[codility] MaxCounters

Task description

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:

def solution(N, 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 an array 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].

Copyright 2009–2021 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

 

O(N*M)

# you can write to stdout for debugging purposes, e.g.
# print("this is a debug message")

def solution(N, A):
    # write your code in Python 3.6
    ans = [0] * N
    for num in A:
        if num <= N :
            ans[num-1] += 1
        else :
            all = max(ans)
            all_list = [all] * N
            ans = all_list
    return ans
    pass

 

O(N + M)

def solution(N, A):
    # write your code in Python 3.6
    ans = [0] * N
    num_max = 0
    for num in A:
        if num <= N :
            ans[num-1] += 1
            if ans[num-1] >  num_max:
                num_max = ans[num-1]
        else :
            ans = [num_max] * N

    return ans
    pass

def solution(N, A):
    # write your code in Python 3.6
    ans = [0] * N
    next_max = num_max = 0
    for num in A:
        if num <= N:
            current = ans[num-1] = max(ans[num-1]+1, num_max + 1)
            next_max = max(current, next_max)
        else :
            num_max = next_max
            

    return [c if c > num_max else num_max for c in ans]
    pass