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
Below is the implementation of finding Delannoy Number:
C++
#include <bits/stdc++.h>
using namespace std;
int dealnnoy(int n, int m)
{
if (m == 0 || n == 0)
return 1;
return dealnnoy(m - 1, n) +
dealnnoy(m - 1, n - 1) +
dealnnoy(m, n - 1);
}
int main()
{
int n = 3, m = 4;
cout << dealnnoy(n, m) << endl;
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
public class GfG{
public static int dealnnoy(int n, int m)
{
if (m == 0 || n == 0)
return 1;
return dealnnoy(m - 1, n) +
dealnnoy(m - 1, n - 1) +
dealnnoy(m, n - 1);
}
public static void main(String args[]){
int n = 3, m = 4;
System.out.println(dealnnoy(n, m));
}
}
|
Python3
def dealnnoy(n, m):
if (m == 0 or n == 0) :
return 1
return dealnnoy(m - 1, n) + dealnnoy(m - 1, n - 1) + dealnnoy(m, n - 1)
n = 3
m = 4;
print( dealnnoy(n, m) )
|
C#
using System;
public class GfG {
public static int dealnnoy(int n, int m)
{
if (m == 0 || n == 0)
return 1;
return dealnnoy(m - 1, n) +
dealnnoy(m - 1, n - 1) +
dealnnoy(m, n - 1);
}
public static void Main()
{
int n = 3, m = 4;
Console.WriteLine(dealnnoy(n, m));
}
}
|
PHP
<?php
function dealnnoy( $n, $m)
{
if ($m == 0 or $n == 0)
return 1;
return dealnnoy($m - 1, $n) +
dealnnoy($m - 1, $n - 1) +
dealnnoy($m, $n - 1);
}
$n = 3;
$m = 4;
echo dealnnoy($n, $m);
?>
|
Javascript
<script>
function dealnnoy(n, m)
{
if (m == 0 || n == 0)
return 1;
return dealnnoy(m - 1, n) +
dealnnoy(m - 1, n - 1) +
dealnnoy(m, n - 1);
}
let n = 3, m = 4;
document.write(dealnnoy(n, m));
</script>
|
Output:
129
Below is the Dynamic Programming program to find nth Delannoy Number:
C++
#include <bits/stdc++.h>
using namespace std;
int dealnnoy(int n, int m)
{
int dp[m + 1][n + 1];
for (int i = 0; i <= m; i++)
dp[i][0] = 1;
for (int i = 0; i <= m; i++)
dp[0][i] = 1;
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
dp[i][j] = dp[i - 1][j] +
dp[i - 1][j - 1] +
dp[i][j - 1];
return dp[m][n];
}
int main()
{
int n = 3, m = 4;
cout << dealnnoy(n, m) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static int dealnnoy(int n, int m)
{
int dp[][]=new int[m + 1][n + 1];
for (int i = 0; i <= m; i++)
dp[i][0] = 1;
for (int i = 0; i < m; i++)
dp[0][i] = 1;
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
dp[i][j] = dp[i - 1][j] +
dp[i - 1][j - 1] +
dp[i][j - 1];
return dp[m][n];
}
public static void main(String args[])
{
int n = 3, m = 4;
System.out.println(dealnnoy(n, m));
}
}
|
Python3
def dealnnoy (n, m):
dp = [[0 for x in range(n+1)] for x in range(m+1)]
for i in range(m):
dp[0][i] = 1
for i in range(1, m + 1):
dp[i][0] = 1
for i in range(1, m + 1):
for j in range(1, n + 1):
dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1] + dp[i][j - 1];
return dp[m][n]
n = 3
m = 4
print(dealnnoy(n, m))
|
C#
using System;
class GFG {
static int dealnnoy(int n, int m)
{
int[, ] dp = new int[m + 1, n + 1];
for (int i = 0; i <= m; i++)
dp[i, 0] = 1;
for (int i = 0; i < m; i++)
dp[0, i] = 1;
for (int i = 1; i <= m; i++)
for (int j = 1; j <= n; j++)
dp[i, j] = dp[i - 1, j]
+ dp[i - 1, j - 1]
+ dp[i, j - 1];
return dp[m, n];
}
public static void Main()
{
int n = 3, m = 4;
Console.WriteLine(dealnnoy(n, m));
}
}
|
PHP
<?php
function dealnnoy($n, $m)
{
$dp[$m + 1][$n + 1] = 0;
for ($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];
}
$n = 3; $m = 4;
echo dealnnoy($n, $m) ;
?>
|
Javascript
<script>
function dealnnoy(n, m)
{
var dp = Array.from(Array(m+1),
() => Array(n+1));
for (var i = 0; i <= m; i++)
dp[i][0] = 1;
for (var i = 0; i <= m; i++)
dp[0][i] = 1;
for (var i = 1; i <= m; i++)
for (var 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];
}
var n = 3, m = 4;
document.write( dealnnoy(n, m));
</script>
|
Output :
129
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