684. Redundant Connection

Problem

Tags: Depth-First Search, Breadth-First Search, Union Find, Graph

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.

Code

JS

// 684. Redundant Connection (12/26/53452)
// Runtime: 80 ms (66.46%) Memory: 41.41 MB (94.86%) 

/**
 * @param {number[][]} edges
 * @return {number[]}
 */
var findRedundantConnection = function(edges) {
    let par = Array.from({length: edges.length + 1}, (_,i) => i);
    const find = x => x === par[x] ? par[x] : par[x] = find(par[x]);
    const union = (x,y) => par[find(y)] = find(x);
    for (let [a,b] of edges) {
        if (find(a) === find(b)) return [a,b];
        else union(a,b);
    }
};