Leetcode 2413: Smallest Even Multiple

Input: The input consists of a single integer 'n'.

Example: n = 7

Constraints:

• 1 <= n <= 150

Output: Return the smallest integer that is a multiple of both 2 and 'n'.

Example: Output: 14

Constraints:

Goal: The goal is to find the smallest integer that is divisible by both 2 and 'n'.

Steps:

• 1. Check if 'n' is already divisible by 2.

• 2. If 'n' is even, return 'n'.

• 3. If 'n' is odd, return 'n * 2' because the next multiple of both 2 and 'n' will be twice 'n'.

Goal: The solution must efficiently handle values of 'n' up to 150.

Steps:

• 1 <= n <= 150

Assumptions:

• The input is a positive integer.

• The problem ensures that 'n' is always between 1 and 150.

• Input: n = 7

• Explanation: For 'n' = 7, the smallest multiple of both 2 and 7 is 14. For 'n' = 6, since 6 is already divisible by 2, the result is 6.

Approach: The solution uses a simple check to determine if 'n' is already divisible by 2. If not, we double 'n' to get the smallest multiple of both 2 and 'n'.

Observations:

• The smallest multiple of both 2 and any integer can be obtained by checking if the number is even.

• If the number is odd, we just need to multiply it by 2.

• This approach is straightforward and runs in constant time, making it very efficient.

Steps:

• 1. Check if 'n' is divisible by 2.

• 2. If 'n' is even, return 'n'.

• 3. If 'n' is odd, return 'n * 2'.

Empty Inputs:

• The input will always be a positive integer as per the problem constraints.

Large Inputs:

• The input is constrained to a maximum of 150, so there are no concerns regarding large inputs.

Special Values:

• If 'n' is 1, the smallest multiple of both 2 and 1 is 2.

Constraints:

• Make sure the solution handles both even and odd numbers correctly.

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

1 : Function Declaration

int smallestEvenMultiple(int n) {

Function declaration for 'smallestEvenMultiple'. It takes an integer 'n' as input.

2 : Return Statement

    return n % 2 == 0? n: n * 2;

The function checks if 'n' is even. If true, it returns 'n'; otherwise, it returns 'n' multiplied by 2 to make it even.

Best Case: O(1)

Average Case: O(1)

Worst Case: O(1)

Description: The time complexity is constant because only a few arithmetic operations are needed.

Best Case: O(1)

Worst Case: O(1)

Description: The space complexity is constant, requiring only a small amount of space to store the result.

Leetcode 2413: Smallest Even Multiple

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