Here is my c++ solution, you can watch the explanation here : https://youtu.be/HTbex2B-Yas

int hurdleRace(int k, vector<int> height) {
    int mx = *max_element(height.begin(), height.end());
    return max(0, mx-k);
}

This is a great problem for thinking through logic step by step. It all comes down to checking the tallest hurdle and seeing if the character can already clear it. If not, just calculate how many extra boosts are needed. Pretty straightforward but a fun way to break down constraints.

It kind of reminds me of video editing—sometimes you have all the tools you need, and other times you need that extra boost to make things work smoothly. Just like how some editing software gives you everything upfront while others require add-ons to get the job done efficiently.

Perl:

sub hurdleRace {
    my ($k, $height) = @_;
   
    return (max(@$height) < $k) ? 0 : max(@$height) - $k;
}

Java one line solution

public static int hurdleRace(int k, List<Integer> height) {
        return Collections.max(height) <= k ? 0 : Collections.max(height) - k;
    }

int hurdleRace(int k, vector height) { auto result = std::max_element(height.begin(), height.end()); if(k > *result) return 0; return *result - k; }