255. MaxCounters

1. JavaScript

정해

/// you can write to stdout for debugging purposes, e.g.
// console.log('this is a debug message');

function solution(N, A) {
    // write your code in JavaScript (Node.js 8.9.4)
    let counters = new Array(N).fill(0);
    let max = 0;
    let lastMax = 0;

    for(let k of A){
        if(k <= N){
            if(counters[k - 1] < lastMax){
                counters[k - 1] = lastMax;
            }
            counters[k - 1]++;
            if(max < counters[k -1]){
                max = counters[k - 1];
            }
        }else{
            lastMax = max;
        }
    }

    for(let i = 0; i < N; i++){
        if(counters[i] < lastMax){
            counters[i] = lastMax;
        }
    }

    return counters
}

• Math.max로 매번 최댓값을 구하기 어렵기 때문에 다음과 같은 방식을 사용하는 것 같다.

50%

function solution(N, A) {
    let count = Array(N).fill(0);

    for (i = 0; i < A.length; i++) {
        if (1 <= A[i] && A[i] < N) {
            count[A[i] - 1] += 1;

        } else {
            count.fill(Math.max(...count));
        }
    }

    return count;

}


console.log(solution(5, [3, 4, 4, 6, 1, 4, 4]))

2. Python

정해

def solution(N,A):
    savemaximum = 0
    maximum = 0
    counter = [0]*N
    for i in range(len(A)):
        if A[i]<=N:
            if counter[A[i]-1]<savemaximum:
            counter[A[i]-1]=savemaximum
            counter[A[i]-1]+=1
            maximum = max(counter[A[i]-1],maximum)
        else:# N보다 큰 수가 들어오면 maximum으로 동기화
            savemaximum = maximum

    for i in range(N):
        if counter[i]<savemaximum:
            counter[i]=savemaximum
    return counter


출처: https://imksh.com/68 [강승현입니다]

60%

def solution(N, A):
    result = [0] * N
    for i in A:
        if 1 <= i <= N:
            result[i-1] += 1
        else:
            result = [max(result)] * N 
    return result