LeetCode 765: Couples Holding Hands Solution

Master LeetCode problem 765 (Couples Holding Hands), a hard challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

765. Couples Holding Hands

Hard

Greedy Depth-First Search Breadth-First Search Union Find Graph

Problem Explanation

Explanation

To solve this problem, we need to count the number of swaps required to ensure that every couple is sitting side by side. We can achieve this by using the concept of graph cycles and matching pairs. We iterate through the given row and for each couple, we create an edge between the two people of the couple. If the couple is not sitting side by side, it means they are not connected in the graph. We then count the number of connected components in the graph, which gives the minimum number of swaps needed.

Algorithm:

Create a graph where each person is a node and each couple is an edge.
Count the number of connected components in the graph.
The minimum number of swaps required is (number of couples) - (number of connected components).

Time Complexity:

The time complexity of this algorithm is O(n), where n is the number of couples.

Space Complexity:

The space complexity of this algorithm is O(n).

Solution Code

class Solution {
    public int minSwapsCouples(int[] row) {
        int n = row.length / 2;
        int[] parent = new int[n];
        
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
        
        for (int i = 0; i < row.length; i += 2) {
            int c1 = row[i] / 2;
            int c2 = row[i + 1] / 2;
            union(parent, c1, c2);
        }
        
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (find(parent, i) == i) {
                count++;
            }
        }
        
        return n - count;
    }
    
    private void union(int[] parent, int x, int y) {
        int rootX = find(parent, x);
        int rootY = find(parent, y);
        if (rootX != rootY) {
            parent[rootX] = rootY;
        }
    }
    
    private int find(int[] parent, int x) {
        if (parent[x] != x) {
            parent[x] = find(parent, parent[x]);
        }
        return parent[x];
    }
}

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