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Home Code Solutions Hackerrank Algorithms

Chief Hopper – HackerRank Solution

Chief Hopper - HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

bhautik bhalala by bhautik bhalala
May 27, 2022
Reading Time: 1 min read
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15 Days to learn SQL Hard SQL(Advanced)-Solution

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Table of Contents

  • Chief Hopper  – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.
  • Solutions of Algorithms Data Structures Hard HackerRank:
    • Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts
  • C++ Chief Hopper HackerRank Solution
  • Java Chief Hopper HackerRank Solution
  • Python 3 Chief Hopper HackerRank Solution
  • Python 2 Chief Hopper HackerRank Solution
  • C Chief Hopper HackerRank Solution
    • Warmup Implementation Strings Sorting Search Graph Theory Greedy Dynamic Programming Constructive Algorithms Bit Manipulation Recursion Game Theory NP Complete Debugging
    • Leave a comment below
      • Related posts:

Chief Hopper  – HackerRank Solution Java , Python 3, Python 2 , C , C++, Best and Optimal Solutions , All you need.

Solutions of Algorithms Data Structures Hard HackerRank:

Here are all the Solutions of Hard , Advanced , Expert Algorithms of Data Structure of Hacker Rank , Leave a comment for similar posts

C++ Chief Hopper HackerRank Solution


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#include <vector>
#include <list>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <algorithm>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <limits>
#include <cstring>
#include <string>
using namespace std;

#define pairii pair<int, int>
#define llong long long
#define pb push_back
#define sortall(x) sort((x).begin(), (x).end())
#define INFI  numeric_limits<int>::max()
#define INFLL numeric_limits<llong>::max()
#define INFD  numeric_limits<double>::max()
#define FOR(i,s,n) for (int (i) = (s); (i) < (n); (i)++)
#define FORZ(i,n) FOR((i),0,(n))

const int MAXN = 100005;
int ar[MAXN];

void solve() {
  int n;
  scanf("%d",&n);
  FORZ(i,n) scanf("%d",ar+i);
  int res = 0;
  for (int i = n-1; i >= 0; i--) {
    int x = res + ar[i];
    res = x/2 + x%2;
  }
  printf("%d",res);
}

int main() {
#ifdef DEBUG
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
#endif
  solve();
  return 0;
}

Java Chief Hopper HackerRank Solution


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import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;

public class Solution {

	private static class InputArray {
		final StringTokenizer strt;
		public final int length;
		
		InputArray(String str, int N){
			strt = new StringTokenizer(str);
			length = N;
		}
		
		int next(){
			return Integer.parseInt(strt.nextToken());
		}
	}
	
	public static long solve(InputArray array) {
		long botEnergy = 0L; // initial bot energy
		long power2 = 1L;
		
		final long MAX_LIMIT = 2 << 18;
		
		long b = botEnergy;
		for (int i = 0; i < array.length && i < 30; ++i) {
			b = 2 * b - array.next();
			
			if(b >= MAX_LIMIT) //after this no amount of input will make energy negative
				return botEnergy;
			
			power2 *= 2;
			if (b < 0) {
				// increase the initial botEnergy by a minimum amount necessary
				double dif = -1 * b;
				long x = Math.round(Math.floor(dif / power2));
				for (long inc = 0; inc <= 10; ++inc) {
					long newb = power2 * (x + inc) + b;
					if (newb >= 0) {
						botEnergy += (x + inc);
						b = newb;
						break;
					}
				}
			}
		}

		for (int i = 30; i < array.length; ++i) {
			b = 2 * b - array.next();
			if (b < 0) {
				botEnergy++; // after this increase bot energy can never become
								// smaller than 0
				break;
			}
		}

		return botEnergy;
	}

	public static void main(String[] args) throws NumberFormatException, IOException {
		BufferedReader bfr = new BufferedReader(new InputStreamReader(System.in));
		final int N = Integer.parseInt(bfr.readLine());
		InputArray array = new InputArray(bfr.readLine(), N);
		
		System.out.println(solve(array));
	}
}



Python 3 Chief Hopper HackerRank Solution


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N = int(input())
H = [int(_) for _ in input().split()]

i,b = len(H),H[-1]
while i > 0 :
    i -= 1
    b = (H[i]+b+1)//2
#    print(b)
    
print(b)



Python 2 Chief Hopper HackerRank Solution


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from math import ceil

def validate(val):
    global n, h
    for i in xrange(n):
        val = ((val * 2) - h[i])
        if val < 0:
            return False
    return True

if __name__ == "__main__":
    n = int(raw_input())
    
    h = map(int, raw_input().split())
    i = 0
    mx = max(h)
    result = mx + 1
    start = 0
    
    while start >= 0 and start <= mx:
        mid = ceil((start+mx)/2.0)
        if validate(mid):
            mx = mid
            if mid < result:
                result = mid
            else:
                break
        else:
            start = mid
            
    print int(result)



C Chief Hopper HackerRank Solution


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

int main() {
  int n, *h, i;
  unsigned long long tot;
  
  scanf("%d",&n);
  h = malloc(n * sizeof(int));
  for (i=0; i<n; i++) scanf("%d",&h[i]);
  
  tot = 0;
  i--;
  while (i>=0) {
    tot += h[i];
    if (tot & 1) tot++;
    tot /= 2;
    i--;
  }
  
  printf("%lld\n",tot);
  
  return 0;
}

 

Warmup
Implementation
Strings
Sorting
Search
Graph Theory
Greedy
Dynamic Programming
Constructive Algorithms
Bit Manipulation
Recursion
Game Theory
NP Complete
Debugging

Leave a comment below

 

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Tags: Cc++14Chief Hopperfull solutionGoHackerRank Solutionjavajava 15java 7java 8java8javascriptpypy 3Python 2python 3
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