# Dyck path

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

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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.

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