684. Redundant Connection

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684. Redundant Connection

Description

In this problem, a tree is an undirected graph that is connected and has no cycles.

You are given a graph that started as a tree with n nodes labeled from 1 to 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 graph is represented as an array edges of length n where edges[i] = [ai, bi] indicates that there is an edge between nodes ai and bi in the graph.

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 input.

Example 1:

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Input: edges = [[1,2],[1,3],[2,3]]
Output: [2,3]

Example 2:

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Input: edges = [[1,2],[2,3],[3,4],[1,4],[1,5]]
Output: [1,4]

Constraints:

  • n == edges.length
  • 3 <= n <= 1000
  • edges[i].length == 2
  • 1 <= ai < bi <= edges.length
  • ai != bi
  • There are no repeated edges.
  • The given graph is connected.

Hints/Notes

  • 2024/12/31
  • union find
  • No solution fromm 0x3F

Solution

Language: C++

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class Solution {
public:
    vector<int> roots;

    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size();
        roots.resize(n + 1);
        iota(roots.begin(), roots.end(), 0);
        for (int i = 0; i < n; i++) {
            auto& e = edges[i];
            int u = e[0], v = e[1];
            int rootU = findRoot(u), rootV = findRoot(v);
            if (rootU == rootV) {
                return e;
            }
            roots[rootU] = rootV;
        }
        return {};
    }

    int findRoot(int root) {
        if (roots[root] != root) {
            root = findRoot(roots[root]);
        }
        return root;
    }
};