#2413

Easy

LeetCode Problem Workspace

Smallest Even Multiple

Find the smallest even multiple of a given integer using math and number theory concepts efficiently.

Problem Statement

Given a positive integer n, return the smallest positive integer that is a multiple of both 2 and n. The solution must handle any n within the given constraints efficiently.

For example, if n = 5, the smallest multiple of 5 and 2 is 10. If n = 6, since 6 is already even, the smallest multiple of 6 and 2 is 6 itself. Constraints: 1 <= n <= 150.

Examples

Example 1

Input: n = 5

Output: 10

The smallest multiple of both 5 and 2 is 10.

Example 2

Input: n = 6

Output: 6

The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

Constraints

1 <= n <= 150

Solution Approach

Check if n is even

If n is even, it is already a multiple of 2, so return n directly. This uses the property that any even number is divisible by 2, minimizing computation.

Multiply by 2 if n is odd

If n is odd, the smallest even multiple cannot be n itself. Multiply n by 2 to guarantee an even multiple that is also divisible by n, using the minimal number theory approach.

Constant time solution

Combine the parity check and conditional multiplication in a single line. This yields an O(1) time and O(1) space solution without loops, ensuring correctness for all 1 <= n <= 150.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity is O(1) because the solution only requires a single parity check and potentially one multiplication. Space complexity is O(1) as no extra storage is needed beyond the input and output.

What Interviewers Usually Probe

Asks if n is even or odd to guide towards a simple formula.
Looks for a constant time solution without loops.
Wants understanding of minimal number theory application for multiples.

Common Pitfalls or Variants

Common pitfalls

Forgetting that even numbers return n itself, leading to unnecessary multiplication.
Overcomplicating by trying to find least common multiple manually.
Not considering constraints, assuming n could be very large and optimizing incorrectly.

Follow-up variants

Find smallest multiple divisible by 3 and n instead of 2.
Compute smallest multiple divisible by a given k and n.
Generalize to find smallest multiple divisible by multiple given numbers.

FAQ

What is the simplest way to find the smallest even multiple of n?

Check if n is even. If yes, return n. If not, return 2*n.

Does this approach work for all n between 1 and 150?

Yes, the parity check covers all values in the constraint range, ensuring correctness.

Why not compute multiples until finding an even one?

Iterating is unnecessary and slower. The parity check provides a direct O(1) solution.

Is there a formula to generalize this to other divisors?

Yes, for divisor k, return n if n is divisible by k, otherwise return n*k divided by gcd(n,k).

Which problem pattern does this follow?

This is a Math plus Number Theory pattern focused on multiples and parity checks.

terminal

Solution

Solution 1: Mathematics

If $n$ is even, then the least common multiple (LCM) of $2$ and $n$ is $n$ itself. Otherwise, the LCM of $2$ and $n$ is $n \times 2$.

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class Solution:
    def smallestEvenMultiple(self, n: int) -> int:
        return n if n % 2 == 0 else n * 2

Continue Practicing

#2427 number of common factors #2447 number of subarrays with gcd equal to k #2470 number of subarrays with lcm equal to k