Check whether a given number is even or odd - Arrays

Problem Statement: Given a number n, check whether a given number is even or odd.

Examples:

Example 1:
Input: n=5
Output: odd
Explanation: 5 is not divisible by 2.
 
Example 2:  
Input: n=6
Output: even
Explanation: 6 is divisible by 2.

Solution

Disclaimer: Don’t jump directly to the solution, try it out yourself first.

Solution1: Using division

Intuition: If we can prove that the given number is divisible by 2 then it will be even, otherwise it will be odd.

Approach:

Find the remainder of the number.
If it is 0 then the number is even otherwise it is odd.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;

int remainder(int n) {
    return (n % 2);
  }

int main() {
    int n = 5;
    if (remainder(n) == 0) {
      cout<<n <<" is even.";
    } else {
      cout<<n<<" is odd.";
    }
  }

Output:

5 is odd.

Time complexity: O(1)

Space Complexity: O(1)

Java Code

import java.util.*;

public class tuf {
  public static void main(String[] args) {
    int n = 5;
    if (remainder(n) == 0) {
      System.out.println(n + " is even.");
    } else {
      System.out.println(n + " is odd.");
    }
  }
  public static int remainder(int n) {
    return (n % 2);
  }
}

Output:

5 is odd.

Time complexity: O(1)

Space Complexity: O(1)

Solution2: Using the bitwise operator

Intuition: As we know that bitwise operations are generally faster than normal operations so we should prefer it over normal operations.

Even Number: The last bit of even number is always 0.

Odd Number: Last bit of the odd number is always 1.

If we can get the last bit of any number then we can say that it is even or odd.

Approach:

Take AND of n with 1.
If it is 0 then print even otherwise odd.

Code:

C++ Code

#include<bits/stdc++.h>
using namespace std;
 int solve(int n) {
    if ((n & 1) == 0)
      return 0;
    else
      return 1;
  }

  int main() {
    int n = 5;
    if (solve(n) == 0) {
      cout<<n<<" is even.";
    } else {
      cout<<n << " is odd.";
    }
  }

Output:

5 is odd.

Time complexity: O(1)

Space Complexity: O(1)

Java Code

import java.util.*;
public class tuf {

  public static void main(String[] args) {

    int n = 5;
    if (solve(n) == 0) {
      System.out.println(n + " is even.");
    } else {
      System.out.println(n + " is odd.");
    }
  }
  public static int solve(int n) {
    if ((n & 1) == 0)
      return 0;
    else
      return 1;
  }
}

Output:

5 is odd.

Time complexity: O(1)

Space Complexity: O(1)

Special thanks to Prashant Sahu for contributing to this article on takeUforward. If you also wish to share your knowledge with the takeUforward fam, please check out this article