# Redundant Connection Problem

## Description

LeetCode Problem 684.

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] = [a_i, b_i] indicates that there is an edge between nodes a_i and b_i 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:

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

Example 2:

``````1
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 <= a_i < b_i <= edges.length
• a_i != b_i
• There are no repeated edges.
• The given graph is connected.

## Sample C++ Code using Union Find

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
class Solution {
public:
vector<int> nodes;

int find(int p) {
while (nodes[p] != p) {
p = nodes[p];
}
return p;
}
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
int n = edges.size();
nodes.push_back(0);
for (int i = 1; i <= n; i ++) {
nodes.push_back(i);
}

int root1, root2;
for (int i = 0; i < n; i ++) {
root1 = find(edges[i]);
root2 = find(edges[i]);

if (root1 == root2)
return edges[i];

nodes[root2] = root1;
}

return edges.back();
}
};
``````