codility 4-4. MaxCounters

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문제출처

문제

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 zero-indexed 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 = 3
A = 4
A = 4
A = 6
A = 1
A = 4
A = 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 zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.

The sequence should be returned as:

a structure Results (in C), or
a vector of integers (in C++), or
a record Results (in Pascal), or
an array of integers (in any other programming language).
For example, given:

A = 3
A = 4
A = 4
A = 6
A = 1
A = 4
A = 4
the function should return [3, 2, 2, 4, 2], as explained above.

Assume that:

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].
Complexity:

expected worst-case time complexity is O(N+M);
expected worst-case space complexity is O(N),
beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

풀이코드

def solution(N, A):
result =  * N
for i in A:
if 1 <= i <= N:
result[i-1] += 1
else:
result = [max(result)] * N # O(N)
return result

다른사람 코드

def solution(N, A):

counters = N * 
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]