Shortest Cycle Containing Target Node - Google Top Interview Questions


Problem Statement :


You are given a two-dimensional list of integers graph representing a directed graph as an adjacency list. You are also given an integer target.

Return the length of a shortest cycle that contains target. If a solution does not exist, return -1.

Constraints

n, m ≤ 250 where n and m are the number of rows and columns in graph


Example 1

Input

Visualize

graph = [

    [1],


    [2],


    [0]


]

target = 0

Output

3

Explanation

The nodes 0 -> 1 -> 2 -> 0 form a cycle




Example 2

Input

Visualize

graph = [

    [1],

    [2],

    [4],

    [],


    [0]

]

target = 3


Output
-1


Solution :



title-img 




                        Solution in C++ :

vector<int> color;
vector<long long int> dist;
vector<vector<int>> adj;
int dest;
long long int dfs(int u) {
    if (u == dest) {
        return 0;
    }
    color[u] = 1;  // gray
    long long int mindist = INT_MAX;
    for (int v : adj[u]) {
        if (color[v] == 0) {
            mindist = min(mindist, 1 + dfs(v));
        } else if (color[v] == 2) {
            mindist = min(mindist, 1 + dist[v]);
        }
    }
    color[u] = 2;  // black
    return dist[u] = mindist;
}

int solve(vector<vector<int>>& graph, int target) {
    adj = graph;
    dest = target;
    color.assign(graph.size(), 0);
    dist.assign(graph.size(), 0);
    long long int res = INT_MAX;
    for (int i : graph[target]) {
        res = min(res, 1 + dfs(i));
    }
    if (res == INT_MAX) return -1;
    return res;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public int solve(int[][] graph, int target) {
        boolean[] visited = new boolean[graph.length];
        Queue<Integer> q = new LinkedList<>();
        q.add(target);

        int level = 0;
        while (!q.isEmpty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                int tmp = q.remove();
                visited[tmp] = true;

                for (int neighbor : graph[tmp]) {
                    if (neighbor == target)
                        return level + 1;
                    else if (!visited[neighbor])
                        q.add(neighbor);
                }
            }

            level++;
        }

        return -1;
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, graph, target):
        visited = set()
        layer = [target]
        length = 0

        while layer:
            length += 1
            next_layer = []
            for u in layer:
                for v in graph[u]:
                    if v == target:
                        return length
                    if v in visited:
                        continue
                    visited.add(v)
                    next_layer.append(v)
            layer = next_layer

        return -1


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