Skip to main content
CodePeet

CodePeet 444 Complete DSA Sheet

Even or Odd

Try it yourself
EasyProblem

Description

Given an integer n, return "Even" if n is even and "Odd" if it is odd.

A number is even when it is a multiple of 2: it splits into pairs with nothing left over. Zero and negative numbers count too, so 0 is even and -3 is odd.

Examples

Example 1

Input: n = 4

Output: "Even"

Explanation: 4 is 2 × 2, so it splits into pairs with nothing left over.

Example 2

Input: n = 7

Output: "Odd"

Explanation: 7 is 2 × 3 + 1: one is left over.

Example 3

Input: n = 0

Output: "Even"

Explanation: 0 is 2 × 0, a multiple of 2, so it's even.

Example 4

Input: n = -3

Output: "Odd"

Explanation: -3 is 2 × (-2) + 1, so it's odd; negative numbers are even or odd just like positive ones.

Constraints

  • -10^9 ≤ n ≤ 10^9

Editorial

Approach - Check the Remainder

Intuition

Even means n splits into pairs with nothing left over, so the remainder after dividing by 2 says it all: n % 2 is 0 when n is even.

Test for 0, not for 1. In Java, C++ and JavaScript the remainder takes the sign of the number, so -3 % 2 is -1 there, while Python gives 1 (run on our judge). A remainder of 0 means even in every language, negatives included.

Step-by-Step Explanation

Let's trace Example 2, n = 7:

  1. Take n = 7: 7 items in a row.
  2. Pair them up two at a time: 3 pairs, 1 with 2, 3 with 4, 5 with 6.
  3. One item is left over: that's the remainder, 7 % 2 = 1.
  4. The remainder isn't 0, so 7 is odd: return "Odd".

Pair Them Up — The items paired two at a time; the one without a partner is the remainder.

Algorithm

  1. Work out n % 2.
  2. If it is 0, return "Even"; otherwise return "Odd".

Code

class Solution {
public:
    string evenOrOdd(int n) {
        return n % 2 == 0 ? "Even" : "Odd";
    }
};

class Solution:
    def evenOrOdd(self, n: int) -> str:
        return "Even" if n % 2 == 0 else "Odd"

class Solution {
    public String evenOrOdd(int n) {
        return n % 2 == 0 ? "Even" : "Odd";
    }
}

var evenOrOdd = function(n) {
    return n % 2 === 0 ? "Even" : "Odd";
};

Complexity Analysis

Time Complexity: O(1)

One remainder, however large n is.

Space Complexity: O(1)

Nothing is stored.

Common Mistakes

Each of these looks right on a quick read, and the tests still reject it. The outputs below come from running the code.

Testing for a remainder of 1. n % 2 == 1 ? "Odd" : "Even"

On n = -3 it returns "Even" in Java; the answer is "Odd". In Java, C++ and JavaScript the remainder takes the sign of the number, so a negative odd number leaves -1, not 1, and the test for 1 calls it even. Test for a remainder of 0 instead, which is the same in every language.

Dividing with /. n / 2 * 2 == n

On n = 7 it returns "Even"; the answer is "Odd". In Python, / always gives a float: 7 / 2 is 3.5, and 3.5 * 2 is 7 again, so every number passes the check and comes out even. Halving and doubling only loses the remainder with //, and n % 2 says it directly.

Reading a remainder as even. return "Even" if n % 2 else "Odd"

On n = 4 it returns "Odd"; the answer is "Even". n % 2 is 1 (true) when n is odd and 0 (false) when it's even, so if n % 2 is the test for odd. Swapping the two answers gets every number wrong.

Where This Pattern Is Taught

This problem practises the remainder operator %, and the sign it gives a negative number in each language. It's taught in Python for DSA, Java for DSA, C++ for DSA and JavaScript for DSA.