[Codility] Lesson 4 - Counting Elements : MaxCounters

Lesson 4 - Counting Elements : MaxCounters

📌 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 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].

📝 풀이

• A[K] = X 의 값이 첫번째 조건을 만족할 때 해당 포지션의 값을 increase 해주고,
• increase 한 값이 현재 제일 큰 값인지 비교해준다.
• A[K] = X 의 값이 두번째 조건을 만족하는 순간부터 첫번째 조건을 만족할 때,
  두번째 조건을 만족했던 순간의 제일 큰 숫자보다 작으면 해당 숫자를 넣어준다.
• 마지막으로 두번째 조건을 만족할 때의 숫자보다 작은 숫자가 있다면 해당 숫자로 넣어준다.

💻 구현

static class Solution {
    public int[] solution(int N, int[] A) {
        int[] answer = new int[N];

        int base = 0;
        int max = 0;
        for(int i = 0; i < A.length; i++){
            if(A[i] >= 1 && A[i] <= N){
                if(answer[A[i] - 1] < base){
                    answer[A[i] - 1] = base;
                }

                if(++answer[A[i] - 1] > max){
                    max = answer[A[i] - 1];
                }
            } else if (A[i] == N + 1){
                base = max;
            }
        }

        for(int i = 0; i < answer.length; i++){
            if(answer[i] < base){
                answer[i] = base;
            }
        }

        return answer;
    }
}