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 Programint main(){ int n = 4, k = 3; cout << zigzag(n, k) << endl; return 0;} |
Java
// JAVA Code For Entringer Numberimport 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 Programn = 4k = 3print(zigzag(n, k))# This code is contributed by# Smitha Dinesh Semwal |
C#
// C# Code For Entringer Numberusing 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 Programint main(){ int n = 4, k = 3; cout << zigzag(n, k) << endl; return 0;} |
Java
// JAVA Code For Entringer Numberimport 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 Programn = 4k = 3print(zigzag(n, k))# This code is contributed by # Prasad Kshirsagar |
C#
// C# Code For Entringer Numberusing 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
Time Complexity: O(n * n)
Auxiliary Space: O(n * k)
Efficient approach : Space optimization
In previous approach the current value dp[i][j] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 1D array to store the computations.
Implementation steps:
- Create a 1D vector dp of size K+1.
- Set a base case by initializing the values of DP .
- Now iterate over subproblems by the help of nested loop and get the current value from previous computations.
- Now Create a temporary 1d vector curr used to store the current values from previous computations.
- After every iteration assign the value of curr to dp for further iteration.
- At last return and print the final answer stored in dp[k].
Implementation:
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){ // vector to store values vector<int>dp(k+1,0); // Base cases dp[0] = 1; // iterate to get current value from previous value for (int i = 1; i <= n; i++) { vector<int>curr(k+1,0); for (int j = 1; j <= i; j++){ curr[j] = curr[j - 1] + dp[i - j]; } // assigning values to iterate further dp = curr; } // return final answer return dp[k];}// Driven Programint main(){ int n = 4, k = 3; cout << zigzag(n, k) << endl; return 0;} |
Java
import java.util.*;public class Main { // Return Entringer Number E(n, k) public static int zigzag(int n, int k) { // array to store values int[] dp = new int[n+1]; Arrays.fill(dp, 0); // Base cases dp[0] = 1; // iterate to get current value from previous value for (int i = 1; i <= n; i++) { int[] curr = new int[n+1]; Arrays.fill(curr, 0); for (int j = 1; j <= i; j++){ curr[j] = curr[j - 1] + dp[i - j]; } // assigning values to iterate further dp = curr; } // return final answer return dp[k]; } // Driven Program public static void main(String[] args) { int n = 4, k = 3; System.out.println(zigzag(n, k)); }} |
C#
using System;using System.Collections.Generic;using System.Linq;using System.Text;using System.Threading.Tasks;class Program { // Return Entringer Number E(n, k) public static int zigzag(int n, int k) { // array to store values int[] dp = new int[n + 1]; // fill array with 0 Array.Fill(dp, 0); // Base cases dp[0] = 1; // iterate to get current value from previous value for (int i = 1; i <= n; i++) { // create a new array to store current values int[] curr = new int[n + 1]; // fill array with 0 Array.Fill(curr, 0); for (int j = 1; j <= i; j++) { // assign values to current array curr[j] = curr[j - 1] + dp[i - j]; } // assigning values to iterate further dp = curr; } // return final answer return dp[k]; } static void Main(string[] args) { int n = 4, k = 3; Console.WriteLine(zigzag(n, k)); }} |
Javascript
// Function to calculate Entringer Number E(n, k)function zigzag(n, k) { // Array to store values let dp = new Array(k + 1).fill(0); // Base cases dp[0] = 1; // Iterate to get current value from previous value for (let i = 1; i <= n; i++) { let curr = new Array(k + 1).fill(0); for (let j = 1; j <= i; j++) { curr[j] = curr[j - 1] + dp[i - j]; } // Assigning values to iterate further dp = curr; } // Return final answer return dp[k];}// Main function to test the zigzag functionfunction main() { let n = 4; // The value of n let k = 3; // The value of k let ans = zigzag(n, k); // Calculate the Entringer Number console.log(ans); // Print the result}main();// This code is contributed by sarojmcy2e |
Output
5
Time complexity: O(N*K)
Auxiliary Space: O(K)
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