Common Reachable Node - Facebook Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers edges representing a directed graph and integers a and b. Each element in edges contains [u, v] meaning there's an edge from u to v.

Return whether there's some node c such that we can go from c to a and c to b.

Constraints

1 ≤ n ≤ 100,000 where n is the length of edges

Example 1

Input



edges = [

    [0, 4],

    [4, 3],

    [1, 2],

    [0, 1],

    [0, 2],

    [1, 1]

]

a = 2

b = 3

Output

True

Explanation

We can reach 2 and 3 from 0

Example 2

Input

edges = [

    [0, 1],

    [1, 2],

    [1, 3],

    [1, 1],

    [4, 5]

]

a = 2

b = 4

Output

False

Explanation

2 and 4 are on different connected components.



Example 3

Input



edges = [

    [0, 0]

]

a = 0

b = 0

Output

True



Solution :



title-img




                        Solution in C++ :

vector<int> adj[100001];
vector<bool> vis(100001, false);
unordered_map<int, int> mp;
bool flag;
void dfs(int curr) {
    vis[curr] = true;
    mp[curr]++;
    if (mp[curr] > 1) {
        flag = true;
        return;
    }
    for (int node : adj[curr])
        if (!vis[node]) dfs(node);
}
bool solve(vector<vector<int>>& edges, int a, int b) {
    int n = edges.size() + 1;
    if (a == b) return true;
    for (int i = 0; i < n; i++) adj[i].clear();
    for (int i = 0; i < n - 1; i++) {
        adj[edges[i][1]].push_back(edges[i][0]);
    }
    mp.clear();
    flag = false;
    for (int i = 0; i < n; i++) vis[i] = false;
    dfs(a);
    for (int i = 0; i < n; i++) vis[i] = false;
    dfs(b);
    return flag;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    static ArrayList<Integer>[] transpose;
    static int N;
    public boolean solve(int[][] edges, int a, int b) {
        N = 0;
        for (int[] e : edges) N = Math.max(N, Math.max(e[0], e[1]));
        N++;

        // build the transpose of the graph
        transpose = new ArrayList[N];
        for (int i = 0; i < N; i++) transpose[i] = new ArrayList<Integer>();
        for (int[] e : edges) transpose[e[1]].add(e[0]);

        // find ancestors of A and B
        boolean[] visA = bfs(a);
        boolean[] visB = bfs(b);

        // cehck if there is any overlap
        for (int i = 0; i < N; i++) {
            if (visA[i] && visB[i]) {
                // node i is a common ancestor!
                return true;
            }
        }
        return false;
    }

    public boolean[] bfs(int root) {
        ArrayDeque<Integer> q = new ArrayDeque<Integer>();
        boolean[] vis = new boolean[N];
        vis[root] = true;
        q.add(root);
        while (!q.isEmpty()) {
            int n = q.pollFirst();
            for (int p : transpose[n]) {
                if (!vis[p]) {
                    vis[p] = true;
                    q.add(p);
                }
            }
        }
        return vis;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, edges, a, b):
        graph = defaultdict(set)
        for u, v in edges:
            graph[v].add(u)

        def explore(root):
            queue = [root]
            seen = {root}
            for node in queue:
                for nei in graph[node] - seen:
                    seen.add(nei)
                    queue.append(nei)
            return seen

        return bool(explore(a) & explore(b))
                    


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