Scale Engineer

Smallest Even Multiple

Easy
#2197Time: O(1). The loop will execute at most twice. If `n` is even, it executes once. If `n` is odd, it executes twice (for `n` and `2n`).Space: O(1). We only use a single extra variable to store the current multiple.
Patterns
MathNumber Theory

Prompt

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

 

Example 1:

Input: n = 5
Output: 10
Explanation: The smallest multiple of both 5 and 2 is 10.

Example 2:

Input: n = 6
Output: 6
Explanation: The smallest multiple of both 6 and 2 is 6. Note that a number is a multiple of itself.

 

Constraints:

  • 1 <= n <= 150

Approaches

3 approaches with complexity analysis and trade-offs.

This approach simulates a search for the smallest common multiple. We start with the number n and check its multiples one by one (n, 2n, 3n, ...) until we find one that is also a multiple of 2 (i.e., an even number).

Algorithm

    1. Initialize a variable multiple to n.
    1. Start a while loop that runs indefinitely.
    1. Inside the loop, check if multiple is even using the modulo operator (multiple % 2 == 0).
    1. If it is even, return multiple as it is the smallest even multiple of n.
    1. If it is odd, update multiple to the next multiple of n by adding n to it (multiple += n).

Walkthrough

The algorithm starts by initializing a candidate number, let's call it multiple, to the input n. It then enters a loop. Inside the loop, it checks if multiple is divisible by 2. If multiple % 2 == 0, it means we have found the smallest positive integer that is a multiple of both n and 2. The loop terminates, and this value is returned. If multiple is not divisible by 2, we need to check the next multiple of n. We update multiple by adding n to it (multiple = multiple + n). The loop continues until an even multiple is found. Since 2 * n is always an even multiple of n, this loop is guaranteed to terminate quickly.

class Solution {    public int smallestEvenMultiple(int n) {        int multiple = n;        while (true) {            if (multiple % 2 == 0) {                return multiple;            }            multiple += n;        }    }}

Complexity

Time

O(1). The loop will execute at most twice. If `n` is even, it executes once. If `n` is odd, it executes twice (for `n` and `2n`).

Space

O(1). We only use a single extra variable to store the current multiple.

Trade-offs

Pros

  • Simple to understand and directly models the process of finding a multiple.

  • Guaranteed to be correct and terminates quickly for this problem's constraints.

Cons

  • Involves a loop, which is slightly less direct and efficient than a mathematical or bitwise solution.

  • For this specific problem, the performance difference is negligible, but as a general approach for finding LCM, it can be slow if the numbers are large.

Solutions

class Solution {public  int smallestEvenMultiple(int n) { return n % 2 == 0 ? n : n * 2; }}

Video walkthrough

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