[codility] lessons 4 MaxCounters

MaxCounters

출처: https://app.codility.com/programmers/lessons

문제 설명

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].

나의 풀이

// 77% max counter의 값으로 set에서 타임아웃
function solution(N, A) {
    let answer = [];

    for (let i = 0; i < N; i++) answer[i] = 0;

    for (let i = 0; i < A.length; i++) {
        if (A[i] <= N) {
            answer[A[i] - 1]++;
        } else {
            answer.fill(Math.max.apply(null, answer));
        }
    }

    return answer;
}

숙련된 조교의 풀이

function solution(N, A) {
    let counters = []
    let maxVal = 0
    let lastMax = 0

    for (var i = 0; i < N; i++) counters[i] = 0;

    for (var j = 0; j < A.length; j++) {
        if (A[j] > N) {
            lastMax = maxVal
        } else {
            let currentMax = Math.max(lastMax, counters[A[j] - 1])
            counters[A[j] - 1] = currentMax + 1
            maxVal = Math.max(counters[A[j] - 1], maxVal)
        }
    }

    for (var l = 0; l < N; l++) {
        counters[l] = Math.max(counters[l], lastMax)
    }

    return counters;
}