CareerCup

Hey Anup,

The idea described by LinhHA05 is the most optimal one. Here's some sample code:

int getHighest(vector<int> numbers, int givenNumber) {
    int bestNumber = numeric_limits<int>::min();
    
    for (int i = 0; i < numbers.size(); i++) {
        if (numbers[i] <= givenNumber && numbers[i] > bestNumber) {
            bestNumber = numbers[i];
        }
    }
    
    return bestNumber;
}

At an interview, you should be very careful with two things:
1) How should you set your initial value for bestNumber (how low can givenNumber be)?
2) What should your code return when there are no numbers smaller than or equal to givenNumber?

Also:

The solution above is optimal assuming the "getHighest" function is called only once. If you need a solution that works for multiple queries over the same set of numbers, let me know.

Cheers,

Slavi
HiredInTech

The O(n) solution is trivial, go through all element and update the maximal value less than the reference.