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:
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; } |
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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. |
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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 = 4k = 3print(zigzag(n, k)) # This code is contributed by # Smitha Dinesh Semwal |
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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. |
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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. ?> |
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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; } |
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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. |
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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 = 4k = 3print(zigzag(n, k)) # This code is contributed by # Prasad Kshirsagar |
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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. |
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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. ?> |
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Output :
5
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