**The Hurdle Race**

### Problem Statement :

A video player plays a game in which the character competes in a hurdle race. Hurdles are of varying heights, and the characters have a maximum height they can jump. There is a magic potion they can take that will increase their maximum jump height by 1 unit for each dose. How many doses of the potion must the character take to be able to jump all of the hurdles. If the character can already clear all of the hurdles, return 0. Example height = [1, 2, 3, 3, 2] k = 1 The character can jump 1 unit high initially and must take 3 - 1 = 2 doses of potion to be able to jump all of the hurdles. Function Description Complete the hurdleRace function in the editor below. hurdleRace has the following parameter(s): int k: the height the character can jump naturally int height[n]: the heights of each hurdle Returns int: the minimum number of doses required, always 0 or more Input Format The first line contains two space-separated integers n and k, the number of hurdles and the maximum height the character can jump naturally. The second line contains n space-separated integers height[i] where 0 <= i < n. Constraints 1 <= n, k <= 100 1 <= height[i] <= 100

### Solution :

` ````
Solution in C :
python 3 :
#!/bin/python3
import sys
n,k = input().strip().split(' ')
n,k = [int(n),int(k)]
height = list(map(int, input().strip().split(' ')))
# your code goes here
print(max(0, max(height) - k))
java :
import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;
public class Solution {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
int[] height = new int[n];
int max = 0;
for(int height_i=0; height_i < n; height_i++){
height[height_i] = in.nextInt();
max = Math.max(max, height[height_i]);
}
System.out.println(Math.max(0, max - k));
// your code goes here
}
}
C ++ :
#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;
int main(){
int n;
int k;
cin >> n >> k;
vector<int> height(n);
int maxall=0;
for(int height_i = 0; height_i < n; height_i++){
cin >> height[height_i];
maxall=max(maxall,height[height_i]);
}
cout<<max(0,maxall-k);
// your code goes here
return 0;
}
C :
#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>
int main(){
int n;
int k;
int max=-1,ans=0;
scanf("%d %d",&n,&k);
int *height = malloc(sizeof(int) * n);
for(int height_i = 0; height_i < n; height_i++){
scanf("%d",&height[height_i]);
if(height[height_i]>max)
max=height[height_i];
}
if(max>k)
ans=max-k;
printf("%d",ans);
// your code goes here
return 0;
}
```

## View More Similar Problems

## Tree: Inorder Traversal

In this challenge, you are required to implement inorder traversal of a tree. Complete the inorder function in your editor below, which has 1 parameter: a pointer to the root of a binary tree. It must print the values in the tree's inorder traversal as a single line of space-separated values. Input Format Our hidden tester code passes the root node of a binary tree to your $inOrder* func

View Solution →## Tree: Height of a Binary Tree

The height of a binary tree is the number of edges between the tree's root and its furthest leaf. For example, the following binary tree is of height : image Function Description Complete the getHeight or height function in the editor. It must return the height of a binary tree as an integer. getHeight or height has the following parameter(s): root: a reference to the root of a binary

View Solution →## Tree : Top View

Given a pointer to the root of a binary tree, print the top view of the binary tree. The tree as seen from the top the nodes, is called the top view of the tree. For example : 1 \ 2 \ 5 / \ 3 6 \ 4 Top View : 1 -> 2 -> 5 -> 6 Complete the function topView and print the resulting values on a single line separated by space.

View Solution →## Tree: Level Order Traversal

Given a pointer to the root of a binary tree, you need to print the level order traversal of this tree. In level-order traversal, nodes are visited level by level from left to right. Complete the function levelOrder and print the values in a single line separated by a space. For example: 1 \ 2 \ 5 / \ 3 6 \ 4 F

View Solution →## Binary Search Tree : Insertion

You are given a pointer to the root of a binary search tree and values to be inserted into the tree. Insert the values into their appropriate position in the binary search tree and return the root of the updated binary tree. You just have to complete the function. Input Format You are given a function, Node * insert (Node * root ,int data) { } Constraints No. of nodes in the tree <

View Solution →## Tree: Huffman Decoding

Huffman coding assigns variable length codewords to fixed length input characters based on their frequencies. More frequent characters are assigned shorter codewords and less frequent characters are assigned longer codewords. All edges along the path to a character contain a code digit. If they are on the left side of the tree, they will be a 0 (zero). If on the right, they'll be a 1 (one). Only t

View Solution →