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# Valera and Plates – Code forces (Round #216) problem

In this post, we discuss a simple implementation problem from the recently concluded Code forces programming round. The problem description is given below.

Valera is a lazy student. He has m clean bowls and k clean plates.
Valera has made an eating plan for the next n days. As Valera is lazy, he will eat exactly one dish per day. At that, in order to eat a dish, he needs exactly one clean plate or bowl. We know that Valera can cook only two types of dishes. He can eat dishes of the first type from bowls and dishes of the second type from either bowls or plates.
When Valera finishes eating, he leaves a dirty plate/bowl behind. His life philosophy doesn’t let him eat from dirty kitchenware. So sometimes he needs to wash his plate/bowl before eating. Find the minimum number of times Valera will need to wash a plate/bowl, if he acts optimally.
Input

The first line of the input contains three integers n, m, k (1 ≤ n, m, k ≤ 1000) — the number of the planned days, the number of clean bowls and the number of clean plates.
The second line contains n integers a1, a2, …, an (1 ≤ ai ≤ 2). If ai equals one, then on day i Valera will eat a first type dish. If ai equals two, then on day i Valera will eat a second type dish.

Output

Print a single integer — the minimum number of times Valera will need to wash a plate/bowl.

Sample test(s)
Input
`3 1 11 2 1`
Output
`1`
Input
`4 3 11 1 1 1`
Output
`1`
Input
`3 1 22 2 2`
Output
`0`
Input
`8 2 21 2 1 2 1 2 1 2`
Output
`4`
Explanation:

In the first sample Valera will wash a bowl only on the third day, so the answer is one.

In the second sample, Valera will have the first type of the dish during all four days, and since there are only three bowls, he will wash a bowl exactly once.

In the third sample, Valera will have the second type of dish for all three days, and as they can be eaten from either a plate or a bowl, he will never need to wash a plate/bowl.

Solution:
We first count how many times he makes type-1 dish and type-2 dish. He eats type-1 dishes only in bowls.

We first calculate how many cleanings to eat first dish. Let us denote cb as the number of clean bowls and cp as the number of clean plates. Let type1 be the number of first dish type2 be the number of second dish in his plan.

If type1 > cb, We need (type1-cb) cleanings and we are left with 0 clean bowls. Otherwise we need no cleanings , we are left with (cb-type1) bowls.

We add the clean bowls from the previous step to number of clean plates.
cbp = cb + cp.
If type2 > cbp then we need typ2-cbp cleanings otherwise zero.

Here is the implementation in C++.