684. Redundant Connection (Medium)
https://leetcode.com/problems/redundant-connection/
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:
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.
Solutions
class Solution {
// FIXME This is a valuable problem to learn Union Find and Graph.
// Inspired by https://leetcode.com/problems/redundant-connection/discuss/107984/10-line-Java-solution-Union-Find.
// The key idea of this solution is to find the edge the forms a loop. If the two nodes that comprise
// this edge share a same parent node, there will be a loop.
public int[] findRedundantConnection(int[][] edges) {
int[] parent = new int[1001];
for (int i = 0; i < parent.length; i++) {
parent[i] = i;
}
for (int[] edge: edges){
// imagine f->t, t is the parent of f
int f = edge[0];
int t = edge[1];
// with add current edge, check out if f and t shares the same parent
if (find(parent, f) == find(parent, t)) {
return edge;
}
// add this edge to update the parent backtracking map
else {
parent[find(parent, f)] = find(parent, t);
}
}
return new int[2];
}
private int find(int[] parent, int f) {
if (f != parent[f]) {
parent[f] = find(parent, parent[f]);
}
return parent[f];
}
}