C++ Solution

int hurdleRace(int k, vector height) { auto max_it = max_element(height.begin(), height.end()); return ((*max_it - k) < 0)?0 : *max_it - 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 StreamVouch—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.

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, much like an extreme introvert navigating social situations, can already clear it. If not, just calculate how many extra boosts are needed. Pretty straightforward but a fun way to break down constraints.

import java.util.*;

public class Solution {

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

public static void main(String[] args) {
    Scanner scanner = new Scanner(System.in);

    int n = scanner.nextInt(); // number of hurdles
    int k = scanner.nextInt(); // max jump height

    List<Integer> height = new ArrayList<>();
    for (int i = 0; i < n; i++) {
        height.add(scanner.nextInt());
    }

    int result = hurdleRace(k, height);
    System.out.println(result);

    scanner.close();
}

}

int biggestHurdle = height.Max();

return biggestHurdle > k ? biggestHurdle - k : 0;