HackerRank Equal problem solution

In this HackerRank Equal Problem solution you have given a starting distribution, calculate the minimum number of operations needed so that every colleague has the same number of pieces.

Problem solution in Python.

def g(diff):
    ans = {0:0, 1:1, 2:1, 3:2, 4:2}
    return diff // 5 + ans[diff % 5]

def f(chocolates, goal):
    return sum(g(chocolate-goal) for chocolate in chocolates)

def get_ans(chocolates):
    min_chocolate = min(chocolates)
    return min(f(chocolates, min_chocolate - dc) for dc in range(4))

T = int(input(""))
for i in range(T):
    N = int(input(""))
    inp = input("").split()
    chocolates = [int(chocolate) for chocolate in inp]
    print (get_ans(chocolates))
        

Problem solution in Java.

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


public class Equal {
    public static ArrayList<Integer> chocoDist ;
    public static int steps ;


    public static void main(String[] args) throws NumberFormatException, IOException
    {    
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        int testCases = Integer.parseInt( in.readLine() );
        //System.out.println("testCases "+testCases);
        int testCount = 0;

        while( testCount < testCases)
        {
            chocoDist = new ArrayList<Integer>();
            int coInterns = Integer.parseInt(in.readLine()) ;
            StringTokenizer st = new StringTokenizer(in.readLine());
            while( st.hasMoreTokens())    chocoDist.add(Integer.parseInt(st.nextToken()));
            //System.out.println("coInterns "+coInterns+" chocoDist "+chocoDist.toString());
            Collections.sort(chocoDist);
            int sum = 0;
            int sum1 = 1;
            int sum2 = 1;
            for(int i = 1 ; i < chocoDist.size() ; i++)	{
            	int diff = chocoDist.get(i) - chocoDist.get(0);
            	sum += diff/5 + (diff%5)/2 + (diff%5)%2/1;
            	diff+=1 ;
            	sum1 += diff/5 + (diff%5)/2 + (diff%5)%2/1;
            	diff+=1 ;
            	sum2 += diff/5 + (diff%5)/2 + (diff%5)%2/1;
            }
            System.out.println(Math.min(Math.min(sum, sum1), sum2));
        }
    }

}

Problem solution in C++.

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>

#define MAX_N 10005
#define INF 1000000000

using namespace std;

int tests,n,answer,s;
int h[MAX_N],minn;

int find_min(int x) {
    int res = INF;
    int tmp = x / 5;
    if ((x - 5 * tmp) % 2 == 0) res = min(res, tmp + (x - 5 * tmp) / 2);
    else res = min(res, tmp + (x - 5 * tmp) / 2 + 1);
    if (tmp >= 1) {
        if ((x - 5 * tmp) % 2 == 1) res = min(res, tmp - 1 + (x + 5 - 5 * tmp) / 2);
        else res = min(res, tmp - 1 + (x + 5 - 5 * tmp) / 2 + 1);
    }
    return res;
}

int main() {
    scanf("%d", &tests);
    for (int test = 0 ; test < tests ; test ++) {
        scanf("%d", &n);
        minn = INF;
        for (int i = 0 ; i < n ; i ++) {
            scanf("%d", &h[i]);
            minn = min(minn, h[i]);
        }
        answer = INF;
        for (int tmp = 0 ; tmp < 20 ; tmp ++) {
            s = 0;
            for (int i = 0 ; i < n ; i ++) {
                s += find_min(h[i] - minn + tmp);
            }
            answer = min(answer, s);
        }
        printf("%dn", answer);
    }
    return 0;
}

Problem solution in C.

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define min(a,b) ((a < b) ? (a) : (b) )

//const int INF = (int)1e9;
int main() {

    /* Enter your code here. Read input from STDIN. Print output to STDOUT */    
    
int testcase,n,array[10000];
    scanf("%d", &testcase);
    
      while(testcase--)
     {
         scanf("%d",&n);
        int sum, min1 = 10000000, sum1 = 100000000;
         
         for(int i=0;i<n;i++)
         {
              scanf("%d",&array[i]);
              min1 = min(min1, array[i]);
         } 

       for(int k = min1; k >= (min1-5);k--)
       {
           sum = 0;
         for(int i=0;i<n;i++)
          {
             int d = array[i]-k;
              sum += d/5;
              d %= 5;
              sum += d/2;
              d %= 2;
              sum += d;
          }
         sum1 = min(sum1,sum);
     }    
        printf("%dn",sum1);
     
     }
    
    return 0;
}