Problem
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.
class Solution { public int[] findRedundantConnection(int[][] edges) { UnionFind uf = new UnionFind(edges.length+1); for (int[] edge: edges) { int a = edge[0], b = edge[1]; if (uf.find(a) == uf.find(b)) return edge; else uf.union(a, b); } return new int[0]; } } class UnionFind { int[] parents; public UnionFind(int n) { parents = new int[n]; for (int i = 0; i < n; i++) parents[i] = i; } public int find(int a) { if (parents[a] != a) return find(parents[a]); else return a; } public void union(int a, int b) { int pA = find(a), pB = find(b); if (pA != pB) parents[pB] = pA; } }
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