## Algorithm

Problem Name: 685. Redundant Connection II

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:

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

Example 2:

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

## Code Examples

### #1 Code Example with C++ Programming

Code - C++ Programming

class Solution {
public:
vector<int> findRedundantDirectedConnection(vector < vector<int>>& edges) {
vector<int>res;
unordered_map < int, int>parent;
unordered_map<int, vector<int>>g;
// c0: the last edge to form a circle
// c1, c2: parents of the node who has two parents
vector<int>c0, c1, c2;
int rootCandidate = 0, root = 0, count = 0;
for(auto x: edges){
int u = x[0], v = x[1];
if(parent[v] > 0){
c1 = x;
c2 = {parent[v], v};
}
if(!parent[u]){
rootCandidate++;
parent[u] = -1;
}
if(parent[v] == -1) rootCandidate--;
parent[v] = u;
if(rootCandidate == 0) c0 = x, rootCandidate = -1; // No valid root can be found, circle detected
g[u].push_back(v);
}
for(auto x: parent) if(x.second == -1) root = x.first;
int nodes = parent.size();
vector<int>visited(nodes + 1);
return !root ? c0 : (!hasCircle(g, -1, root, c1, visited, count) && count == nodes) ? c1 : c2;
}

bool hasCircle(unordered_map<int, vector < int>>& g, int from, int root, vector<int>c, vector<int>& visited, int& count){
if(from == c[0] && root == c[1]) return false;
if(visited[root]) return true;
count++;
visited[root] = 1;
bool foundCircle = false;
for(auto x: g[root]) foundCircle |= hasCircle(g, root, x, c, visited, count>;
return foundCircle;
}
};
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Input

cmd
edges = [[1,2],[1,3],[2,3]]

Output

cmd
[2,3]

### #2 Code Example with Javascript Programming

Code - Javascript Programming

const findRedundantDirectedConnection = function (edges) {
const parent = []
//detect circle
for (let i = 1; i  < = edges.length; i++) {
parent[i] = i
}
let circleEdge, removedEdge, candidateEdge
for (let i = 0; i  <  edges.length; i++) {
const [u, v] = edges[i]
const pu = findParent(parent, u)
const pv = findParent(parent, v)
if (pv !== v) {
removedEdge = [u, v] // node with 2 parents
} else {
if (pv === pu) {
circleEdge = [u, v] // circle edge
}
parent[v] = pu
}
}
if (!removedEdge) {
return circleEdge
}
if (circleEdge) {
return edges.find((d) => d[1] === removedEdge[1] && d[0] !== removedEdge[0])
} else {
return removedEdge
}
}
const findParent = function (parent, i) {
if (parent[i] !== i) {
parent[i] = findParent(parent, parent[i])
}
return parent[i]
}
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Input

cmd
edges = [[1,2],[1,3],[2,3]]

Output

cmd
[2,3]

### #3 Code Example with Python Programming

Code - Python Programming

class Solution:
def findRedundantDirectedConnection(self, edges):
def root(i):
return parent[i] == i and i or root(parent[i])

parent, a, b, c = [0] * (len(edges) + 1), None, None, None
for i, edge in enumerate(edges):
if parent[edge[1]]:
a, b, c, edges[i][0]= parent[edge[1]], edge[0], edge[1], 0
else:
parent[edge[1]] = edge[0]

parent = [i for i in range(len(edges) + 1)]
for u, v in edges:
if u:
if root(u) == v:
return a and [a, c] or [u, v]
parent[v] = u
return [b, c]
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Input

cmd
edges = [[1,2],[2,3],[3,4],[4,1],[1,5]]

Output

cmd
[4,1]