# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to find if there is a ` `// path between two vertices in a ` `// directed graph using Dynamic Programming ` ` `  `#include ` `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; ` `} `

## Java

 `// 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 `

Output:

```Yes
```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : princi singh