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 :



title-img 


                            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;
}








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