MaxCounters

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 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:

    vector<int> solution(int N, vector<int> &A);

that, given an integer N and a non-empty array A consisting of M integers, returns a sequence of integers representing the values of the counters.

Result array should be returned as a vector of integers.

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.

Write an efficient algorithm for the following assumptions:

        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].

vector<int> solution(int N, vector<int> &A) {
    // write your code in C++11 (g++ 4.8.2)
    vector<int> res(N, 0);
    
    int base = 0;
    int maximum = 0;
    for (int i = 0; i < int(A.size()); i++) {
        if (A[i] != N + 1) {
            res[A[i] - 1] = max(base, res[A[i] - 1]) + 1;
            maximum = max(maximum, res[A[i] - 1]);
        } else {
            base = maximum;
        }
    }
    
    for (int i = 0; i < N; i++) {
        if (res[i] < base) {
            res[i] = base;
        }
    }
    return res;
}