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[0] = 3
    A[1] = 4
    A[2] = 4
    A[3] = 6
    A[4] = 1
    A[5] = 4
    A[6] = 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[0] = 3
    A[1] = 4
    A[2] = 4
    A[3] = 6
    A[4] = 1
    A[5] = 4
    A[6] = 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