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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:

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

Example 2:

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.

Solutions

  • class Solution {
    private int[] p;

    public int[] findRedundantConnection(int[][] edges) {
        p = new int[1010];
        for (int i = 0; i < p.length; ++i) {
            p[i] = i;
        }
        for (int[] e : edges) {
            int a = e[0], b = e[1];
            if (find(a) == find(b)) {
                return e;
            }
            p[find(a)] = find(b);
        }
        return null;
    }

    private int find(int x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
}

  • class Solution {
public:
    vector<int> p;

    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        p.resize(1010);
        for (int i = 0; i < p.size(); ++i) p[i] = i;
        for (auto& e : edges) {
            int a = e[0], b = e[1];
            if (find(a) == find(b)) return e;
            p[find(a)] = find(b);
        }
        return {};
    }

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

  • class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        def find(x):
            if p[x] != x:
                p[x] = find(p[x])
            return p[x]

        p = list(range(1010))
        for a, b in edges:
            if find(a) == find(b):
                return [a, b]
            p[find(a)] = find(b)
        return []


  • func findRedundantConnection(edges [][]int) []int {
	p := make([]int, 1010)
	for i := range p {
		p[i] = i
	}
	var find func(x int) int
	find = func(x int) int {
		if p[x] != x {
			p[x] = find(p[x])
		}
		return p[x]
	}
	for _, e := range edges {
		a, b := e[0], e[1]
		if find(a) == find(b) {
			return e
		}
		p[find(a)] = find(b)
	}
	return []int{}
}

  • /**
 * @param {number[][]} edges
 * @return {number[]}
 */
var findRedundantConnection = function (edges) {
    let p = Array.from({ length: 1010 }, (_, i) => i);
    function find(x) {
        if (p[x] != x) {
            p[x] = find(p[x]);
        }
        return p[x];
    }
    for (let [a, b] of edges) {
        if (find(a) == find(b)) {
            return [a, b];
        }
        p[find(a)] = find(b);
    }
    return [];
};


  • function findRedundantConnection(edges: number[][]): number[] {
    const n = edges.length;
    const p: number[] = Array.from({ length: n }, (_, i) => i);
    const find = (x: number): number => {
        if (p[x] !== x) {
            p[x] = find(p[x]);
        }
        return p[x];
    };
    for (let i = 0; ; ++i) {
        const pa = find(edges[i][0] - 1);
        const pb = find(edges[i][1] - 1);
        if (pa === pb) {
            return edges[i];
        }
        p[pa] = pb;
    }
}

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