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


View More Similar Problems

Delete a Node

Delete the node at a given position in a linked list and return a reference to the head node. The head is at position 0. The list may be empty after you delete the node. In that case, return a null value. Example: list=0->1->2->3 position=2 After removing the node at position 2, list'= 0->1->-3. Function Description: Complete the deleteNode function in the editor below. deleteNo

View Solution →

Print in Reverse

Given a pointer to the head of a singly-linked list, print each data value from the reversed list. If the given list is empty, do not print anything. Example head* refers to the linked list with data values 1->2->3->Null Print the following: 3 2 1 Function Description: Complete the reversePrint function in the editor below. reversePrint has the following parameters: Sing

View Solution →

Reverse a linked list

Given the pointer to the head node of a linked list, change the next pointers of the nodes so that their order is reversed. The head pointer given may be null meaning that the initial list is empty. Example: head references the list 1->2->3->Null. Manipulate the next pointers of each node in place and return head, now referencing the head of the list 3->2->1->Null. Function Descriptio

View Solution →

Compare two linked lists

You’re given the pointer to the head nodes of two linked lists. Compare the data in the nodes of the linked lists to check if they are equal. If all data attributes are equal and the lists are the same length, return 1. Otherwise, return 0. Example: list1=1->2->3->Null list2=1->2->3->4->Null The two lists have equal data attributes for the first 3 nodes. list2 is longer, though, so the lis

View Solution →

Merge two sorted linked lists

This challenge is part of a tutorial track by MyCodeSchool Given pointers to the heads of two sorted linked lists, merge them into a single, sorted linked list. Either head pointer may be null meaning that the corresponding list is empty. Example headA refers to 1 -> 3 -> 7 -> NULL headB refers to 1 -> 2 -> NULL The new list is 1 -> 1 -> 2 -> 3 -> 7 -> NULL. Function Description C

View Solution →

Get Node Value

This challenge is part of a tutorial track by MyCodeSchool Given a pointer to the head of a linked list and a specific position, determine the data value at that position. Count backwards from the tail node. The tail is at postion 0, its parent is at 1 and so on. Example head refers to 3 -> 2 -> 1 -> 0 -> NULL positionFromTail = 2 Each of the data values matches its distance from the t

View Solution →