## среда, 25 декабря 2013 г.

### Codility. Train. Max-Counters ★★★

You are given N counters, initially set to 0, and you have two possible operations on them:
• increase(X) − counter X is increased by 1,
• max_counter − all counters are set to the maximum value of any counter.
A non-empty zero-indexed array A of M integers is given. This array represents consecutive operations:
• if A[K] = X, such that 1 ≤ X ≤ N, then operation K is increase(X),
• if A[K] = N + 1 then operation K is max_counter.
For example, given integer N = 5 and array A such that:
```    A = 3
A = 4
A = 4
A = 6
A = 1
A = 4
A = 4```
the values of the counters after each consecutive operation will be:
```    (0, 0, 1, 0, 0)
(0, 0, 1, 1, 0)
(0, 0, 1, 2, 0)
(2, 2, 2, 2, 2)
(3, 2, 2, 2, 2)
(3, 2, 2, 3, 2)
(3, 2, 2, 4, 2)```
The goal is to calculate the value of every counter after all operations.
Write a function:
def solution(N, A)
that, given an integer N and a non-empty zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.
The sequence should be returned as:
• a structure Results (in C), or
• a vector of integers (in C++), or
• a record Results (in Pascal), or
• an array of integers (in any other programming language).
For example, given:
```    A = 3
A = 4
A = 4
A = 6
A = 1
A = 4
A = 4```
the function should return [3, 2, 2, 4, 2], as explained above.
Assume that:
• N and M are integers within the range [1..100,000];
• each element of array A is an integer within the range [1..N + 1].
Complexity:
• expected worst-case time complexity is O(N+M);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

Solution:

```def solution(N, A):
iMin, iMax = 0, 0
iMas = [iMax] * N
for i in A:
if i == N + 1:
iMin = iMax
else:
if iMas[i - 1] < iMin:
iMas[i - 1] = iMin + 1
else:
iMas[i - 1] = iMas[i - 1] + 1
if iMas[i - 1] > iMax:
iMax = iMas[i - 1]
for i in xrange(len(iMas)):
if iMas[i] < iMin:
iMas[i] = iMin
return iMas
```