Given an Octal number N, check whether it is even or odd.
Examples:
Input: N = 7234 Output: Even Input: N = 333333333 Output: Odd
Naive Approach:
- Convert the number from Octal base to Decimal base.
- Then check if the number is even or odd, which can be easily checked by dividing by 2.
Time Complexity: O(N)
Efficient approach: Since Octal numbers contain digits from 0 to 7, therefore we can simply check if the last digit is either ‘0’, ‘2’, ‘4’ or ‘6’ . If it is, then the given Octal number will be Even, else Odd.
Below is the implementation of the above approach.
C++
// C++ code to check if a Octal // number is Even or Odd #include <bits/stdc++.h> using namespace std;
// Check if the number is odd or even string even_or_odd(string N) { int len = N.size();
// Check if the last digit
// is either '0', '2', '4',
// or '6'
if (N[len - 1] == '0'
|| N[len - 1] == '2'
|| N[len - 1] == '4'
|| N[len - 1] == '6' )
return ( "Even" );
else
return ( "Odd" );
} // Driver code int main()
{ string N = "735" ;
cout << even_or_odd(N);
return 0;
} |
Java
// Java code to check if a Octal // number is Even or Odd import java.io.*;
class GFG{
// Check if the number is odd or even static String even_or_odd(String N)
{ int len = N.length();
// Check if the last digit
// is either '0', '2', '4',
// or '6'
if (N.charAt(len - 1 ) == '0'
|| N.charAt(len - 1 ) == '2'
|| N.charAt(len - 1 ) == '4'
|| N.charAt(len - 1 ) == '6' )
return ( "Even" );
else
return ( "Odd" );
} // Driver code public static void main(String[] args)
{ String N = "735" ;
System.out.print(even_or_odd(N));
} } // This code is contributed by Rajput-Ji |
Python 3
# Python 3 code to check if a Octal # number is Even or Odd # Check if the number is odd or even def even_or_odd( N):
l = len (N);
# Check if the last digit
# is either '0', '2', '4',
# or '6'
if (N[l - 1 ] = = '0' or N[l - 1 ] = = '2' or
N[l - 1 ] = = '4' or N[l - 1 ] = = '6' ):
return ( "Even" )
else :
return ( "Odd" )
# Driver code N = "735"
print (even_or_odd(N))
# This code is contributed by ANKITKUMAR34 |
C#
// C# code to check if a Octal // number is Even or Odd using System;
public class GFG{
// Check if the number is odd or even static String even_or_odd(String N)
{ int len = N.Length;
// Check if the last digit
// is either '0', '2', '4',
// or '6'
if (N[len - 1] == '0'
|| N[len - 1] == '2'
|| N[len - 1] == '4'
|| N[len - 1] == '6' )
return ( "Even" );
else
return ( "Odd" );
} // Driver code public static void Main(String[] args)
{ String N = "735" ;
Console.Write(even_or_odd(N));
} } // This code contributed by Princi Singh |
Javascript
<script> // Javascript code to check if a Octal // number is Even or Odd // Check if the number is odd or even function even_or_odd(N)
{ var len = N.length;
// Check if the last digit
// is either '0', '2', '4',
// or '6'
if (N[len - 1] == '0'
|| N[len - 1] == '2'
|| N[len - 1] == '4'
|| N[len - 1] == '6' )
return ( "Even" );
else
return ( "Odd" );
} // Driver code var N = "735" ;
document.write(even_or_odd(N));
// This code is contributed by Mayank Tyagi </script> |
Output:
Odd
Time Complexity: O(1)
Auxiliary Space: O(1)