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Solution Review: Check If a Given Number is Even/Odd

This is a solution to another Bitwise question. Think of the rightmost bit of a number and think of some logic that could solve this. We solve using AND operator with O(1) time Complexity.

Gopi Gorantala
Gopi Gorantala
3 min read

Table of Contents

Here, we see the solution for checking if a number is even/odd.

Solution review

We must review all the elements to check if any are even/Odd.

It would be best to see the pattern of bits present in odd numbers. If you take a closer look at each of them, you can see the right-most significant bit is set to 1 (for 20 place).

So, we do an AND bitwise operation with 1 decimal number to see if the resultant output is 1. If not, it is an even number else odd.

Algorithm

  • Iterate over all the elements to check
  • if (element & 1) == 1
    • if true, print "Odd"
    • else, print "Even"

Explanation

Table illustrating values at various levels of each operation.

n n & 1 ((n & 1) == 1) Output
1 1 True Odd
2 0 False Even
3 1 True Odd
4 0 False Even
5 1 True Odd
6 0 False Even
7 1 True Odd
8 0 False Even
9 1 True Odd

Solution

Here is the logic behind this problem and its solution.

Java

class IsEven {
    private static String helper(int n) {
        return (n & 1) == 0 ? "Even" : "Odd";
    }

    public static void main(String[] args) {
        System.out.println("Number '" + 1 + "' is : " + helper(1));
        System.out.println("Number '" + 2 + "' is : " + helper(2));
        System.out.println("Number '" + 3 + "' is : " + helper(3));
        System.out.println("Number '" + 4 + "' is : " + helper(4));
        System.out.println("Number '" + 5 + "' is : " + helper(5));
    }
}

Python

def CheckEvenOdd(n):
    return "even" if (n & 1)==0 else "odd"

print("Number ",1," is : " ,CheckEvenOdd(1))
print("Number ",2," is : " ,CheckEvenOdd(2))
print("Number ",3," is : " ,CheckEvenOdd(3))
print("Number ",4," is : " ,CheckEvenOdd(4))
print("Number ",5," is : " ,CheckEvenOdd(5))

JavaScript

const IsEven = n => {
    return (n & 1) === 0 ? 'Even' : 'Odd';
}

console.log (`Number '1' is : ${IsEven (1)}`);
console.log (`Number '2' is : ${IsEven (2)}`);
console.log (`Number '3' is : ${IsEven (3)}`);
console.log (`Number '4' is : ${IsEven (4)}`);
console.log (`Number '5' is : ${IsEven (5)}`);

C++

#include <iostream>
#include <string>

using namespace std;

string helper(int n) {
    return (n & 1) == 0 ? "even" : "odd";
}

int main() {
    int firstNumber = 125;
    int secondNumber = 8;
    cout << "Number '" << 1 << "' is : " << helper(1) << endl;
    cout << "Number '" << 2 << "' is : " << helper(2) << endl;
    cout << "Number '" << 3 << "' is : " << helper(3) << endl;
    cout << "Number '" << 4 << "' is : " << helper(4) << endl;
    cout << "Number '" << 5 << "' is : " << helper(5) << endl;
    return 0;
}

TypeScript

export const IsEven = (n: number): string => {
    return (n & 1) === 0 ? 'Even' : 'Odd';
}

console.log(`Number '1' is : ${IsEven(1)}`);
console.log(`Number '2' is : ${IsEven(2)}`);
console.log(`Number '3' is : ${IsEven(3)}`);
console.log(`Number '4' is : ${IsEven(4)}`);
console.log(`Number '5' is : ${IsEven(5)}`);

Array of outputs

Let’s return an array of string outputs based on the input values. A code snippet of each language follows.

Java

import java.util.Arrays;

class CheckEvenOdd {
    private static String[] checkEvenOdd(int[] nums) {
        String[] ans = new String[nums.length];

        int k = 0;
        for (int n : nums) {
            ans[k++] = ((n & 1) == 1) ? "Odd" : "Even"; 
        }
        return ans;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        System.out.println(Arrays.toString(checkEvenOdd(nums)));
    }
}

Python

def IsEven(array):
    result = []

    def helper(array):
        k = 0
        for n in array:
            result.append("Odd" if (n & 1) == 1 else "Even")
            k += 1
        return result

    return helper(array)

print(IsEven([1, 2, 3, 4, 5, 6, 7, 8, 9]))

JavaScript

const IsEven = array => {
    const result = [];

    function helper (array) {
        let k = 0;
        for(let n of array) {
            result[k++] = ((n & 1) === 1) ? "Odd" : "Even";
        }
        return result;
    }

    return helper (array);
}

console.log (IsEven ([1, 2, 3, 4, 5, 6, 7, 8, 9]));

TypeScript

export const IsEven = (array: number[]): string[] => {
    const result: string[] = [];

    function helper(array: number[]): string[] {
        let k: number = 0;
        for (let n of array) {
            result[k++] = ((n & 1) === 1) ? "Odd" : "Even";
        }
        return result;
    }

    return helper(array);
}

console.log(IsEven([1, 2, 3, 4, 5, 6, 7, 8, 9]));

Complexity analysis

Time complexity: O(n)

We need to review the entire array to check each and see if they are even/odd. So, the time complexity is directly proportional to the number of elements present in the nums array.

Space complexity: O(n)

We do not need to create an array here. We can print the Odd and Even directly on the console.

To make things easier to represent, we used an extra space array of Strings to store the output.

Coding Interview QuestionsData Structures and AlgorithmsBit Manipulation

Gopi Gorantala Twitter

Gopi is an engineering leader with 12+ of experience in full-stack development—a specialist in Java technology stack. He worked for multiple startups, the European govt, and FAANG in India and Europe.

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