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.
// OJ: https://leetcode.com/problems/redundant-connection/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(N)
class UnionFind {
private:
vector<int> id, rank;
int cnt;
public:
UnionFind(int cnt) : cnt(cnt) {
id = vector<int>(cnt);
rank = vector<int>(cnt, 0);
for (int i = 0; i < cnt; ++i) id[i] = i;
}
int find(int p) {
if (id[p] == p) return p;
return id[p] = find(id[p]);
}
int getCount() { return cnt; }
bool connected(int p, int q) { return find(p) == find(q); }
void connect(int p, int q) {
int i = find(p), j = find(q);
if (i == j) return;
if (rank[i] < rank[j]) id[i] = j;
else {
id[j] = i;
if (rank[i] == rank[j]) rank[j]++;
}
--cnt;
}
};
class Solution {
public:
vector<int> findRedundantConnection(vector<vector<int>>& edges) {
UnionFind uf(edges.size());
for (auto e : edges) {
if (uf.connected(e[0] - 1, e[1] - 1)) return e;
uf.connect(e[0] - 1, e[1] - 1);
}
}
};