# 684. Redundant Connection (Medium)

https://leetcode.com/problems/redundant-connection/

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:
1
/ \
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
```

Note:

• 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.

## Solutions

``````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];
}
}
``````