684. Redundant Connection

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LeetCode

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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 $edges[i] = [a_i, b_i]$ indicates that there is an edge between nodes $a_i$ and $b_i$ 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
• $1 <= a_i < b_i <= edges.length$
• $a_i != b_i$
• There are no repeated edges.
• The given graph is connected.

Code

261. Graph Valid Tree

class Solution {
        public class UnionFind {
        int[] parents;
        int[] ranks;

        public UnionFind(int n) {
            parents = new int[n];
            ranks = new int[n];

            for (int i = 0; i < n; i++) {
                parents[i] = i;
                ranks[i] = 1;
            }
        }

        public boolean union(int x, int y) {
            int xParent = find(x);
            int yParent = find(y);

            if (xParent == yParent) return false;

            if (ranks[xParent] > ranks[yParent]) {
                parents[yParent] = xParent;
            } else if (ranks[xParent] < ranks[yParent]) {
                parents[xParent] = yParent;
            } else {
                parents[yParent] = xParent;
                ranks[xParent]++;
            }

            return true;
        }

        public int find(int node) {
            if(node == parents[node]) return node;

            return find(parents[node]);
        }
    }

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

        for(int[] edge : edges){
            if(!node.union(edge[0], edge[1])){
                return edge;
            }
        }

        return new int[]{-1, -1};
    }
}

class Solution {
    public int[] findRedundantConnection(int[][] edges) {
        HashMap<Integer, List<Integer>> graph = new HashMap<>();

        for(int[] edge : edges) {
            HashSet<Integer> visited = new HashSet<>();
            // 添加边之前看是否已经有连接
            if(helper(graph, visited, edge[0], edge[1])) {
                return edge;
            } else {
                if(!graph.containsKey(edge[0])) {
                    graph.put(edge[0], new ArrayList<>());
                }

                graph.get(edge[0]).add(edge[1]);

                if(!graph.containsKey(edge[1])) {
                    graph.put(edge[1], new ArrayList<>());
                }

                graph.get(edge[1]).add(edge[0]);
            }
        }

        return new int[]{-1, -1};
    }

    private boolean helper(
        HashMap<Integer, List<Integer>> graph,
        HashSet<Integer> visited,
        int x,
        int y
    ) {
        if(x == y) return true;

        if(!graph.containsKey(x)) return false;

        for(int sub : graph.get(x)) {
            if(!visited.add(sub)) continue;

            if(helper(graph, visited, sub, y)) return true;
        }

        return false;
    }
}