Number of Operations to Make Network Connected

There are n computers numbered from 0 to n-1 connected by ethernet cables connections forming a network where connections[i] = [a, b] represents a connection between computers a and b. Any computer can reach any other computer directly or indirectly through the network.

Given an initial computer network connections. You can extract certain cables between two directly connected computers, and place them between any pair of disconnected computers to make them directly connected. Return the minimum number of times you need to do this in order to make all the computers connected. If it's not possible, return -1. 

 

Example 1:

Input: n = 4, connections = [[0,1],[0,2],[1,2]]
Output: 1
Explanation: Remove cable between computer 1 and 2 and place between computers 1 and 3.

Example 2:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2],[1,3]]
Output: 2

Example 3:

Input: n = 6, connections = [[0,1],[0,2],[0,3],[1,2]]
Output: -1
Explanation: There are not enough cables.

Example 4:

Input: n = 5, connections = [[0,1],[0,2],[3,4],[2,3]]
Output: 0

 

Constraints:


Solution:

class Solution {
    static class UF {
        private int[] parent;
        private int[] size;
        private int count;
        
        public UF(int n) {
            parent = new int[n];
            size = new int[n];
            count = n;
            for (int i = 0; i < n; i ++) {
                parent[i] = i;
                size[i] = 1;
            }
        }
        
        public void union(int x, int y) {
            int rootX = find(x);
            int rootY = find(y);
            if (rootX == rootY) return;
            if (size[rootX] < size[rootY]) {
                parent[rootX] = rootY;
                size[rootY] += size[rootX];
            } else {
                parent[rootY] = rootX;
                size[rootX] += size[rootY];
            }
            count --;
        }
        
        public int find(int x) {
            while (x != parent[x]) {
                parent[x] = parent[parent[x]];
                x = parent[x];
            }
            return x;
        }
        
        public int count() {
            return count;
        }
    }
    
    
    public int makeConnected(int n, int[][] connections) {
        int m = connections.length;
        if (m + 1 < n) return -1;
        UF computers = new UF(n);
        for (int[] con : connections) {
            computers.union(con[0], con[1]);
        }
        return computers.count() - 1;
    }
}