684. Redundant Connection (Medium)


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

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., 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 resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

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 given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
 / \
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
    |   |
    4 - 3


  • The size of the input 2D-array will be between 3 and 1000.
  • Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

  • Update (2017-09-26):
    We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.


    class Solution {
        // FIXME This is a valuable problem to learn Union Find and Graph.
        // Inspired by https://leetcode.com/problems/redundant-connection/discuss/107984/10-line-Java-solution-Union-Find.
        // The key idea of this solution is to find the edge the forms a loop. If the two nodes that comprise
        // this edge share a same parent node, there will be a loop.
        public int[] findRedundantConnection(int[][] edges) {
            int[] parent = new int[1001];
            for (int i = 0; i < parent.length; i++) {
                parent[i] = i;
            for (int[] edge: edges){
                // imagine f->t, t is the parent of f
                int f = edge[0];
                int t = edge[1];
                // with add current edge, check out if f and t shares the same parent
                if (find(parent, f) == find(parent, t)) {
                    return edge;
                // add this edge to update the parent backtracking map
                else {
                    parent[find(parent, f)] = find(parent, t);
            return new int[2];
        private int find(int[] parent, int f) {
            if (f != parent[f]) {
                parent[f] = find(parent, parent[f]);
            return parent[f];

    Incorrect Solutions


    Copyright © iovi.com 2017 all right reserved,powered by GitbookLast Modification: 2019-12-03 11:01:18

    results matching ""

      No results matching ""