Redundant Connection

Problem

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.

Solution

DFS

Union Find

class Solution {
    int[] root;
    int[] rank;

    public int[] findRedundantConnection(int[][] edges) {
        root = new int[edges.length + 1];
        rank = new int[edges.length + 1];

        for (var i = 0; i < edges.length; i++) {
            root[i] = i;
            rank[i] = 1;
        }

        for (var e : edges) {
            if (find(e[0]) == find(e[1])) {
                return e;
            }
            union(e[0], e[1]);
        }

        return null;
    }

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

    void union(int x, int y) {
        var rootX = find(x);
        var rootY = find(y);

        if (rank[rootX] > rank[rootY]) {
            rootX ^= rootY;
            rootY ^= rootX;
            rootX ^= rootY;
        }

        root[rootX] = rootY;
        rank[rootY] += rank[rootX];
    }
}

Recent posts from blogs that I like

An Introduction to Google’s Approach to AI Agent Security

via Simon Willison

Notes on Cramer's rule

Cramer's rule is a clever solution to the classical system of linear equations Ax=b: \[\begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \\ \end{bmatrix} \begin{bmatrix}x_1 \\ x_2 \\ x_3\end{bmatrix} = \begin{bmatrix}b_1 \\ b_2 \\ b_3\end{bmatrix}\] Usi...

via Eli Bendersky

Brandjes: Paintings as witnesses to fires 1640-1813

Dramatic paintings of towns and cities on fire, usually at night, were popular during the Dutch Golden Age, and known as brandjes. Examples to well into the 19th century.

via The Eclectic Light Company