Redundant Connection

Union Find, Depth-first Search

Medium

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 ofedges. Each element ofedgesis a pair[u, v]withu < v, that represents anundirectededge connecting nodesuandv.

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, withu < 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 anundirectedgraph. For thedirectedgraph follow up please seeRedundant Connection II). We apologize for any inconvenience caused.

Solution

Union Find - Path Compression + Union-By-Rank

2 ms, faster than 100.00%

class Solution {
class UnionFind {
int[] parent;
int[] rank;
int size;

public UnionFind(int size) {
parent = new int[size];
for (int i = 0; i < size; i++) {
parent[i] = i;
}
rank = new int[size];
this.size = size;
}

public int find(int x) {
if (parent[x] != x) {
parent[x] = find(parent[x]);
}
return parent[x];
}

public boolean union(int x, int y) {
int xp = find(x);
int yp = find(y);
if (xp == yp) {
return false;
}
if (rank[xp] < rank[yp]) {
parent[xp] = yp;
} else if (rank[xp] > rank[yp]) {
parent[yp] = xp;
} else {
parent[yp] = xp;
rank[xp]++;
}

return true;
}
}
public int[] findRedundantConnection(int[][] edges) {
//  int N = 1001;
int N = edges.length;
UnionFind uf = new UnionFind(N);
for (int[] edge: edges) {
if (!uf.union(edge - 1, edge - 1)) {
return edge;
}
}
return new int;
}

}

Reference

https://leetcode.com/problems/redundant-connection/solution/