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Leetcode 2413. Smallest Even Multiple

Problem Statement

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

Clarifying Questions

1. Q: What is the range of n?
   • A: n is a positive integer, and for this problem, you can assume typical constraints such as 1 <= n <= 1000.
2. Q: Can n be 1?
   • A: Yes, n can be 1.
3. Q: Should we handle large numbers or only typical input ranges?
   • A: Typical ranges suffices, typically up to 1000 or similar.

Strategy

The task involves finding the smallest positive integer that is a multiple of both n and 2.

• Note that the smallest even multiple of a number is either the number itself if it is even, or twice the number if it is odd.
• Therefore, if n is already even, the smallest multiple is n. If n is odd, the smallest even multiple will be 2 * n.

Code

#include <iostream>

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

int main() {
    Solution solution;
    std::cout << solution.smallestEvenMultiple(5) << std::endl;  // Outputs 10
    std::cout << solution.smallestEvenMultiple(6) << std::endl;  // Outputs 6
    return 0;
}

Time Complexity

The time complexity of this solution is (O(1)) because:

• The solution involves only a simple conditional check and a potential multiplication, both of which are constant time operations.

Space Complexity

The space complexity is (O(1)) because:

• No extra space is used apart from the input and a few variables.

Feel free to ask further questions if any arise!

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