# 684. Redundant Connection

## Description

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.

## Solutions

• class Solution {
private int[] p;

public int[] findRedundantConnection(int[][] edges) {
p = new int[1010];
for (int i = 0; i < p.length; ++i) {
p[i] = i;
}
for (int[] e : edges) {
int a = e[0], b = e[1];
if (find(a) == find(b)) {
return e;
}
p[find(a)] = find(b);
}
return null;
}

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

• class Solution {
public:
vector<int> p;

vector<int> findRedundantConnection(vector<vector<int>>& edges) {
p.resize(1010);
for (int i = 0; i < p.size(); ++i) p[i] = i;
for (auto& e : edges) {
int a = e[0], b = e[1];
if (find(a) == find(b)) return e;
p[find(a)] = find(b);
}
return {};
}

int find(int x) {
if (p[x] != x) p[x] = find(p[x]);
return p[x];
}
};

• class Solution:
def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
def find(x):
if p[x] != x:
p[x] = find(p[x])
return p[x]

p = list(range(1010))
for a, b in edges:
if find(a) == find(b):
return [a, b]
p[find(a)] = find(b)
return []


• func findRedundantConnection(edges [][]int) []int {
p := make([]int, 1010)
for i := range p {
p[i] = i
}
var find func(x int) int
find = func(x int) int {
if p[x] != x {
p[x] = find(p[x])
}
return p[x]
}
for _, e := range edges {
a, b := e[0], e[1]
if find(a) == find(b) {
return e
}
p[find(a)] = find(b)
}
return []int{}
}

• /**
* @param {number[][]} edges
* @return {number[]}
*/
var findRedundantConnection = function (edges) {
let p = Array.from({ length: 1010 }, (_, i) => i);
function find(x) {
if (p[x] != x) {
p[x] = find(p[x]);
}
return p[x];
}
for (let [a, b] of edges) {
if (find(a) == find(b)) {
return [a, b];
}
p[find(a)] = find(b);
}
return [];
};