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

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

Time complexity: O(n).

Auxiliary space: O(1).

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.

Approach 2 :-approach to count the number of Dyck paths –In this implementation, we generate all possible Dyck paths of length n by generating all binary numbers with n bits. We then traverse through each bit in the binary representation of the number and update the depth accordingly. If at any point the depth becomes negative, then the path is not a Dyck path, so we break out of the loop. If we reach the end of the path and the depth is zero, then the path is a Dyck path, so we increment the count. Finally, we return the count of Dyck paths.

C++

#include <iostream>

using namespace std;

// Function to calculate the factorial of a given number
int factorial(int n) {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
        fact *= i;
    }
    return fact;
}

// Function to calculate the number of Dyck paths of length n using the 2 approach
int dyck_paths_2(int n) {
    int numerator = factorial(2 * n);
    int denominator = factorial(n + 1) * factorial(n);
    return numerator / denominator;
}

int main() {
    int n = 4;

    cout << "Number of Dyck paths is " << n << ": " << dyck_paths_2(n) << endl;

    return 0;
}

Java

import java.util.*;

public class DyckPaths
{

  // Function to calculate the factorial of a given number
  public static int factorial(int n)
  {
    int fact = 1;
    for (int i = 1; i <= n; i++) {
      fact *= i;
    }
    return fact;
  }

  // Function to calculate the number of Dyck paths of
  // length n using the 2 approach
  public static int dyck_paths_2(int n)
  {
    int numerator = factorial(2 * n);
    int denominator = factorial(n + 1) * factorial(n);
    return numerator / denominator;
  }

  public static void main(String[] args)
  {
    int n = 4;

    System.out.println("Number of Dyck paths is " + n
                       + ": " + dyck_paths_2(n));
  }
}

// This code is contributed by Prajwal Kandekar

Python3

# Function to calculate the factorial of a given number
def factorial(n):
    fact = 1
    for i in range(1, n + 1):
        fact *= i
    return fact

# Function to calculate the number of Dyck paths of length n using the 2 approach
def dyck_paths_2(n):
    numerator = factorial(2 * n)
    denominator = factorial(n + 1) * factorial(n)
    return numerator // denominator

if __name__ == '__main__':
    n = 4
    print("Number of Dyck paths is {}: {}".format(n, dyck_paths_2(n)))

Output

Number of Dyck paths is 4: 14

Time complexity: O(n).

Auxiliary space: O(1).

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