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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 = 1
Output : 1

Input : n = 2
Output : 2

Input : n = 3
Output : 5

Input : n = 4
Output : 14

dyckpaths

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



C_n=frac{(2n)!}{(n+1)!n1}=prod_{k=2}^{n}frac{n+k}{k}  for ngeq 0

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.
?>


Output :

Number of Dyck Paths is 14

Exercise :

  1. 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.
  2. .

  3. 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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This article is attributed to GeeksforGeeks.org

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