Krdytkyu

PRAMP Award Budget Cuts

The awards committee of your alma mater (i.e. your college/university) asked for your assistance with a budget allocation problem they’re facing. Originally, the committee planned to give N research grants this year. However, due to spending cutbacks, the budget was reduced to newBudget dollars and now they need to reallocate the grants. The committee made a decision that they’d like to impact as few grant recipients as possible by applying a maximum cap on all grants. Every grant initially planned to be higher than cap will now be exactly cap dollars. Grants less or equal to cap, obviously, won’t be impacted.

Given an array grantsArray of the original grants and the reduced budget newBudget, write a function findGrantsCap that finds in the most efficient manner a cap such that the least number of recipients is impacted and that the new budget constraint is met (i.e. sum of the N reallocated grants equals to newBudget).

Example:

input:  grantsArray = [2, 100, 50, 120, 1000], newBudget = 190



output: 47 # and given this cap the new grants array would be

           # [2, 47, 47, 47, 47]. Notice that the sum of the

           # new grants is indeed 190

My approach:

import java.util.Arrays;



class Solution {



  static double findGrantsCap(double grantsArray, double newBudget) {

    // your code goes here

    int len = grantsArray.length;



    Arrays.sort(grantsArray);



    //Calculate the newBudget

    double modBudget = newBudget;

    double cap;

    double sum = 0;

    double avg; 

    for (double elem: grantsArray)

      sum += elem;



    avg = sum/len;



    if( avg == (int)newBudget)

        return (newBudget)/len;

    else

    {

      Arrays.sort(grantsArray);

      modBudget -= grantsArray[0];

      cap = modBudget/(len - 1);

      for( int i = 1; i < len-1; i++ )

      {

          System.out.println(cap);

           if( (int)cap > grantsArray[i] )

           {

             modBudget -= grantsArray[i];

             cap = modBudget/(len - i - 1);

           }

      }

    }

    return cap;  

  }



  public static void main(String args) {

    double  grantsArray = {2,100,50,120,167};

    double newBudget = 400;



    System.out.println(findGrantsCap(grantsArray,newBudget));

  }



}

I have the following questions with regards to the above code:

1. Is there a better way to attempt this question?




2. Is my solution satisfying all the test cases and why is it failing if any?




3. How can I further improve the algorithm, time or space wise?




4. Have I made any gross violations to the Java Coding conventions?

Reference

I believe that the problem in the first place is not well defined (or we are trying to solve a different problem). And I mean...

The problem states the following:

write a function findGrantsCap that finds in the most efficient manner a cap such that the least number of recipients is impacted and that the new budget constraint is met

So it focuses on the recipients impacted and not on the amounts.

e.g

Input:  grantsArray = [2, 100, 50, 120, 1000], newBudget = 190

If I pick for cap the 100 value the array becomes like this

Input:  grantsArray = [2, 100, 50, 1, 37], newBudget = 190

Now only 2 recipients affected (of course the sum of the array elements meets the newBudget constraint), but it's totally unfair.

1) Code from https://repl.it/@sixxta/Award-Budget-Cuts suggests you could sort array in descending order and keep decreasing maximum single grant until there is no overspending. I don't know if you version, starting with smallest value, works correctly.

2) for cases:

{14,15,16,17,18,19}, 97

{14,15,16,17,18,19}, 98

{14,15,16,17,18,19}, 99

your code returns:

17.33

17.66

18.5

While the proper answer is:

17

18

19

3) the algorithm is optimal in terms of time and space, unless the is some way to remove sorting that I am not aware of.

4) you are looking for exact solution, using floating point variables isn't a good idea

 if( avg == (int)newBudget)

    return (newBudget)/len;

This check doesn't make any sense. You probably wanted to compare with newBudget/len but even then you would return 2 for {1,2,3},6 when right answer is 3.

There is a better approach for that problem.

First start by sorting the array, next traverse the array, for every element that is lower than the cap (your cap starts as the avg of budget/array.length) reduce it from your current funds and recalculate the cap (i.e. now the cap equals to budget/(array.length - (i+1))).

Once you got to the end of the array or to the first element that is higher than your current cap you just return the last cap you have in hand.

So all in all this is your solution:

  static double findGrantsCap(double arr, double newBudget) {

    // your code goes here

    Arrays.sort(arr); // O(N*LogN)

    int n = arr.length;

    double cap = newBudget/n;

    for(int i = 0; i < n; i++) { // O(N)

      if(arr[i] < cap) {

        newBudget -= arr[i];

        cap = (newBudget/(n-(1+i)));

      }else {

        arr[i] = cap;

      }

    }



    return cap;

  }

Adding all the numbers on the array will be the newBudget since for every lower number we subtract it from the total and everything above just gets capped (essentially gets to be the average from what we got left with divided with the total elements we got left with - adding those will be the remains of the budget we got up to them).

This is my solution with complexity: time $O(N*log(N))$, space $O(1)$.

static double findGrantsCap(double grantsArray, double newBudget) {

if (grantsArray == null || 

    grantsArray.length == 0 || 

    (grantsArray.length == 1 && grantsArray[0] <= newBudget)) {

  return newBudget;

}



Arrays.sort(grantsArray);

double cap = newBudget / grantsArray.length;

double allocated = 0.0;



for (int i=0; i<grantsArray.length; i++) {

 double currentRequestValue = grantsArray[i];

 if (currentRequestValue <= cap) {

   allocated += currentRequestValue;

   int divisor = (grantsArray.length - i - 1);

   // If last index in the array is below the cap divisor will be "0"

   // It means that all the items in the carousel can be allocated as they are

   cap = divisor != 0 ? (newBudget - allocated) / divisor : currentRequestValue;

 }

}



return cap;

}