LeetCode-in-Java

684. Redundant Connection

Medium

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to n, with one additional edge added. The added edge has two different vertices chosen from 1 to n, and was not an edge that already existed. The graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

Return an edge that can be removed so that the resulting graph is a tree of n nodes. If there are multiple answers, return the answer that occurs last in the input.

Example 1:

Input: edges = [[1,2],[1,3],[2,3]]

Output: [2,3]

Example 2:

Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]

Output: [1,4]

Constraints:

• n == edges.length
• 3 <= n <= 1000
• edges[i].length == 2
• 1 <= ai < bi <= edges.length
• ai != bi
• There are no repeated edges.
• The given graph is connected.

Solution

public class Solution {
    private int[] par;

    public int[] findRedundantConnection(int[][] edges) {
        int[] ans = new int[2];
        int n = edges.length;
        par = new int[n + 1];
        for (int i = 0; i < n; i++) {
            par[i] = i;
        }
        for (int[] edge : edges) {
            int lx = find(edge[0]);
            int ly = find(edge[1]);
            if (lx != ly) {
                par[lx] = ly;
            } else {
                ans[0] = edge[0];
                ans[1] = edge[1];
            }
        }
        return ans;
    }

    private int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return find(par[x]);
    }
}

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