getCouple(row[i]));
            swap(row, i + 1, target);
            ++count;
        }
        
        return count;
    }
    
    private boolean isCouple(int x, int y) {
        return y == getCouple(x);
    }
    
    private int getCouple(int x) {
        // return x % 2 > 0 ? x - 1 : x + 1;
        return x ^ 1;   // interesting
    }
    
    private int search(int[] arr, int start, int target) {
        while (start < arr.length && arr[start] != target) {
            ++start;
        }
        return start;
    }
    
    private void swap(int[] arr, int i, int j) {
        if (i == j) {
            return;
        }
        
        arr[i] ^= arr[j];
        arr[j] ^= arr[i];
        arr[i] ^= arr[j];
    }
}