Delannoy Number - GeeksforGeeks

In mathematics, a Delannoy number D describes the number of paths from the southwest corner (0, 0) of a rectangular grid to the northeast corner (m, n), using only single steps north, northeast, or east.
For Example, D(3, 3) equals 63.

Delannoy Number can be calculated by:

Delannoy number can be used to find:

Counts the number of global alignments of two sequences of lengths m and n.
Number of points in an m-dimensional integer lattice that are at most n steps from the origin.
In cellular automata, the number of cells in an m-dimensional von Neumann neighborhood of radius n.
Number of cells on a surface of an m-dimensional von Neumann neighborhood of radius n.

Examples :

Input : n = 3, m = 3
Output : 63

Input : n = 4, m = 5
Output : 681

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below is the implementation of finding Delannoy Number:

C++

// CPP Program of finding nth Delannoy Number. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Return the nth Delannoy Number. 
int dealnnoy(int n, int m) 
{ 
    // Base case 
    if (m == 0 || n == 0) 
        return 1; 
  
    // Recursive step. 
    return dealnnoy(m - 1, n) +  
           dealnnoy(m - 1, n - 1) + 
           dealnnoy(m, n - 1); 
} 
  
// Driven Program 
int main() 
{ 
    int n = 3, m = 4; 
    cout << dealnnoy(n, m) << endl; 
    return 0; 
} 

Java

// Java Program for finding nth Delannoy Number. 
import java.util.*; 
import java.lang.*; 
  
public class GfG{ 
      
    // Return the nth Delannoy Number. 
    public static int dealnnoy(int n, int m) 
    { 
        // Base case 
        if (m == 0 || n == 0) 
            return 1; 
  
        // Recursive step. 
        return dealnnoy(m - 1, n) +  
            dealnnoy(m - 1, n - 1) + 
            dealnnoy(m, n - 1); 
    } 
      
    // driver function 
    public static void main(String args[]){ 
        int n = 3, m = 4; 
        System.out.println(dealnnoy(n, m)); 
    } 
} 
  
/* This code is contributed by Sagar Shukla. */

Python3

# Python3 Program for finding  
# nth Delannoy Number. 
  
# Return the nth Delannoy Number. 
def dealnnoy(n, m): 
      
    # Base case 
    if (m == 0 or n == 0) : 
        return 1
  
    # Recursive step. 
    return dealnnoy(m - 1, n) + dealnnoy(m - 1, n - 1) + dealnnoy(m, n - 1) 
  
# Driven code 
n = 3
m = 4; 
print( dealnnoy(n, m) ) 
  
# This code is contributed by "rishabh_jain". 

C#

// C# Program for finding nth Delannoy Number. 
using System; 
  
public class GfG { 
  
    // Return the nth Delannoy Number. 
    public static int dealnnoy(int n, int m) 
    { 
  
        // Base case 
        if (m == 0 || n == 0) 
            return 1; 
  
        // Recursive step. 
        return dealnnoy(m - 1, n) +  
               dealnnoy(m - 1, n - 1) +  
                     dealnnoy(m, n - 1); 
    } 
  
    // driver function 
    public static void Main() 
    { 
        int n = 3, m = 4; 
        Console.WriteLine(dealnnoy(n, m)); 
    } 
} 
  
/* This code is contributed by vt_m. */

PHP

<?php 
// PHP Program of finding nth 
// Delannoy Number. 
  
// Return the nth Delannoy Number. 
function dealnnoy( $n, $m) 
{ 
      
    // Base case 
    if ($m == 0 or $n == 0) 
        return 1; 
  
    // Recursive step. 
    return dealnnoy($m - 1, $n) +  
           dealnnoy($m - 1, $n - 1) + 
           dealnnoy($m, $n - 1); 
} 
  
    // Driver Code 
    $n = 3;  
    $m = 4; 
    echo dealnnoy($n, $m); 
  
// This code is contributed by anuj_67. 
?> 

Output:

Below is the Dynamic Programming program to find nth Delannoy Number:

C++

// CPP Program of finding nth Delannoy Number. 
#include <bits/stdc++.h> 
using namespace std; 
  
// Return the nth Delannoy Number. 
int dealnnoy(int n, int m) 
{ 
    int dp[m + 1][n + 1]; 
  
    // Base cases 
    for (int i = 0; i <= m; i++)  
        dp[i][0] = 1; 
    for (int i = 0; i <= m; i++)  
        dp[0][i] = 1;     
  
    for (int i = 1; i <= m; i++)  
        for (int j = 1; j <= n; j++)  
            dp[i][j] = dp[i - 1][j] + 
                       dp[i - 1][j - 1] +  
                       dp[i][j - 1]; 
  
    return dp[m][n]; 
} 
  
// Driven Program 
int main() 
{ 
    int n = 3, m = 4; 
    cout << dealnnoy(n, m) << endl; 
    return 0; 
} 

Java

// Java Program of finding nth Delannoy Number. 
  
import java.io.*; 
  
class GFG { 
      
    // Return the nth Delannoy Number. 
    static int dealnnoy(int n, int m) 
    { 
        int dp[][]=new int[m + 1][n + 1]; 
       
        // Base cases 
        for (int i = 0; i <= m; i++)  
            dp[i][0] = 1; 
        for (int i = 0; i < m; i++)  
            dp[0][i] = 1;     
       
        for (int i = 1; i <= m; i++)  
            for (int j = 1; j <= n; j++)  
                dp[i][j] = dp[i - 1][j] + 
                           dp[i - 1][j - 1] +  
                           dp[i][j - 1]; 
       
        return dp[m][n]; 
    } 
       
    // Driven Program 
    public static void main(String args[]) 
    { 
        int n = 3, m = 4; 
        System.out.println(dealnnoy(n, m)); 
          
    } 
} 
  
// This code is contributed by Nikita Tiwari. 

Python3

# Python3 Program for finding nth 
# Delannoy Number. 
  
# Return the nth Delannoy Number. 
def dealnnoy (n, m): 
    dp = [[0 for x in range(n+1)] for x in range(m+1)] 
  
    # Base cases 
    for i in range(m): 
        dp[0][i] = 1
      
    for i in range(1, m + 1): 
        dp[i][0] = 1
  
    for i in range(1, m + 1): 
        for j in range(1, n + 1): 
            dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + dp[i][j - 1]; 
  
    return dp[m][n] 
  
# Driven code 
n = 3
m = 4
print(dealnnoy(n, m)) 
  
# This code is contributed by "rishabh_jain". 

C#

// C# Program of finding nth Delannoy Number. 
using System; 
  
class GFG { 
  
    // Return the nth Delannoy Number. 
    static int dealnnoy(int n, int m) 
    { 
          
        int[, ] dp = new int[m + 1, n + 1]; 
  
        // Base cases 
        for (int i = 0; i <= m; i++) 
            dp[i, 0] = 1; 
        for (int i = 0; i < m; i++) 
            dp[0, i] = 1; 
  
        for (int i = 1; i <= m; i++) 
            for (int j = 1; j <= n; j++) 
                dp[i, j] = dp[i - 1, j] 
                     + dp[i - 1, j - 1] 
                         + dp[i, j - 1]; 
  
        return dp[m, n]; 
    } 
  
    // Driven Program 
    public static void Main() 
    { 
        int n = 3, m = 4; 
          
        Console.WriteLine(dealnnoy(n, m)); 
    } 
} 
  
// This code is contributed by vt_m. 

PHP

<?php 
// PHP Program of finding 
// nth Delannoy Number. 
  
// Return the nth Delannoy Number. 
function dealnnoy($n, $m) 
{ 
    $dp[$m + 1][$n + 1] = 0; 
  
    // Base cases 
    for ($i = 0; $i <= $m; $i++)  
        $dp[$i][0] = 1; 
    for ( $i = 0; $i <= $m; $i++)  
        $dp[0][$i] = 1;  
  
    for ($i = 1; $i <= $m; $i++)  
        for ($j = 1; $j <= $n; $j++)  
            $dp[$i][$j] = $dp[$i - 1][$j] + 
                      $dp[$i - 1][$j - 1] +  
                           $dp[$i][$j - 1]; 
  
    return $dp[$m][$n]; 
} 
  
// Driven Code 
$n = 3; $m = 4; 
echo dealnnoy($n, $m) ; 
  
// This code is contributed by SanjuTomar  
?> 

Output :

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

PHP

C++

Java

Python3

C#

PHP

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