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Entringer Number

  • Difficulty Level : Medium
  • Last Updated : 05 May, 2021

The Entringer Number E(n, k) are the number of permutations of {1, 2, …, n + 1}, starting with k + 1, which, after initially falling, alternatively fall then rise. The Entringer are given by: 
 

Image

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For example, for n = 4 and k = 2, E(4, 2) is 4. 
They are: 
3 2 4 1 5 
3 2 5 1 4 
3 1 4 2 5 
3 1 5 2 4
Examples : 
 



Input : n = 4, k = 2
Output : 4

Input : n = 4, k = 3
Output : 5

 

Below is program to find Entringer Number E(n, k). The program is based on above simple recursive formula. 
 

C++




// CPP Program to find Entringer Number E(n, k)
#include <bits/stdc++.h>
using namespace std;
 
// Return Entringer Number E(n, k)
int zigzag(int n, int k)
{
    // Base Case
    if (n == 0 && k == 0)
        return 1;
 
    // Base Case
    if (k == 0)
        return 0;
 
    // Recursive step
    return zigzag(n, k - 1) +
           zigzag(n - 1, n - k);
}
 
// Driven Program
int main()
{
    int n = 4, k = 3;
    cout << zigzag(n, k) << endl;
    return 0;
}
Java



// JAVA Code For Entringer Number
import java.util.*;
 
class GFG {
     
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        // Base Case
        if (n == 0 && k == 0)
            return 1;
      
        // Base Case
        if (k == 0)
            return 0;
      
        // Recursive step
        return zigzag(n, k - 1) +
               zigzag(n - 1, n - k);
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        System.out.println(zigzag(n, k));
         
    }
}
 
// This code is contributed by Arnav Kr. Mandal.
Python3



# Python Program to find Entringer Number E(n, k)
 
# Return Entringer Number E(n, k)
def zigzag(n, k):
 
    # Base Case
    if (n == 0 and k == 0):
        return 1
 
    # Base Case
    if (k == 0):
        return 0
 
    # Recursive step
    return zigzag(n, k - 1) + zigzag(n - 1, n - k);
 
# Driven Program
n = 4
k = 3
print(zigzag(n, k))
 
# This code is contributed by
# Smitha Dinesh Semwal   
C#



// C# Code For Entringer Number
using System;
 
class GFG {
 
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        // Base Case
        if (n == 0 && k == 0)
            return 1;
 
        // Base Case
        if (k == 0)
            return 0;
 
        // Recursive step
        return zigzag(n, k - 1) +
               zigzag(n - 1, n - k);
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 4, k = 3;
        Console.WriteLine(zigzag(n, k));
    }
}
 
// This code is contributed by vt_m.
PHP



<?php
// PHP Program to find
// Entringer Number E(n, k)
 
// Return Entringer Number E(n, k)
function zigzag($n, $k)
{
    // Base Case
    if ($n == 0 and $k == 0)
        return 1;
 
    // Base Case
    if ($k == 0)
        return 0;
 
    // Recursive step
    return zigzag($n, $k - 1) +
        zigzag($n - 1,$n - $k);
}
 
// Driven Code
$n = 4; $k = 3;
echo zigzag($n, $k) ;
 
// This code is contributed by anuj_67.
?>
Javascript



<script>
//  Program to find Entringer Number E(n, k)
 
// Return Entringer Number E(n, k)
function zigzag( n, k)
{
    // Base Case
    if (n == 0 && k == 0)
        return 1;
  
    // Base Case
    if (k == 0)
        return 0;
  
    // Recursive step
    return zigzag(n, k - 1) +
           zigzag(n - 1, n - k);
}
  
// Driven Program
 
     n = 4;
     k = 3;
    document.write( zigzag(n, k));
//This code is contributed by sweetyty
</script>
Output : 
 
5
Below is the implementation of finding Entringer Number using Dynamic Programming: 
 
C++



// CPP Program to find Entringer Number E(n, k)
#include <bits/stdc++.h>
using namespace std;
 
// Return Entringer Number E(n, k)
int zigzag(int n, int k)
{
    int dp[n + 1][k + 1];
    memset(dp, 0, sizeof(dp));
 
    // Base cases
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++)
        dp[i][0] = 0;   
 
    // Finding dp[i][j]
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= i; j++)
            dp[i][j] = dp[i][j - 1] +
                       dp[i - 1][i - j];
 
    return dp[n][k];
}
 
// Driven Program
int main()
{
    int n = 4, k = 3;
    cout << zigzag(n, k) << endl;
    return 0;
}
Java



// JAVA Code For Entringer Number
import java.util.*;
 
class GFG {
     
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        int dp[][] = new int[n + 1][k + 1];
        
        // Base cases
        dp[0][0] = 1;
        for (int i = 1; i <= n; i++)
            dp[i][0] = 0;   
      
        // Finding dp[i][j]
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Math.min(i, k);
                                          j++)
                dp[i][j] = dp[i][j - 1] +
                           dp[i - 1][i - j];
            }
      
        return dp[n][k];
    }
     
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int n = 4, k = 3;
        System.out.println(zigzag(n, k));
    }
}
     
// This code is contributed by Arnav Kr. Mandal.   
Python3



# Python3 Program to find Entringer
# Number E(n, k)
 
# Return Entringer Number E(n, k)
def zigzag(n, k):
    dp = [[0 for x in range(k+1)]
             for y in range(n+1)]
 
    # Base cases
    dp[0][0] = 1
    for i in range(1, n+1):
        dp[i][0] = 0
 
    # Finding dp[i][j]
    for i in range(1, n+1):
        for j in range(1, k+1):
            dp[i][j] = (dp[i][j - 1]
                 + dp[i - 1][i - j])
                         
    return dp[n][k]
 
# Driven Program
n = 4
k = 3
print(zigzag(n, k))
 
# This code is contributed by
# Prasad Kshirsagar
C#



// C# Code For Entringer Number
using System;
 
class GFG {
 
    // Return Entringer Number E(n, k)
    static int zigzag(int n, int k)
    {
        int[, ] dp = new int[n + 1, k + 1];
 
        // Base cases
        dp[0, 0] = 1;
        for (int i = 1; i <= n; i++)
            dp[i, 0] = 0;
 
        // Finding dp[i][j]
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= Math.Min(i, k);
                j++)
                dp[i, j] = dp[i, j - 1] + dp[i - 1, i - j];
        }
 
        return dp[n, k];
    }
 
    /* Driver program to test above function */
    public static void Main()
    {
        int n = 4, k = 3;
        Console.WriteLine(zigzag(n, k));
    }
}
 
// This code is contributed by vt_m.
PHP



<?php
// PHP Program to find
// Entringer Number E(n, k)
 
// Return Entringer Number E(n, k)
function zigzag($n, $k)
{
    $dp = array(array());
     
 
    // Base cases
    $dp[0][0] = 1;
    for ($i = 1; $i <= $n; $i++)
        $dp[$i][0] = 0;
 
    // Finding dp[i][j]
    for ($i = 1; $i <= $n; $i++)
    {
        for ($j = 1; $j <= $i; $j++)
            $dp[$i][$j] = $dp[$i][$j - 1] +
                          $dp[$i - 1][$i - $j];
    }
    return $dp[$n][$k];
}
 
// Driven Code
$n = 4; $k = 3;
echo zigzag($n, $k);
 
// This code is contributed by anuj_67.
?>
Javascript



<script>
 
// JavaScript program For Entringer Number
 
    // Return Entringer Number E(n, k)
    function zigzag(n, k)
    {
        let dp = new Array(n+1);
         
        // Loop to create 2D array using 1D array
        for (var i = 0; i < dp.length; i++) {
            dp[i] = new Array(2);
        }
          
        // Base cases
        dp[0][0] = 1;
        for (let i = 1; i <= n; i++)
            dp[i][0] = 0;   
        
        // Finding dp[i][j]
        for (let i = 1; i <= n; i++) {
            for (let j = 1; j <= Math.min(i, k);
                                          j++)
                dp[i][j] = dp[i][j - 1] +
                           dp[i - 1][i - j];
            }
        
        return dp[n][k];
    }
 
// Driver code
 
        let n = 4, k = 3;
        document.write(zigzag(n, k));
                             
</script>
Output : 
5
 



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Article Contributed By :
https://media.geeksforgeeks.org/auth/avatar.png
anuj0503
@anuj0503
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