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 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



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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Improved By : vt_m, Prasad_Kshirsagar

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