Monday, August 17, 2020

684. Redundant Connection -----------M ~~~

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). We apologize for any inconvenience caused. 


    A:

    因为u<v,  很自然想到union 的思路。  

    可是, merge的时候,我们还需要 再map back 回来。 因为不能简单地用指针都指向同一个地址。 (会给后续的修改 造成missing ends)

    为了简单快捷,我这里用了for loop

    class Solution {
    public:
        vector<int> findRedundantConnection(vector<vector<int>>& edges) {
            int N = edges.size();
            vector<int> V(N+1, INT_MAX); // node value as index, V[i] is their smallest parent
            for (auto edge : edges) {
                int u = edge[0];
                int v = edge[1];
                int smallest = min(u, min(V[u], V[v]) );
                
                if(V[u] != INT_MAX && V[u] == V[v]) // if already same parent
                    return vector<int>{u,v};
                
                int preUsmallest = V[u];
                int preVsmallest = V[v];
                for(int i =1;i<=N;i++){
                    if(i==u || i==v || (V[i] != INT_MAX && (V[i] == preUsmallest || V[i] == preVsmallest) )){// if parent 
                        V[i] = smallest;
                    }
                }
            }
            return vector<int>();// shall not hit
        }
    };




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