programming developer hiring challenge problem

Question

There are N teams in a software company. The i^th team has B_i employees in it and a total budget of A_i units of money. Each team has to divide their budget within employees equally. But for some teams it's not possible. Therefore the company has to perform revisions in their teams' budget. In one revision, to revise the budget of i^th team, the budget of first i teams has to be increased by 1. Your task is to find the minimum number of revisions needed so that for each team, equal distribution of their budget among the employees are possible.

Sample case: (A₁ B₁), (A₂ B₂), (A₃ B₃) : (1 1), (3 7), (5 4).

Solution is 4. Initial budget (1,3,5) -> (2,4,5) -> (5,7,8)

Answer 1

The given problem can be solved by a constructive solution. To start off, let's inc_i denote the amount to be incremented for every A_i such that the sum (A_i + inc_i) can be distributed equally within B_i employees. The two observations to be made here are:

• (A_i + inc_i) mod B_i should be equal to 0 for the amount to be distributed equally
• inc_i <= inc_i-1, since we're always incrementing the budget in a prefix fashion starting from index 1 and hence it's impossible to have inc_i > inc_i-1 since inc_i-1 will always be incremented before inc_i

Using these two observations, we can start iterating over the given array from right-to-left and at every step we determine the value for inc_i. The value of inc_i should be (c * B_i - A_i), where c is a smallest integral value such that c * B_i >= A_i and (c * B_i - A_i) >= inc_i+1 the value for which can be calculated in O(1), since we can reform the equation to estimate the value of c, -

• c >= (A_i / B_i)
• c >= (A_i + inc_i+1) / B_i

Hence c = max(ceil(A_i / B_i), ceil((A_i + inc_i+1) / B_i)

Hence the overall complexity of the given solution is O(n) and the final solution is the value of inc₁ (assuming 1-indexing)