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 evenstring 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 codeint main(){ string N = "735"; cout << even_or_odd(N); return 0;} |
Java
// Java code to check if a Octal// number is Even or Oddimport java.io.*;class GFG{ // Check if the number is odd or evenstatic 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 codepublic 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 evendef 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 codeN = "735"print(even_or_odd(N))# This code is contributed by ANKITKUMAR34 |
C#
// C# code to check if a Octal// number is Even or Oddusing System;public class GFG{ // Check if the number is odd or evenstatic 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 codepublic 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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