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Dyck path
  • Difficulty Level : Medium
  • Last Updated : 01 Apr, 2021
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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\ n\geq 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.
?>

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 : 
 

  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. 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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