[Codility] Lesson4. MaxCounters - 문제풀이

Description

주어진 배열에서 아래와 같이 경우에 따라 2가지 연산을 한 뒤 결과배열을 반환하는 문제입니다.

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].

Solution 1. Iteration

public int[] solution(int N, int[] A) {
    int len = A.length;

    int max = 0;
    int[] result = new int[N];

    for(int i=0; i<len; i++){
        if(A[i] == N+1){
            Arrays.fill(result,max);
        }else{
            max = Math.max(max,++result[A[i]-1]);
        }
    }
    return result;
}

반복을 통해 A[i] == N+1일 경우 요소중에 가장 큰 값으로 배열을 채워주고(fill) 그게 아닐 경우 해당 위치의 요소를 +1해주고 max를 계산해 둡니다.

Reference

MaxCounters coding task - Learn to Code - Codility