# 0685. Redundant Connection II

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

## Description

In this problem, a rooted tree is a **directed** graph such that, there is exactly one node (the root) for which all other nodes are descendants of this node, plus every node has exactly one parent, except for the root node which has no parents.

The given input is a directed graph that started as a rooted tree with `n` nodes (with distinct values from `1` to `n`), with one additional directed 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 `[ui, vi]` that represents a **directed** edge connecting nodes `ui` and `vi`, where `ui` is a parent of child `vi`.

Return *an edge that can be removed so that the resulting graph is a rooted tree of* `n` *nodes*. If there are multiple answers, return the answer that occurs last in the given 2D-array.

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/12/20/graph1.jpg)

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

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/12/20/graph2.jpg)

```
**Input:** edges = [[1,2],[2,3],[3,4],[4,1],[1,5]]
**Output:** [4,1]
```

**Constraints:**

* `n == edges.length`
* `3 <= n <= 1000`
* `edges[i].length == 2`
* `1 <= ui, vi <= n`
* `ui != vi`

## ac

```java
class Solution {
    public int[] findRedundantDirectedConnection(int[][] edges) {
        // edge cases

        int n = edges.length;
        int[] fathers = new int[n+1];


        int[] candi1 = null, candi2 = null;
        // find 2 candidates if exist
        for (int[] e : edges) {
            if (fathers[e[1]] != 0) { // e[1] has parent already
                candi1 = new int[]{fathers[e[1]], e[1]};
                candi2 = new int[]{e[0], e[1]};
                e[1] = 0; // set 2nd candidate invalid
            } else { // set fathers to represent direction
                fathers[e[1]] = e[0];
            }
        }

        // build union find
        for (int i = 0; i < fathers.length; i++) fathers[i] = i; // reset
        for (int[] e : edges) {
            int f1 = find(fathers, e[0]);
            int f2 = find(fathers, e[1]);
            if (f1 != f2) { // normal connect
                fathers[f1] = f2;
            } else { // find cycle
                if (candi1 == null) return e; // 1 parent
                else return candi1; // 2 parent
            }
        }

        return candi2;
    }

    private int find(int[] fathers, int i) {
        int res = i;
        while (fathers[res] != res) {
            res = fathers[res];
        }
        return res;
    }
}

/*
3 situations: loop 1 find 2 candidates -> union find -> 1) no cycle, return 2nd candidate; 2) cycle, 1 parent, return current edge; 3) cycle, 2 parent, return 1st candidate.
*/
```


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