##### 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; }

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