# 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:

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

Example 2:

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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

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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][0]);
root2 = find(edges[i][1]);
if (root1 == root2)
return edges[i];
nodes[root2] = root1;
}
return edges.back();
}
};