Find if there is a path between two vertices in a directed graph | Set 2

Given a Directed Graph and two vertices in it, check whether there is a path from the first given vertex to second.

Example:

Consider the following Graph:

Input : (u, v) = (1, 3)
Output: Yes
Explanation:
There is a path from 1 to 3, 1 -> 2 -> 3

Input : (u, v) = (3, 6)
Output: No
Explanation:
There is no path from 3 to 6



A BFS or DFS based solution of this problem is discuss here.

Approach : Here we will discuss a Dynamic Programming based solution using Floyd Warshall Algorithm.

  • Create a boolean 2D matrix mat where mat[i][j] will be true if there is a path from vertex i to j.
  • For every starting vertex i and ending vertex j iterate over all intermediate vertex k and do check if there is a path for i to j through k then mark mat[i][j] as true.
  • Finally check if mat[u][v] is true then return true else return false.

Below is the implementation of the above approach :

C++

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// C++ program to find if there is a
// path between two vertices in a
// directed graph using Dynamic Programming
  
#include <bits/stdc++.h>
using namespace std;
#define X 6
#define Z 2
  
// function to find if there is a
// path between two vertices in a
// directed graph
bool existPath(int V, int edges[X][Z],
               int u, int v)
{
    // dp matrix
    bool mat[V][V];
    memset(mat, false, sizeof(mat));
  
    // set dp[i][j]=true if there is
    // edge between i to j
    for (int i = 0; i < X; i++)
        mat[edges[i][0]][edges[i][1]] = true;
  
    // check for all intermediate vertex
    for (int k = 0; k < V; k++) {
        for (int i = 0; i < V; i++) {
            for (int j = 0; j < V; j++) {
  
                mat[i][j] = mat[i][j]
                            || mat[i][k]
                                   && mat[k][j];
            }
        }
    }
  
    // if vertex is invalid
    if (u >= V || v >= V) {
        return false;
    }
  
    // if there is a path
    if (mat[u][v])
        return true;
    return false;
}
  
// Driver function
int main()
{
    int V = 4;
    int edges[X][Z]
        = { { 0, 2 }, { 0, 1 },
            { 1, 2 }, { 2, 3 },
            { 2, 0 }, { 3, 3 } };
    int u = 1, v = 3;
  
    if (existPath(V, edges, u, v))
        cout << "Yes\n";
    else
        cout << "No\n";
    return 0;
}

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Java

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// Java program to find if there is a path 
// between two vertices in a directed graph
// using Dynamic Programming
import java.util.*;
  
class GFG{
      
static final int X = 6;
static final int Z = 2;
  
// Function to find if there is a
// path between two vertices in a
// directed graph
static boolean existPath(int V, int edges[][],
                         int u, int v)
{
      
    // mat matrix
    boolean [][]mat = new boolean[V][V];
  
    // set mat[i][j]=true if there is
    // edge between i to j
    for (int i = 0; i < X; i++)
        mat[edges[i][0]][edges[i][1]] = true;
  
    // Check for all intermediate vertex
    for(int k = 0; k < V; k++) 
    {
        for(int i = 0; i < V; i++) 
        {
            for(int j = 0; j < V; j++)
            {
                mat[i][j] = mat[i][j] || 
                            mat[i][k] &&
                            mat[k][j];
            }
        }
    }
  
    // If vertex is invalid
    if (u >= V || v >= V)
    {
        return false;
    }
  
    // If there is a path
    if (mat[u][v])
        return true;
    return false;
}
  
// Driver code
public static void main(String[] args)
{
    int V = 4;
    int edges[][] = { { 0, 2 }, { 0, 1 },
                      { 1, 2 }, { 2, 3 },
                      { 2, 0 }, { 3, 3 } };
    int u = 1, v = 3;
  
    if (existPath(V, edges, u, v))
        System.out.print("Yes\n");
    else
        System.out.print("No\n");
}
}
  
// This code is contributed by Princi Singh

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Output:

Yes

Time Complexity : O ( V 3)
Auxiliary Space : O ( V 2)

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Improved By : princi singh