Problem Link:

Practice
Contest

Difficulty:

Simple

Pre-requisites:

None

Problem:

Given a list of N integers A[1], A[2], ... A[N], and an integer X, find the minimum cost of making all elements non-negative. There are two kinds of operations that are allowed:

1. Increase all elements in the array by 1. Cost of this operation is X.
2. Choose an element and increase it by 1. Cost is 1.

Short explanation:

It is clear that we will apply operation 2 only on negative array elements. Therefore, if there are K negative elements in the array, they can all be incremented by 1 at a total cost of K. In this way, operation 2 can simulate operation 1 at a non-fixed cost of K(=number of negative elements in current array). Therefore

• Let K = number of negative elements in current array
• If K ≥ X apply operation 1.
• else apply operation 2 on all negative entries once.
• repeat the above if there exists a negative entry.

Optimality of this strategy will be proved in next section.

Long Explanation:

Consider the following very naive algorithm:
while(some A[i] is negative) apply operation 1.

Lets try to simulate this algorithm for a small list of numbers:
A = {-4, -1, 0, -2, -3, -2}
and let X = 3

after successive steps, it becomes:
A = {-4, -1, 0, -2, -3, -2}, cost so far = 0
A = {-3, 0, 1, -1, -2, -1}, cost so far = 3
A = { -2, 1, 2, 0, -1, 0} , cost so far = 6
A = { -1, 2, 3, 1, 0, 1 } , cost so far = 9
A = { 0 , 3, 4, 2, 1, 2 } , cost so far = 12

As one can see, it makes sense to apply operation 1 initially, as we have lots of negative entries. The cost(3) paid for the first step makes perfect sense because the only alternative we had was to increase all the element by 1 by applying operation 2. That would cost us at least 5 units. Likewise, immediately after first step, A has 4(>X=3) negative entries so it again makes sense to apply operation 1. But after that(in step 3), there are only two negative entries, and we could have as well applied operation 2 twice(costing 2*1=2) at the third step as compared to operation 1.

The above example illustrates that we can
1. Apply operation 1 as long as there are X or more negative entries.
2. Apply operation 2 from there on to increase only the negative entries by one at a time.

In fact this algorithm is optimal and its proof of correctness follows shortly. However, we cant simulate the entire process because we may need to apply the operation 1 upto 10⁹ times(and operation 2 upto 10⁹ x 10⁵ times).

To find the total cost of the above, the following observation is handy:

When operation 1 is applied, the relative order of elements in the array do not change.

Therefore if the array A was sorted initially, it will be sorted after applying operation 1 any number of times. Hence operation 1 is applicable as long as A[X] is negative(so the number of negative elements is at least X). Operation 1 is applied a total of op1 times costing op1 * X where

op1 = max(0, -A[x])

The total number of times operation 2 is applied is

max(-(A[1] + op1), 0) + max(-(A[2] + op1), 0) + ... + max(-(A[X-1] + op1), 0)

Proof of correctness

Consider an optimum strategy, and let it use operation 1 op1 times, operation 2 is applied n₁ times on a₁^th element, n₂ times on a₂^th element .... n_K times on a_K^th element, with n₁, n₂ ... n_K> 0.

The total cost of these operations is op1 * X + n₁ + n₂ + ... + n_K.

If K ≥ X then we could as well apply operation 1 one more time (total of op1+1 times) and apply operation 2 to all the elements one less time, which costs a total of

(op1+1) * X + n₁-1 + n₂-1 + ... + n_K-1 = op1 * X + n₁ + n₂ + ... + n_K + X-K

Which is no more than original cost.

Therefore, If the array has X or more negative entries, we need to apply operation 1 at least once to this array. This is because otherwise operation 2 will be applied to all the negative elements leading to K ≥ X. Thus we have justified the first point about Applying operation 1 if there are X or more negative entries.

Justification of the second point is relatively easy because any application of operation 1 can be replaced by an application of operation 2 on K(< X) array elements, and in the process we reduce the total cost.

Setter's Solution:

Can be found here

Tester's Solution:

1. Mahbub's code
2. Sergey's code

Editorialist's Solution:

Can be found here