Consider a n x n grid with indexes of top left corner as (0, 0). Dyck path is a staircase walk from bottom left, i.e., (n-1, 0) to top right, i.e., (0, n-1) that lies above the diagonal cells (or cells on line from bottom left to top right).

The task is to count the number of Dyck Paths from (n-1, 0) to (0, n-1).**Examples :**

Input :n = 1Output : 1 Input :n = 2Output : 2 Input :n = 3Output : 5 Input :n = 4Output : 14

The number of Dyck paths from (n-1, 0) to (0, n-1) can be given by the Catalan numberC(n).

## We strongly recommend that you click here and practice it, before moving on to the solution.

Below are the implementations to find count of Dyck Paths (or n’th Catalan number).

## C++

`// C++ program to count` `// number of Dyck Paths` `#include<iostream>` `using` `namespace` `std;` `// Returns count Dyck` `// paths in n x n grid` `int` `countDyckPaths(unsigned ` `int` `n)` `{` ` ` `// Compute value of 2nCn` ` ` `int` `res = 1;` ` ` `for` `(` `int` `i = 0; i < n; ++i)` ` ` `{` ` ` `res *= (2 * n - i);` ` ` `res /= (i + 1);` ` ` `}` ` ` `// return 2nCn/(n+1)` ` ` `return` `res / (n+1);` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `n = 4;` ` ` `cout << ` `"Number of Dyck Paths is "` ` ` `<< countDyckPaths(n);` ` ` `return` `0;` `}` |

## Java

`// Java program to count` `// number of Dyck Paths` `class` `GFG` `{` ` ` `// Returns count Dyck` ` ` `// paths in n x n grid` ` ` `public` `static` `int` `countDyckPaths(` `int` `n)` ` ` `{` ` ` `// Compute value of 2nCn` ` ` `int` `res = ` `1` `;` ` ` `for` `(` `int` `i = ` `0` `; i < n; ++i)` ` ` `{` ` ` `res *= (` `2` `* n - i);` ` ` `res /= (i + ` `1` `);` ` ` `}` ` ` `// return 2nCn/(n+1)` ` ` `return` `res / (n + ` `1` `);` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `main(String args[])` ` ` `{` ` ` `int` `n = ` `4` `;` ` ` `System.out.println(` `"Number of Dyck Paths is "` `+` ` ` `countDyckPaths(n));` ` ` `}` `}` |

## Python3

`# Python3 program to count` `# number of Dyck Paths` `# Returns count Dyck` `# paths in n x n grid` `def` `countDyckPaths(n):` ` ` ` ` `# Compute value of 2nCn` ` ` `res ` `=` `1` ` ` `for` `i ` `in` `range` `(` `0` `, n):` ` ` `res ` `*` `=` `(` `2` `*` `n ` `-` `i)` ` ` `res ` `/` `=` `(i ` `+` `1` `)` ` ` `# return 2nCn/(n+1)` ` ` `return` `res ` `/` `(n` `+` `1` `)` `# Driver Code` `n ` `=` `4` `print` `(` `"Number of Dyck Paths is "` `,` ` ` `str` `(` `int` `(countDyckPaths(n))))` `# This code is contributed by` `# Prasad Kshirsagar` |

## C#

`// C# program to count` `// number of Dyck Paths` `using` `System;` `class` `GFG {` ` ` ` ` `// Returns count Dyck` ` ` `// paths in n x n grid` ` ` `static` `int` `countDyckPaths(` `int` `n)` ` ` `{` ` ` ` ` `// Compute value of 2nCn` ` ` `int` `res = 1;` ` ` `for` `(` `int` `i = 0; i < n; ++i)` ` ` `{` ` ` `res *= (2 * n - i);` ` ` `res /= (i + 1);` ` ` `}` ` ` `// return 2nCn/(n+1)` ` ` `return` `res / (n + 1);` ` ` `}` ` ` `// Driver code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `n = 4;` ` ` `Console.WriteLine(` `"Number of "` ` ` `+ ` `"Dyck Paths is "` `+` ` ` `countDyckPaths(n));` ` ` `}` `}` `// This code is contributed by anuj_67.` |

## PHP

`<?php` `// PHP program to count` `// number of Dyck Paths` `// Returns count Dyck` `// paths in n x n grid` `function` `countDyckPaths( ` `$n` `)` `{` ` ` `// Compute value of 2nCn` ` ` `$res` `= 1;` ` ` `for` `( ` `$i` `= 0; ` `$i` `< ` `$n` `; ++` `$i` `)` ` ` `{` ` ` `$res` `*= (2 * ` `$n` `- ` `$i` `);` ` ` `$res` `/= (` `$i` `+ 1);` ` ` `}` ` ` `// return 2nCn/(n+1)` ` ` `return` `$res` `/ (` `$n` `+ 1);` `}` `// Driver Code` `$n` `= 4;` `echo` `"Number of Dyck Paths is "` `,` ` ` `countDyckPaths(` `$n` `);` `// This code is contributed by anuj_67.` `?>` |

## Javascript

`<script>` `// JavaScript program to count` `// number of Dyck Paths` ` ` `// Returns count Dyck` ` ` `// paths in n x n grid` ` ` `function` `countDyckPaths(n)` ` ` `{` ` ` ` ` `// Compute value of 2nCn` ` ` `let res = 1;` ` ` `for` `(let i = 0; i < n; ++i)` ` ` `{` ` ` `res *= (2 * n - i);` ` ` `res /= (i + 1);` ` ` `}` ` ` ` ` `// return 2nCn/(n+1)` ` ` `return` `res / (n + 1);` ` ` `}` `// Driver Code` ` ` `let n = 4;` ` ` `document.write(` `"Number of Dyck Paths is "` `+` ` ` `countDyckPaths(n));` ` ` ` ` `// This code is contributed by target_2.` `</script>` |

**Output :**

Number of Dyck Paths is 14

**Exercise :**

- Find number of sequences of 1 and -1 such that every sequence follows below constraints :

a) The length of a sequence is 2n

b) There are equal number of 1’s and -1’s, i.e., n 1’s, n -1s

c) Sum of prefix of every sequence is greater than or equal to 0. For example, 1, -1, 1, -1 and 1, 1, -1, -1 are valid, but -1, -1, 1, 1 is not valid. - Number of paths of length m + n from (m-1, 0) to (0, n-1) that are restricted to east and north steps.

This article is contributed by **Aditya Chatterjee**. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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.