[algorithms] max counters

mgm-dev·2021년 4월 19일
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codility algorithms

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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].

my solution

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

faster solution

def solution(N, A):
    # write your code in Python 3.6
    counters = [0] * N
    max_counter = 0
    last_update = 0

    for K,X in enumerate(A): 
        if 1 <= X <= N:
            counters[X-1] = max(counters[X-1], last_update)
            counters[X-1] += 1
            max_counter = max(counters[X-1], max_counter)
        elif A[K] == (N + 1):
            last_update = max_counter

    for i in range(N):
        counters[i] = max(counters[i], last_update)

    return counters
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