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.

#2413

Easy

auto_awesome数学·结合·number·theory

LeetCode 题解工作台

最小偶倍数

给你一个正整数 n ，返回 2 和 n 的最小公倍数（正整数）。示例 1：输入： n = 5 输出： 10 解释： 5 和 2 的最小公倍数是 10 。示例 2：输入： n = 6 输出： 6 解释： 6 和 2 的最小公倍数是 6 。注意数字会是它自身的倍数。提示： 1

数学数论

题目描述

给你一个正整数 n ，返回 2 和 n 的最小公倍数（正整数）。

示例 1：

输入：n = 5
输出：10
解释：5 和 2 的最小公倍数是 10 。

示例 2：

输入：n = 6
输出：6
解释：6 和 2 的最小公倍数是 6 。注意数字会是它自身的倍数。

提示：

1 <= n <= 150

lightbulb

解题思路

方法一：数学

如果 $n$ 为偶数，那么 $2$ 和 $n$ 的最小公倍数就是 $n$ 本身。否则， $2$ 和 $n$ 的最小公倍数就是 $n \times 2$ 。

时间复杂度 $O(1)$ 。

1

2

3

4

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

speed

复杂度分析

指标	值
时间	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.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

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

warning

常见陷阱

外企场景

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

swap_horiz

进阶变体

外企场景

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

help

常见问题

外企场景

继续练习

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