# MaxCounters

**Problema:**

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.

**Solución:**

def solution(N, A): counters = [0] * N for x in A: if 1 <= x <= N: counters[x-1] += 1 else: counters[:] = [max(counters)] * N return counters

Aquí esta el repo con todas mis soluciones de codility:https://github.com/OmarIbannez/codility-lessons