Check if an Octal number is Even or Odd

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)