LeetCode 685. Redundant Connection II

原题链接在这里：https://leetcode.com/problems/redundant-connection-ii/

题目：

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.

题解：

If it is a invalid tree, there could be 2 cases:

• one node has 2 parents. [i, j], [k, j], two edges both point to j.
• there is cyrcle.

If remove one redundant edge could make it a valid tree.

If case 1 happens, then redundant edge must be either [i, j] or [k, j]. Otherwise, even you remove the redundant edge, [i, j] and [k, j] still point to the same node j and it is still invalid. Thus make them candidate 1 and candidate 2.

We check if cycle exists, if it exists, we check if case 1 happens or not. If no, then edge contecting 2 nodes already within the same union is the redundant edge likeRedundant Connection.

If yes, redundant edge is either candiate 1 or 2. First remove candidate 2 and check if cycle still exists, if no then answer is candidate 2, otherwise it is candidate 1.

Note: only update parent if parent[edge[1]]  == 0.

Time Complexity: O(nlogn). find takes O(logn).

Space: O(n).

AC Java:

class Solution {
    int [] parent;
    
    public int[] findRedundantDirectedConnection(int[][] edges) {
        int n = edges.length;
        parent = new int[n+1];
        
        int [] can1 = new int[]{-1, -1};
        int [] can2 = new int[]{-1, -1};
        for(int i = 0; i<n; i++){
            if(parent[edges[i][1]] == 0){
                parent[edges[i][1]] = edges[i][0];
            }else{
                can2 = new int[]{edges[i][0], edges[i][1]};
                can1 = new int[]{parent[edges[i][1]], edges[i][1]};
                edges[i][1] = 0;
            }
        }
        
        for(int i = 0; i<=n; i++){
            parent[i] = i;
        }
        
        for(int [] edge : edges){
            if(find(edge[0]) == find(edge[1])){
                if(can1[0] == -1){
                    return edge;
                }
                
                return can1;
            }
            
            union(edge[0], edge[1]);
        }
        
        return can2;
    }
    
    private int find(int i){
        if(i != parent[i]){
            parent[i] = find(parent[i]);
        }
        
        return parent[i];
    }
    
    private void union(int i, int j){
        int p = find(i);
        int q = find(j);
        parent[q] = p;
    }
}

类似Redundant Connection.