MaxCounters

MaxCounters 문제 링크
https://app.codility.com/c/run/trainingP5XD3A-G9T/

Summary
효율적인 개수 세기

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].

Checking List

• 혼자서 문제를 해결
• 힌트를 보고 해결
• 답을 보고 해결

My Answer

• Total Score: 88%
  • extreme_large : TIMEOUT ERROR
• 시간복잡도 : O(N+M)
  

def solution(N, A):
    lis = [0] * N
    max_val = 0
    for n in A:
        if n <= N:
            lis[n-1] +=1
            if max_val < lis[n-1]:
                max_val = lis[n-1]
        else:
            lis = [max_val]*N
    return lis

Answer Sheet

def solution(N, A):

    counters = N * [0]
    next_max_counter =  max_counter = 0

    for oper in A:
        if oper <= N:
            current_counter = counters[oper-1] = max(counters[oper-1] +1, max_counter+1)
            next_max_counter = max(current_counter, next_max_counter)
        else:
            max_counter = next_max_counter

    return [c if c > max_counter else max_counter for c in counters]

출처

Trial & Error

• Codility는 효율성 통과가 정말 힘들다.....

• 마지막 테스트 케이스를 통과하기 위해 어떻게 해야할지 정말 고민이 많았다. max_val로 다시 리스트를 만드는 부분을 수정해야한다는 생각이 들었지만 결국 해결하지 못했다.

• 다른 사람 풀이를 보고 마지막에 최대값을 저장해두고 한번에 처리하면 되겠구나 하는 인사이트를 얻을 수 있었다.

Takeaway

• 연산할 때 필요한 값을 한번에 처리할 수 있도록 코드 구성을 해보자.

• 값을 비교할 때, 깔끔한 코드 표현을 위해 연산자보다 내장함수(max, min, ...)를 사용해보자.