Award Budget Cuts

Question

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

Solution

First, we need to sort the grants array in ascending order.

Then, we use a variable to track the left budget and traverse the array to apply for the grant.

For every grant, assume it is the cap, so all the other grants will be the same as this one. Required money is equal to the cap multiply the number of left people.

If the Required money is less than the left budget, then we can just subtract the current grant from the left budget and move to the next. Otherwise; the cap should be less than the current grant. And the cap is equal to the left budget evenly divided by the number of left people.

Code

python

def solution(grants_array, new_budget):
    # sort the grants array
    grants_array.sort()

    left_budget = new_budget
    left_grant_num = len(grants_array)
    for i in range(len(grants_array)):
        cap = grants_array[i]
        if cap * left_grant_num >= left_budget:
            cap = left_budget // left_grant_num
            return cap
        left_budget -= grants_array[i]
        left_grant_num -= 1
    return grants_array[-1]


def main():
    assert solution(grants_array=[2, 100, 50, 120, 1000], new_budget=190) == 47
    assert solution(grants_array=[2, 100, 50, 120, 1000], new_budget=10000000) == 1000


if __name__ == '__main__':
    main()

Time& Space Complexity

Time Complexity: O(NlogN)

Sorting cost O(N LogN) Time, and traverse the array cost O(N) time, So O(NLogN)

Space Complexity: O(1)

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