# Redundant Connection

`Union Find`, `Depth-first Search`

**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:
  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**](https://leetcode.com/problems/redundant-connection-ii/description/)). We apologize for any inconvenience caused.

## Solution

### Union Find - *Path Compression* + *Union-By-Rank*

2 ms, faster than 100.00%

```java
class Solution {
    class UnionFind {
        int[] parent;
        int[] rank;
        int size;

        public UnionFind(int size) {
            parent = new int[size];
            for (int i = 0; i < size; i++) {
                parent[i] = i;
            }
            rank = new int[size];
            this.size = size;
        }

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

        public boolean union(int x, int y) {
            int xp = find(x);
            int yp = find(y);
            if (xp == yp) {
                return false;
            }
            if (rank[xp] < rank[yp]) {
                parent[xp] = yp;
            } else if (rank[xp] > rank[yp]) {
                parent[yp] = xp;
            } else {
                parent[yp] = xp;
                rank[xp]++;
            }

            return true;
        }
    }
    public int[] findRedundantConnection(int[][] edges) {
        //  int N = 1001;
        int N = edges.length;
        UnionFind uf = new UnionFind(N);
        for (int[] edge: edges) {
            if (!uf.union(edge[0] - 1, edge[1] - 1)) {
                return edge;
            }
        }
        return new int[0];
    }

}
```

## Reference

<https://leetcode.com/problems/redundant-connection/solution/>


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