[codillity] kotlin - lesson4. MaxCounters

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:

class Solution { public int[] solution(int N, int[] 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].

var list = IntArray(N)

for(i in A.indices){
    if(A[i] > N){
        //최대값으로 통일 어쨋든 포문이라 2중포문이 되어 자주걸리면 난감해짐.
        Arrays.fill(list, list.maxOrNull()?:0)
    }else{
        list[A[i]-1] += 1
    }
    println(list.contentToString())
}
return list

88점. 리스트 크기 최대값이 전부 counter보다 큰경우 6sec 이상.

 

이런문제는 보통 15분 안짝으로 풀어야되서 그냥 넘어감.

Arrays.fill를 for문 밖에 둬야 all pass될듯.