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 1, 2, ..., 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 [u, v] that represents a directed edge connecting nodes u and v, where u is a parent of child v.

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: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given directed graph will be like this:
  1
 / \
v   v
2-->3

Example 2:

Input: [[1,2], [2,3], [3,4], [4,1], [1,5]]
Output: [4,1]
Explanation: The given directed graph will be like this:
5 <- 1 -> 2
     ^    |
     |    v
     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.

// Union find 
int getRoot(vector<int>& root, int i) {
    // Recursion
    return root[i] == i ? i : root[i] = getRoot(root, root[i]);
    
    // // Iteration
    // while (root[i] != i) {
    //     root[i] = root[root[i]];
    //     i = root[i];
    // }
    // return i;
}
vector<int> findRedundantDirectedConnection(vector<vector<int>>& edges) { // time: O(n lg*(n)); space: O(n)
    // There are 3 types of violation:
    // 1. A node with indegree = 2
    // 2. Cycle exists
    // 3. Cycle and a node with indegree = 2 both exist
    
    int n = edges.size();
    vector<int> root(n + 1, 0), sz(n + 1, 1), first, second;
    
    // Check if there is any node with two parents, which means indegree = 2
    for (auto& edge : edges) {
        if (root[edge[1]] == 0) { // no parent yet
            root[edge[1]] = edge[0];
        } else {
            first = {root[edge[1]], edge[1]};
            second = edge;
            edge[1] = 0; // skip to find cycle later
            break;
        }
    }
    
    // Union find to check if there is any cycle
    
    // Set each root with a different value at first
    for (int i = 0; i <= n; ++i) root[i] = i;
    
    // Check cycle
    for (auto& edge : edges) {
        // skip if find an edge causing indegree = 2
        if (edge[1] == 0) continue;
        
        int r1 = getRoot(root, edge[0]), r2 = getRoot(root, edge[1]);
        if (r1 == r2) return first.empty() ? edge : first;
        if (sz[r1] < sz[r2]) {
            root[r1] = r2;
            sz[r2] += sz[r1];
        } else {
            root[r2] = r1;
            sz[r1] += sz[r2];
        }
    }
    return second;
}

684. Redundant Connection