Smallest Even Multiple - Practice Coding Problems

Smallest Even Multiple - Problem

Given a positive integer n, return the smallest positive integer that is a multiple of both 2 and n.

In other words, find the smallest number that is divisible by both 2 and n.

Input & Output

Example 1 — Odd Number

$ Input: n = 5

› Output: 10

💡 Note: Since 5 is odd, the smallest multiple of both 2 and 5 is 2×5 = 10. We can verify: 10÷2 = 5 and 10÷5 = 2.

Example 2 — Even Number

$ Input: n = 6

› Output: 6

💡 Note: Since 6 is even, it's already divisible by 2. The smallest multiple of both 2 and 6 is 6 itself. We can verify: 6÷2 = 3 and 6÷6 = 1.

Example 3 — Small Odd

$ Input: n = 1

› Output: 2

💡 Note: Since 1 is odd, the smallest multiple of both 2 and 1 is 2×1 = 2. We can verify: 2÷2 = 1 and 2÷1 = 2.

Constraints

1 ≤ n ≤ 150

Visualization

Tap to expand

Understanding the Visualization

Input

Given positive integer n

Process

Find LCM of 2 and n

Output

Return smallest multiple

Key Takeaway

🎯 Key Insight: The LCM of 2 and n is simply n if n is even, or 2n if n is odd

Asked in

a Amazon 15 Apple 8

The key insight is recognizing this as a Least Common Multiple (LCM) problem. Since we need the smallest number divisible by both 2 and n, we use the fact that LCM(2,n) = n if n is even, or 2n if n is odd. Best approach is the mathematical solution with Time: O(1), Space: O(1).

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force	O(n)	O(1)	Check every positive integer until finding one divisible by both 2 and n
Mathematical Approach	O(1)	O(1)	Use the fact that LCM(2,n) = n if n is even, or 2n if n is odd

Brute Force — Algorithm Steps

Start with number 1
Check if current number is divisible by both 2 and n
If yes, return it; otherwise increment and repeat

Visualization

Tap to expand

Step-by-Step Walkthrough

Start

Begin checking from number 1

Check

Test divisibility by both 2 and n

Found

Return first number that works

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int n) {
    int current = 1;
    while (1) {
        if (current % 2 == 0 && current % n == 0) {
            return current;
        }
        current++;
    }
}

int main() {
    int n;
    scanf("%d", &n);
    int result = solution(n);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n)

In worst case (when n is odd), we check up to 2n numbers

✓ Linear Growth

Space Complexity

O(1)

Only using a few variables for iteration

✓ Linear Space

31.5K Views

Low Frequency

~5 min Avg. Time

850 Likes

Ln 1, Col 1

Smart Actions

💡 Explanation

AI Ready

💡 Suggestion Tab to accept Esc to dismiss

// Output will appear here after running code

Code Editor Closed

Click the red button to reopen

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Select Compiler