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

Solution: Union Find

• Time complexity: O(n)
• Space complexity: O(n)

 

JavaScript

/**
 * @param {number[][]} edges
 * @return {number[]}
 */
const findRedundantConnection = function (edges) {
  const n = edges.length;
  const groups = Array.from({ length: n + 1 }, (_, index) => index);
  const ranks = Array.from({ length: n + 1 }, () => 1);

  const find = node => {
    const group = groups[node];

    return group === node ? node : find(group);
  };

  const union = (a, b) => {
    const x = find(a);
    const y = find(b);

    if (x === y) return false;
    if (ranks[x] > ranks[y]) {
      groups[y] = x;
    } else if (ranks[x] < ranks[y]) {
      groups[x] = y;
    } else {
      groups[y] = x;
      ranks[x] += 1;
    }
    return true;
  };

  for (const [a, b] of edges) {
    if (!union(a, b)) return [a, b];
  }
  return [];
};