leetcode_684

In this problem, a tree is an undirected graph that is connected and has no cycles.

The given input is a graph that started as a tree with N nodes (with distinct values 1, 2, ..., 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 resulting graph is given as a 2D-array of edges. Each element of edges is a pair [u, v] with u < v, that represents an undirected edge connecting nodes u and v.

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 given 2D-array. The answer edge [u, v] should be in the same format, with u < v.

Example 1:

Input: [[1,2], [1,3], [2,3]]
Output: [2,3]
Explanation: The given undirected graph will be like this:
  1
 / \
2 - 3

Example 2:

Input: [[1,2], [2,3], [3,4], [1,4], [1,5]]
Output: [1,4]
Explanation: The given undirected graph will be like this:
5 - 1 - 2
    |   |
    4 - 3

Note:

The size of the input 2D-array will be between 3 and 1000. Every integer represented in the 2D-array will be between 1 and N, where N is the size of the input array.

Update (2017-09-26):

We have overhauled the problem description + test cases and specified clearly the graph is an undirected graph. For the directed graph follow up please see Redundant Connection II). We apologize for any inconvenience caused.

Solutions

1. Union Find O(n)

2. Return the first edge whose nodes can be merged. ie: they are already connected.

class UnionFind {
private:
    int * nodes;
    int * sizes;
public:
    UnionFind(int size) {
        nodes = new int[size];
        sizes = new int[size];
        for (int i = 0; i < size; i++) {
            nodes[i] = i;
            sizes[i] = 1;
        }
    }
    ~UnionFind() {
        delete [] nodes; delete [] sizes;
    }
    int find(int node) {
        while (nodes[node] != node)
            node = nodes[node] = nodes[nodes[node]];
        return node;
    }
    bool merge(int node1, int node2) {
        int f1 = find(node1);
        int f2 = find(node2);
        if (f1 == f2)
            return false;
        else {
            if (sizes[f1] > sizes[f2])
                swap(f1, f2);
            nodes[f1] = f2;
            sizes[f2] += sizes[f1];
            return true;
        }
    }
};

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        // the maxinum i will be edges.size(), plus 1, otherwise will cause out of bounds
        UnionFind uf(edges.size() + 1);
        for (auto & e : edges)
            if (!uf.merge(e[0], e[1]))
                return e;
        return {0, 0};
    }
};

Or a python version

class Solution:
    def findRedundantConnection(self, edges: List[List[int]]) -> List[int]:
        def find(node):
            return find(tree[node]) if node in tree else node

        tree = {}
        for i, edge in enumerate(edges):
            f1, f2 = map(find, edge)
            if f1 == f2:
                return edge
            else:
                tree[f1] = f2

1. dfs O(n2)

2. Use dfs to check if there is a cycle between two nodes after added a new edges into the graph.

3. We can also use bfs traversal with queue.

class Solution {
public:
    bool hascycle(int src, int target, int pre, vector<vector<int>> & adj) {
        if (src == target)
            return true;
        for (auto & outnode : adj[src]) {
            if (outnode == pre) continue;
            if (hascycle(outnode, target, src, adj))
                return true;
        }
        return false;
    }
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size() + 1;
        vector<vector<int>> adj(n, vector<int>());

        for (auto & e : edges) {
            if (hascycle(e[0], e[1], 0, adj))
                return e;
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
        }

        return {0, 0};
    }
};

1. Topological sort

2. Reference: https://leetcode-cn.com/problems/redundant-connection/comments/179407
3. The idea is to itratively remove nodes with 1 degree, then start checking from the last edge(nodes pair), find the first pair with degree larger than 1.(degrees of both nodes are greater than 1).

class Solution {
public:
    vector<int> findRedundantConnection(vector<vector<int>>& edges) {
        int n = edges.size() + 1;
        vector<vector<int>> adj(n, vector<int>());
        vector<int> degree(n);

        for (auto & e : edges) {
            adj[e[0]].push_back(e[1]);
            adj[e[1]].push_back(e[0]);
            degree[e[0]]++; degree[e[1]]++;
        }

        queue<int> q;
        vector<bool> visited(n);
        for (int i = 1; i < n; i++)
            if (degree[i] == 1) {
                q.push(i);
                visited[i] = true;
            }

        while (!q.empty()) {
            auto cur = q.front(); q.pop();
            for (auto outnode : adj[cur]) {
                if (visited[outnode]) continue;
                degree[cur]--;
                if (--degree[outnode] == 1) {
                    q.push(outnode);
                    visited[cur] = true;
                }
            }
        }
        // check from the back, find the first nodes pair with degree(minimum of two nodes) greater 1.
        for (int i = edges.size() - 1; i >= 0; i--) {
            auto & e = edges[i];
            if (min(degree[e[0]], degree[e[1]]) > 1)
                return e;
        }

        return {0, 0};
    }
};