# Delannoy Number

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 ` `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

 ` `

Output:

```129
```

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

## C++

 `// CPP Program of finding nth Delannoy Number. ` `#include ` `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] = 1; ` `    ``for` `(``int` `i = 0; i <= m; i++)  ` `        ``dp[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

 ` `

Output :

```129
```

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