Given a number N, the task is to print the below pattern upto N terms:
2, 1, 4, 3, 6, 5, …. N terms
Examples:
Input: N = 4 Output: 2, 1, 4, 3 Explanation: Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1) 1st term = (1 % 2 == 0) ? (1 - 1) : (1 + 1) = (1 + 1) = 2 2nd term = (2 % 2 == 0) ? (2 - 1) : (2 + 1) = (2 - 1) = 1 3rd term = (3 % 2 == 0) ? (3 - 1) : (3 + 1) = (3 + 1) = 4 4th term = (4 % 2 == 0) ? (4 - 1) : (4 + 1) = (4 - 1) = 3 Therefore, Series = 2, 1, 4, 3 Input: N = 7 Output: 2, 1, 4, 3, 6, 5, 8
Formula:
Nth Term = (N % 2 == 0) ? (N - 1) : (N + 1)
Below is the solution to above problem:
C++
// C++ program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms #include <iostream> using namespace std;
// Function to print the series void printPattern( int N)
{ for ( int i = 1; i <= N; i++) {
// Find and print the ith term
cout << " " <<((i % 2 == 0) ? (i - 1) : (i + 1));
}
} // Driver code int main()
{ // Get the value of N
int N = 10;
// Print the Series
printPattern(N);
return 0;
} |
Java
// Java Program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{ // Function to print the series static void printPattern( int N)
{ for ( int i = 1 ; i <= N; i++) {
// Find and print the ith term
System.out.print( " " +((i % 2 == 0 ) ? (i - 1 ) : (i + 1 )));
}
} // Driver code public static void main(String args[])
{ // Get the value of N
int N = 10 ;
// Print the Series
printPattern(N);
} } |
Python3
# Python program to print the series # 2, 1, 4, 3, 6, 5, …. up to N terms # Function to print the series def printPattern(N):
for i in range ( 1 , N + 1 ):
# Find and print the ith term
print (i - 1 if i % 2 = = 0
else i + 1 , end = " " )
# Driver code N = 10
printPattern(N) # This code is contributed by Shrikant13 |
C#
// C# program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms using System;
class GFG
{ // Function to print the series public void printPattern( int N)
{ for ( int i = 1; i <= N; i++)
{
// Find and print the ith term
int a = ((i % 2 == 0) ?
(i - 1) : (i + 1));
Console.Write( " {0}" , a);
}
} // Driver code public static void Main()
{ GFG g = new GFG();
// Get the value of N
int N = 10;
// Print the Series
g.printPattern(N);
} } // This code is contributed // by Soumik |
PHP
<?php // PHP program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms // Function to print the series function printPattern( $N )
{ for ( $i = 1; $i <= $N ; $i ++)
{
// Find and print the ith term
echo " " . (( $i % 2 == 0) ?
( $i - 1) : ( $i + 1));
}
} // Driver code // Get the value of N $N = 10;
// Print the Series printPattern( $N );
// This code is contributed // by ChitraNayal ?> |
Javascript
<script> // javascript program to print the series // 2, 1, 4, 3, 6, 5, …. up to N terms // Function to print the series function printPattern( N)
{ for (let i = 1; i <= N; i++)
{
// Find and print the ith term
document.write( " " + ((i % 2 == 0) ? (i - 1) : (i + 1)));
}
} // Driver code // Get the value of N
let N = 10;
// Print the Series
printPattern(N);
// This code is contributed by Rajput-Ji </script> |
Output:
2 1 4 3 6 5 8 7 10 9
Time complexity: O(n) because using a for loop
Auxiliary Space: O(1), since no extra space has been taken.
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