In this HackerRank Chief Hopper problem solution, Chief's bot is playing an old DOS-based game. There is a row of buildings of different heights arranged at each index along a number line. The bot starts at building 0 and at a height of 0. You must determine the minimum energy his bot needs at the start so that he can jump to the top of each building without his energy going below zero.

Problem solution in Python.

n = int(input())
high = low = 0
arr = []

for i in input().split():
arr.append(int(i))
if(int(i) > high):
high = int(i)
if(int(i) < low):
low = int(i)

def tester(x, arr):
energy = x
for height in arr:
if(height > energy):
energy -= (height - energy)
if(energy < 0):
return(False)
else:
energy += (energy - height)

cont = True
i = (high + low) // 2
dic = {}
while(cont):
if(dic.get(i-1, tester(i-1, arr))):
# i is too high
high = i
i = (high + low) // 2
else:
if(dic.get(i, tester(i, arr))):
cont = False
else:
# i is too low, double it and set low to current i
low = i
i = ((i + high + 1) // 2)
if(i == 0):
i = 1

print(i)

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Problem solution in Java.

import java.io.*;
import java.util.*;

public class Solution {

public static void main(String[] args) throws Exception {
/* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */
String s;
StringTokenizer st;
st=new StringTokenizer(s);
int height[]=new int[N];
for(int i=0;i<N;i++)
{
height[i]=Integer.parseInt(st.nextToken());
}
int min=0;
for(int i=N-1;i>=0;i--)
{
min= (int) Math.ceil((double)(min+height[i])/2);

}
System.out.println(min);
}
}

{"mode":"full","isActive":false}

Problem solution in C++.

#include <iostream>
#include <algorithm>
#include <stdio.h>

#define LL long long
#define getcx getchar//_unlocked

inline void inp( int &n )
{ n=0;int ch=getcx();
while(ch<'0'||ch>'9') ch=getcx();
while(  ch >= '0' && ch <= '9' )
n = (n<<3)+(n<<1) + ch-'0', ch=getcx(); }

using namespace std;

int n,i,H;
LL ans;

int main(){inp(n);
for (i=1;i<=n;i++) inp(H[i]);
reverse(H+1,H+n+1);
for (i=1;i<=n;i++)
ans=(ans+(LL)H[i]+1)/2LL;
cout<<ans<<endl;}

{"mode":"full","isActive":false}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {

/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int b,n,i,ans;
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d",&b[i]);
}
ans=0;
for(i=n-1;i>=0;i--){
ans=(ans+b[i])/2+(ans+b[i])%2;
}
printf("%d",ans);
return 0;
}

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