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:

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++

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// 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;
}

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Java

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// Java code to check if a Octal
// number is Even or Odd
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

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Python 3

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# 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

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C#

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// 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

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Output:

Odd

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