##### Short Problem Definition:

Calculate the values of counters after applying all alternating operations: increase counter by 1; set value of all counters to current maximum.

##### Link

##### Complexity:

expected worst-case time complexity is `O(N+M)`

expected worst-case space complexity is `O(N)`

##### Execution:

The idea is to perform the specified operation as stated. It is not required to iterate over the whole array if a new value is set for all the values. Just save the value and check it when an increase on that position is performed.

##### Solution:

```
#include <algorithm>
vector<int> solution(int N, vector<int> &A) {
vector<int> sol;
int current_max = 0;
int last_increase = 0;
for(int i=0; i<N;i++){
sol.push_back(0);
}
for(unsigned int i=0; i<A.size();i++){
if (A[i] > N) {
last_increase = current_max;
} else {
sol[A[i]-1] = max(sol[A[i]-1], last_increase);
sol[A[i]-1]++;
current_max = max(current_max, sol[A[i]-1]);
}
}
for(int i=0; i<N;i++){
sol[i] = max(sol[i], last_increase);
}
return sol;
}
```