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>
usingnamespacestd;
// Return Entringer Number E(n, k)
intzigzag(intn, intk)
{
// Base Case
if(n == 0 && k == 0)
return1;
// Base Case
if(k == 0)
return0;
// Recursive step
returnzigzag(n, k - 1) +
zigzag(n - 1, n - k);
}
// Driven Program
intmain()
{
intn = 4, k = 3;
cout << zigzag(n, k) << endl;
return0;
}
Java
// JAVA Code For Entringer Number
importjava.util.*;
classGFG {
// Return Entringer Number E(n, k)
staticintzigzag(intn, intk)
{
// Base Case
if(n == 0&& k == 0)
return1;
// Base Case
if(k == 0)
return0;
// Recursive step
returnzigzag(n, k - 1) +
zigzag(n - 1, n - k);
}
/* Driver program to test above function */
publicstaticvoidmain(String[] args)
{
intn = 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)
defzigzag(n, k):
# Base Case
if(n ==0andk ==0):
return1
# Base Case
if(k ==0):
return0
# Recursive step
returnzigzag(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
usingSystem;
classGFG {
// Return Entringer Number E(n, k)
staticintzigzag(intn, intk)
{
// Base Case
if(n == 0 && k == 0)
return1;
// Base Case
if(k == 0)
return0;
// Recursive step
returnzigzag(n, k - 1) +
zigzag(n - 1, n - k);
}
/* Driver program to test above function */
publicstaticvoidMain()
{
intn = 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)
functionzigzag($n, $k)
{
// Base Case
if($n== 0 and$k== 0)
return1;
// Base Case
if($k== 0)
return0;
// Recursive step
returnzigzag($n, $k- 1) +
zigzag($n- 1,$n- $k);
}
// Driven Code
$n= 4; $k= 3;
echozigzag($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)
functionzigzag( n, k)
{
// Base Case
if(n == 0 && k == 0)
return1;
// Base Case
if(k == 0)
return0;
// Recursive step
returnzigzag(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>
usingnamespacestd;
// Return Entringer Number E(n, k)
intzigzag(intn, intk)
{
intdp[n + 1][k + 1];
memset(dp, 0, sizeof(dp));
// Base cases
dp[0][0] = 1;
for(inti = 1; i <= n; i++)
dp[i][0] = 0;
// Finding dp[i][j]
for(inti = 1; i <= n; i++) {
for(intj = 1; j <= i; j++)
dp[i][j] = dp[i][j - 1] + dp[i - 1][i - j];
}
returndp[n][k];
}
// Driven Program
intmain()
{
intn = 4, k = 3;
cout << zigzag(n, k) << endl;
return0;
}
Java
// JAVA Code For Entringer Number
importjava.util.*;
classGFG {
// Return Entringer Number E(n, k)
staticintzigzag(intn, intk)
{
intdp[][] = newint[n + 1][k + 1];
// Base cases
dp[0][0] = 1;
for(inti = 1; i <= n; i++)
dp[i][0] = 0;
// Finding dp[i][j]
for(inti = 1; i <= n; i++) {
for(intj = 1; j <= Math.min(i, k);
j++)
dp[i][j] = dp[i][j - 1] +
dp[i - 1][i - j];
}
returndp[n][k];
}
/* Driver program to test above function */
publicstaticvoidmain(String[] args)
{
intn = 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)
defzigzag(n, k):
dp =[[0forx inrange(k+1)]
fory inrange(n+1)]
# Base cases
dp[0][0] =1
fori inrange(1, n+1):
dp[i][0] =0
# Finding dp[i][j]
fori inrange(1, n+1):
forj inrange(1, k+1):
dp[i][j] =(dp[i][j -1]
+dp[i -1][i -j])
returndp[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
usingSystem;
classGFG {
// Return Entringer Number E(n, k)
staticintzigzag(intn, intk)
{
int[, ] dp = newint[n + 1, k + 1];
// Base cases
dp[0, 0] = 1;
for(inti = 1; i <= n; i++)
dp[i, 0] = 0;
// Finding dp[i][j]
for(inti = 1; i <= n; i++) {
for(intj = 1; j <= Math.Min(i, k);
j++)
dp[i, j] = dp[i, j - 1] + dp[i - 1, i - j];
}
returndp[n, k];
}
/* Driver program to test above function */
publicstaticvoidMain()
{
intn = 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)
functionzigzag($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; $k= 3;
echozigzag($n, $k);
// This code is contributed by anuj_67.
?>
Javascript
<script>
// JavaScript program For Entringer Number
// Return Entringer Number E(n, k)
functionzigzag(n, k)
{
let dp = newArray(n+1);
// Loop to create 2D array using 1D array
for(vari = 0; i < dp.length; i++) {
dp[i] = newArray(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];
}
returndp[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>
usingnamespacestd;
// Return Entringer Number E(n, k)
intzigzag(intn, intk)
{
// vector to store values
vector<int>dp(k+1,0);
// Base cases
dp[0] = 1;
// iterate to get current value from previous value
for(inti = 1; i <= n; i++) {
vector<int>curr(k+1,0);
for(intj = 1; j <= i; j++){
curr[j] = curr[j - 1] + dp[i - j];
}
// assigning values to iterate further
dp = curr;
}
// return final answer
returndp[k];
}
// Driven Program
intmain()
{
intn = 4, k = 3;
cout << zigzag(n, k) << endl;
return0;
}
Java
importjava.util.*;
publicclassMain {
// Return Entringer Number E(n, k)
publicstaticintzigzag(intn, intk) {
// array to store values
int[] dp = newint[n+1];
Arrays.fill(dp, 0);
// Base cases
dp[0] = 1;
// iterate to get current value from previous value
for(inti = 1; i <= n; i++) {
int[] curr = newint[n+1];
Arrays.fill(curr, 0);
for(intj = 1; j <= i; j++){
curr[j] = curr[j - 1] + dp[i - j];
}
// assigning values to iterate further
dp = curr;
}
// return final answer
returndp[k];
}
// Driven Program
publicstaticvoidmain(String[] args) {
intn = 4, k = 3;
System.out.println(zigzag(n, k));
}
}
C#
usingSystem;
usingSystem.Collections.Generic;
usingSystem.Linq;
usingSystem.Text;
usingSystem.Threading.Tasks;
classProgram {
// Return Entringer Number E(n, k)
publicstaticintzigzag(intn, intk)
{
// array to store values
int[] dp = newint[n + 1];
// fill array with 0
Array.Fill(dp, 0);
// Base cases
dp[0] = 1;
// iterate to get current value from previous value
for(inti = 1; i <= n; i++) {
// create a new array to store current values
int[] curr = newint[n + 1];
// fill array with 0
Array.Fill(curr, 0);
for(intj = 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
returndp[k];
}
staticvoidMain(string[] args)
{
intn = 4, k = 3;
Console.WriteLine(zigzag(n, k));
}
}
Javascript
// Function to calculate Entringer Number E(n, k)
functionzigzag(n, k) {
// Array to store values
let dp = newArray(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 = newArray(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
returndp[k];
}
// Main function to test the zigzag function
functionmain() {
let n = 4; // The value of n
let k = 3; // The value of k
let ans = zigzag(n, k); // Calculate the Entringer Number
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