Euler Zigzag numbers is a sequence of integers which is a number of arrangements of those numbers so that each entry is alternately greater or less than the preceding entry.
c1, c2, c3, c4 is Alternating permutation where
c1 < c2
c3 < c2
c3 < c4…
zigzag numbers are as follows 1, 1, 1, 2, 5, 16, 61, 272, 1385, 7936, 50521 ……
For a given integer N. The task is to print sequence up to N terms.
Examples:
Input : N = 10
Output : 1 1 1 2 5 16 61 272 1385 7936
Input : N = 14
Output : 1 1 1 2 5 16 61 272 1385 7936 50521 353792 2702765 22368256
Approach :
The (n+1)th Zigzag number is :
We will find the factorial upto n and store them in an array and also create a second array to store the i th zigzag number and apply the formula stated above to find all the n zigzag numbers.
Below is the implementation of the above approach :
// CPP program to find zigzag sequence #include <bits/stdc++.h> using namespace std;
// Function to print first n zigzag numbers void ZigZag( int n)
{ // To store factorial and n'th zig zag number
long long fact[n + 1], zig[n + 1] = { 0 };
// Initialize factorial upto n
fact[0] = 1;
for ( int i = 1; i <= n; i++)
fact[i] = fact[i - 1] * i;
// Set first two zig zag numbers
zig[0] = 1;
zig[1] = 1;
cout << "zig zag numbers: " ;
// Print first two zig zag number
cout << zig[0] << " " << zig[1] << " " ;
// Print the rest zig zag numbers
for ( int i = 2; i < n; i++)
{
long long sum = 0;
for ( int k = 0; k <= i - 1; k++)
{
// Binomial(n, k)*a(k)*a(n-k)
sum += (fact[i - 1]/(fact[i - 1 - k]*fact[k]))
*zig[k] * zig[i - 1 - k];
}
// Store the value
zig[i] = sum / 2;
// Print the number
cout << sum / 2 << " " ;
}
} // Driver code int main()
{ int n = 10;
// Function call
ZigZag(n);
return 0;
} |
// Java program to find zigzag sequence import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{ // Function to print first n zigzag numbers static void ZigZag( int n)
{ // To store factorial and n'th zig zag number
long [] fact= new long [n + 1 ];
long [] zig = new long [n + 1 ];
for ( int i = 0 ; i < n + 1 ; i++)
zig[i] = 0 ;
// Initialize factorial upto n
fact[ 0 ] = 1 ;
for ( int i = 1 ; i <= n; i++)
fact[i] = fact[i - 1 ] * i;
// Set first two zig zag numbers
zig[ 0 ] = 1 ;
zig[ 1 ] = 1 ;
System.out.print( "zig zag numbers: " );
// Print first two zig zag number
System.out.print(zig[ 0 ] + " " + zig[ 1 ] + " " );
// Print the rest zig zag numbers
for ( int i = 2 ; i < n; i++)
{
long sum = 0 ;
for ( int k = 0 ; k <= i - 1 ; k++)
{
// Binomial(n, k)*a(k)*a(n-k)
sum += (fact[i - 1 ] / (fact[i - 1 - k] *
fact[k])) * zig[k] * zig[i - 1 - k];
}
// Store the value
zig[i] = sum / 2 ;
// Print the number
System.out.print(sum / 2 + " " );
}
} // Driver code public static void main (String[] args)
throws java.lang.Exception
{ int n = 10 ;
// Function call
ZigZag(n);
} } // This code is contributed by nidhiva |
# Python3 program to find zigzag sequence # Function to print first n zigzag numbers def ZigZag(n):
# To store factorial and
# n'th zig zag number
fact = [ 0 for i in range (n + 1 )]
zig = [ 0 for i in range (n + 1 )]
# Initialize factorial upto n
fact[ 0 ] = 1
for i in range ( 1 , n + 1 ):
fact[i] = fact[i - 1 ] * i
# Set first two zig zag numbers
zig[ 0 ] = 1
zig[ 1 ] = 1
print ( "zig zag numbers: " , end = " " )
# Print first two zig zag number
print (zig[ 0 ], zig[ 1 ], end = " " )
# Print the rest zig zag numbers
for i in range ( 2 , n):
sum = 0
for k in range ( 0 , i):
# Binomial(n, k)*a(k)*a(n-k)
sum + = ((fact[i - 1 ] / /
(fact[i - 1 - k] * fact[k])) *
zig[k] * zig[i - 1 - k])
# Store the value
zig[i] = sum / / 2
# Print the number
print ( sum / / 2 , end = " " )
# Driver code n = 10
# Function call ZigZag(n) # This code is contributed by Mohit Kumar |
// C# program to find zigzag sequence using System;
class GFG
{ // Function to print first n zigzag numbers static void ZigZag( int n)
{ // To store factorial and n'th zig zag number
long [] fact= new long [n + 1];
long [] zig = new long [n + 1];
for ( int i = 0; i < n + 1; i++)
zig[i] = 0;
// Initialize factorial upto n
fact[0] = 1;
for ( int i = 1; i <= n; i++)
fact[i] = fact[i - 1] * i;
// Set first two zig zag numbers
zig[0] = 1;
zig[1] = 1;
Console.Write( "zig zag numbers: " );
// Print first two zig zag number
Console.Write(zig[0] + " " + zig[1] + " " );
// Print the rest zig zag numbers
for ( int i = 2; i < n; i++)
{
long sum = 0;
for ( int k = 0; k <= i - 1; k++)
{
// Binomial(n, k)*a(k)*a(n-k)
sum += (fact[i - 1] / (fact[i - 1 - k] *
fact[k])) * zig[k] * zig[i - 1 - k];
}
// Store the value
zig[i] = sum / 2;
// Print the number
Console.Write(sum / 2 + " " );
}
} // Driver code public static void Main (String[] args)
{ int n = 10;
// Function call
ZigZag(n);
} } // This code is contributed by 29AjayKumar |
<script> // Javascript program to find zigzag sequence // Function to print first n zigzag numbers function ZigZag(n)
{ // To store factorial and n'th zig zag number
var fact = Array(n+1).fill(0);
var zig = Array(n+1).fill(0);
// Initialize factorial upto n
fact[0] = 1;
for ( var i = 1; i <= n; i++)
fact[i] = fact[i - 1] * i;
// Set first two zig zag numbers
zig[0] = 1;
zig[1] = 1;
document.write( "zig zag numbers: " );
// Print first two zig zag number
document.write( zig[0] + " " + zig[1] + " " );
// Print the rest zig zag numbers
for ( var i = 2; i < n; i++)
{
var sum = 0;
for ( var k = 0; k <= i - 1; k++)
{
// Binomial(n, k)*a(k)*a(n-k)
sum += parseInt(fact[i - 1]/(fact[i - 1 - k]*fact[k]))
*zig[k] * zig[i - 1 - k];
}
// Store the value
zig[i] = parseInt(sum / 2);
// Print the number
document.write( parseInt(sum / 2) + " " );
}
} // Driver code var n = 10;
// Function call ZigZag(n); // This code is contributed by rutvik_56. </script> |
Output:
zig zag numbers: 1 1 1 2 5 16 61 272 1385 7936
Time Complexity: O(n2)
Auxiliary Space: O(n)
Reference
https://en.wikipedia.org/wiki/Alternating_permutation
https://oeis.org/A000111