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Check if an Octal number is Even or Odd

Last Updated : 12 Jul, 2025
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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: 
 


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)


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