Smallest Even Multiple

The task is to find the smallest positive integer that is a multiple of both 2 and the given number n.

A key observation is that for an integer to be divisible by 2, it must be an even number. Therefore, if n is an even number, it’s already the smallest possible number that is a multiple of both 2 and n. However, if n is an odd number, the smallest number that is a multiple of both 2 and n would be 2*n.

We can write this logic in Python as follows:

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

In this function, if n is an even number (i.e., n % 2 == 0), it simply returns n. If n is odd, it returns 2 * n. This ensures that the smallest even multiple of n is returned.

This problem contradicts itself.

The problem asks us to find the smallest positive integer that is a multiple of both 2 and n.

Example 2 notes that a number is a multiple of itself, and then gives the answer for 6 as 6.

However, if a number is a multiple of itself, then the correct answer is 2, as 2 is a multiple of itself and 6.

When the problem statement says that the result needs to be a multiple of both 2 and n, it implies that the result needs to be a multiple of n, but it also has to be an even number.

Let’s take the example where n is 6.

6 is an even number and a multiple of itself. So, it satisfies the conditions mentioned in the problem statement - it’s an even number and a multiple of n.

If you consider 2 as a result, it is indeed a multiple of 2 but it’s not a multiple of n (6 in this case).

The key point is that the number we’re looking for has to be a multiple of n (and hence could be n itself if n is even), but it also has to be an even number.

This is why, if n is already an even number, n is returned. If n is an odd number, 2*n is returned to ensure the result is even.

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int smallestEvenMultiple(int n) {
    return n * (n % 2 + 1);
}

The C++ function smallestEvenMultiple calculates the smallest multiple of n that is even. It takes an integer n as an argument and returns its smallest even multiple.

Here’s how the function works:

It calculates n % 2 + 1. If n is even, n % 2 is 0, and this expression results in 1. If n is odd, n % 2 is 1, and this expression results in 2.
It then multiplies n by the result of the above calculation. If n was even, this results in n * 1, which is n itself. If n was odd, this results in n * 2, which is the smallest even multiple of n.

The function uses simple arithmetic operations and conditional logic to find the smallest even multiple of n, resulting in a time complexity of O(1) and a space complexity of O(1).

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int smallestEvenMultiple(int n) {
    return lcm(2, n);
}

The C++ function smallestEvenMultiple calculates the smallest even multiple of n. It does this by calculating the least common multiple (LCM) of 2 and n using the lcm function.

Here’s how the function works:

The function lcm(2, n) calculates the least common multiple of 2 and n. The least common multiple of two numbers is the smallest number that is a multiple of both numbers. Since 2 is the smallest even number, the LCM of 2 and any number n will be the smallest even multiple of n.

This function uses the lcm function to find the smallest even multiple of n, resulting in a time complexity of O(log n), where n is the input number. The space complexity is O(1).

The lcm function is not shown here, and it should be implemented separately. The function typically utilizes the mathematical relationship lcm(a, b) = |a*b| / gcd(a, b), where gcd(a, b) is the greatest common divisor of a and b.