Given an unweighted graph and a boolean array A[ ], where if the ith index of array A[ ] denotes if that node can be visited (0) or not (1). The task is to find the shortest path to reach (N – 1)th node from the 0th node. If it is not possible to reach, print -1.
Examples :
Input : N = 5, A[] = {0, 1, 0, 0, 0}, Edges = {{0, 1}, {0, 2}, {1, 4}, {2, 3}, {3, 4}}
Output: 3
Explanation: There are two paths from 0th house to 4th house
- 0 ? 1 ? 4
- 0 ?2 ? 3 ? 4
Since a policeman is present at the 1st house, the only path that can be chosen is the 2nd path.
Input : N = 4, A[] = {0, 1, 1, 0}, Edges = {{0, 1}, {0, 2}, {1, 3}, {2, 3}}
Output : -1
Approach: This problem is similar to finding the shortest path in an unweighted graph. Therefore, the problem can be solved using BFS.
Follow the steps below to solve the problem:
- Initialize an unordered_map, say adj to store the edges. The edge (a, b) must be excluded if there is a policeman either at node a or at node b.
- Initialize a variable, pathLength = 0.
- Initialize a vector of the boolean data type, say visited, to store whether a node is visited or not.
- Initialize an arraydist[0, 1, …., v-1] such that dist[i] stores the distance of vertex i from the root vertex
- Initialize a queue and push node 0 in it. Also, mark node 0 as visited.
- Iterate while the queue is not empty and the node N – 1 is not visited, pop the front element from the queue, and push all the elements into the queue that have an edge from the front element of the queue and are not visited and increase the distance of all these nodes by 1 + dist[q.top()].
- If the node (N – 1) is not visited, then print -1.
- Otherwise, print the distance of (N – 1)th node from the root node.
Below is the implementation of the above approach:
C++
// C++ program for the above approach #include <bits/stdc++.h> using namespace std; // Function to create graph edges // where node A and B can be visited void createGraph(unordered_map< int , vector< int > >& adj, int paths[][2], int A[], int N, int E) { // Visit all the connections for ( int i = 0; i < E; i++) { // If a policeman is at any point of // connection, leave that connection. // Insert the connect otherwise. if (!A[paths[i][0]] && !A[paths[i][1]]) { adj[paths[i][0]].push_back(paths[i][1]); } } } // Function to find the shortest path int minPath( int paths[][2], int A[], int N, int E) { // If police is at either at // the 1-st house or at N-th house if (A[0] == 1 || A[N - 1] == 1) // The thief cannot reach // the N-th house return -1; // Stores Edges of graph unordered_map< int , vector< int > > adj; // Function call to store connections createGraph(adj, paths, A, N, E); // Stores whether node is // visited or not vector< int > visited(N, 0); // Stores distances // from the root node int dist[N]; dist[0] = 0; queue< int > q; q.push(0); visited[0] = 1; // Visit all nodes that are // currently in the queue while (!q.empty()) { int temp = q.front(); q.pop(); for ( auto x : adj[temp]) { // If current node is // not visited already if (!visited[x]) { q.push(x); visited[x] = 1; dist[x] = dist[temp] + 1; } } } if (!visited[N - 1]) return -1; else return dist[N - 1]; } // Driver Code int main() { // N : Number of houses // E: Number of edges int N = 5, E = 5; // Given positions int A[] = { 0, 1, 0, 0, 0 }; // Given Paths int paths[][2] = { { 0, 1 }, { 0, 2 }, { 1, 4 }, { 2, 3 }, { 3, 4 } }; // Function call cout << minPath(paths, A, N, E); return 0; } |
Java
// Java program for the above approach import java.io.*; import java.lang.*; import java.util.*; public class GFG { // Function to create graph edges // where node A and B can be visited static void createGraph(HashMap<Integer, ArrayList<Integer> > adj, int paths[][], int A[], int N, int E) { // Visit all the connections for ( int i = 0 ; i < E; i++) { // If a policeman is at any point of // connection, leave that connection. // Insert the connect otherwise. if (A[paths[i][ 0 ]] != 1 && A[paths[i][ 1 ]] != 1 ) { ArrayList<Integer> list = adj.getOrDefault( paths[i][ 0 ], new ArrayList<>()); list.add(paths[i][ 1 ]); adj.put(paths[i][ 0 ], list); } } } // Function to find the shortest path static int minPath( int paths[][], int A[], int N, int E) { // If police is at either at // the 1-st house or at N-th house if (A[ 0 ] == 1 || A[N - 1 ] == 1 ) // The thief cannot reach // the N-th house return - 1 ; // Stores Edges of graph HashMap<Integer, ArrayList<Integer> > adj = new HashMap<>(); // Function call to store connections createGraph(adj, paths, A, N, E); // Stores whether node is // visited or not boolean visited[] = new boolean [N]; // Stores distances // from the root node int dist[] = new int [N]; dist[ 0 ] = 0 ; ArrayDeque<Integer> q = new ArrayDeque<>(); q.addLast( 0 ); visited[ 0 ] = true ; // Visit all nodes that are // currently in the queue while (!q.isEmpty()) { int temp = q.removeFirst(); for ( int x : adj.getOrDefault( temp, new ArrayList<>())) { // If current node is // not visited already if (!visited[x]) { q.addLast(x); visited[x] = true ; dist[x] = dist[temp] + 1 ; } } } if (!visited[N - 1 ]) return - 1 ; else return dist[N - 1 ]; } // Driver Code public static void main(String[] args) { // N : Number of houses // E: Number of edges int N = 5 , E = 5 ; // Given positions int A[] = { 0 , 1 , 0 , 0 , 0 }; // Given Paths int paths[][] = { { 0 , 1 }, { 0 , 2 }, { 1 , 4 }, { 2 , 3 }, { 3 , 4 } }; // Function call System.out.print(minPath(paths, A, N, E)); } } // This code is contributed by Kingash. |
Python3
# Python program for the above approach # Stores Edges of graph adj = {}; # Function to create graph edges # where node A and B can be visited def createGraph(paths, A, N, E): # Visit all the connections for i in range (E): # If a policeman is at any point of # connection, leave that connection. # Insert the connect otherwise. if ( not A[paths[i][ 0 ]] and not A[paths[i][ 1 ]]) : if (paths[i][ 0 ] in adj): tmp = adj[paths[i][ 0 ]]; tmp.append(paths[i][ 1 ]); adj[paths[i][ 0 ]] = tmp; else : tmp = []; tmp.append(paths[i][ 1 ]); adj[paths[i][ 0 ]] = tmp; # Function to find the shortest path def minPath(paths, A, N, E): # If police is at either at # the 1-st house or at N-th house if (A[ 0 ] = = 1 or A[N - 1 ] = = 1 ): # The thief cannot reach # the N-th house return - 1 ; # Function call to store connections createGraph(paths, A, N, E); # Stores whether node is # visited or not visited = [ 0 ] * N # Stores distances # from the root node dist = [ 0 ] * N dist[ 0 ] = 0 ; q = []; q.append( 0 ); visited[ 0 ] = 1 ; # Visit all nodes that are # currently in the queue while ( len (q) ! = 0 ): temp = q[ 0 ]; q.pop(); if (temp in adj): for x in adj[temp] : # If current node is # not visited already if ( not visited[x]): q.append(x); visited[x] = 1 ; dist[x] = dist[temp] + 1 ; if ( not visited[N - 1 ]): return - 1 ; else : return dist[N - 1 ]; # Driver Code # N : Number of houses # E: Number of edges N = 5 E = 5 ; # Given positions A = [ 0 , 1 , 0 , 0 , 0 ]; # Given Paths paths = [ [ 0 , 1 ], [ 0 , 2 ], [ 1 , 4 ], [ 2 , 3 ], [ 3 , 4 ] ]; # Function call print (minPath(paths, A, N, E)); # This code is contributed by Saurabh Jaiswal |
C#
using System; using System.Collections.Generic; using System.Linq; namespace GFG { class Program { static void Main( string [] args) { // N : Number of houses // E: Number of edges int N = 5, E = 5; // Given positions int [] A = { 0, 1, 0, 0, 0 }; // Given Paths int [][] paths = { new int [] { 0, 1 }, new int [] { 0, 2 }, new int [] { 1, 4 }, new int [] { 2, 3 }, new int [] { 3, 4 } }; // Function call Console.WriteLine(MinPath(paths, A, N, E)); } // Function to create graph edges // where node A and B can be visited static void CreateGraph(Dictionary< int , List< int > > adj, int [][] paths, int [] A, int N, int E) { // Visit all the connections for ( int i = 0; i < E; i++) { // If a policeman is at any point of // connection, leave that connection. // Insert the connect otherwise. if (A[paths[i][0]] != 1 && A[paths[i][1]] != 1) { List< int > list = adj.ContainsKey(paths[i][0]) ? adj[paths[i][0]] : new List< int >(); list.Add(paths[i][1]); adj[paths[i][0]] = list; } } } // Function to find the shortest path static int MinPath( int [][] paths, int [] A, int N, int E) { // If police is at either at // the 1-st house or at N-th house if (A[0] == 1 || A[N - 1] == 1) // The thief cannot reach // the N-th house return -1; // Stores Edges of graph Dictionary< int , List< int > > adj = new Dictionary< int , List< int > >(); // Function call to store connections CreateGraph(adj, paths, A, N, E); // Stores whether node is // visited or not bool [] visited = new bool [N]; // Stores distances // from the root node int [] dist = new int [N]; dist[0] = 0; Queue< int > q = new Queue< int >(); q.Enqueue(0); visited[0] = true ; // Visit all nodes that are // currently in the queue while (q.Count > 0) { int temp = q.Dequeue(); if (adj.ContainsKey(temp)) { foreach ( int x in adj[temp]) { // If current node is // not visited already if (!visited[x]) { q.Enqueue(x); visited[x] = true ; dist[x] = dist[temp] + 1; } } } } if (!visited[N - 1]) return -1; else return dist[N - 1]; } } } // This code is contributed by phasing17. |
Javascript
<script> // Javascript program for the above approach // Stores Edges of graph var adj = new Map(); // Function to create graph edges // where node A and B can be visited function createGraph(paths, A, N, E) { // Visit all the connections for ( var i = 0; i < E; i++) { // If a policeman is at any point of // connection, leave that connection. // Insert the connect otherwise. if (!A[paths[i][0]] && !A[paths[i][1]]) { if (adj.has(paths[i][0])) { var tmp = adj.get(paths[i][0]); tmp.push(paths[i][1]); adj.set(paths[i][0], tmp); } else { var tmp = new Array(); tmp.push(paths[i][1]); adj.set(paths[i][0], tmp); } } } } // Function to find the shortest path function minPath(paths, A, N, E) { // If police is at either at // the 1-st house or at N-th house if (A[0] == 1 || A[N - 1] == 1) // The thief cannot reach // the N-th house return -1; // Function call to store connections createGraph(paths, A, N, E); // Stores whether node is // visited or not var visited = Array(N).fill(0); // Stores distances // from the root node var dist = Array(N).fill(0); dist[0] = 0; var q = []; q.push(0); visited[0] = 1; // Visit all nodes that are // currently in the queue while (q.length!=0) { var temp = q[0]; q.pop(); if (adj.has(temp)) { for ( var x of adj.get(temp)) { // If current node is // not visited already if (!visited[x]) { q.push(x); visited[x] = 1; dist[x] = dist[temp] + 1; } } } } if (!visited[N - 1]) return -1; else return dist[N - 1]; } // Driver Code // N : Number of houses // E: Number of edges var N = 5, E = 5; // Given positions var A = [0, 1, 0, 0, 0 ]; // Given Paths var paths = [ [ 0, 1 ], [ 0, 2 ], [ 1, 4 ], [ 2, 3 ], [ 3, 4 ] ]; // Function call document.write(minPath(paths, A, N, E)); // This code is contributed by itsok. </script> |
3
Time complexity: O (N + E)
Auxiliary Space: O (N + E)
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