Codility 9번 MaxCounters

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].

 

가장 큰 값을 기억하고 있다가 배열의 크기보다 큰 숫자가 나왔을 때에 기본 값을 max로 초기화한다.

기본 값보다 작은 값이 등장했을 때에는 해당 값을 기본 값으로 초기화 한 이후 1을 더해준다.

값을 모두 더한 이후 기본 값보다 낮은 값을 가지고 있다면 기본 값으로 초기화 해준다.

public class Solution {
    public int[] solution(int N, int[] A) {
        // write your code in Java SE 8
        int defaultNum = 0;
        int maxNum = 0;
        
        int[] ans = new int[N];
        for (int value : A) {
            if (value > N) {
                defaultNum = maxNum;
            } else {
                if(ans[value - 1] < defaultNum){
                    ans[value - 1] = defaultNum;
                }
                ans[value - 1] += 1;
                if (ans[value - 1] > maxNum) {
                    maxNum = ans[value - 1];
                }
            }
        }

        for (int i = 0; i < ans.length; i++) {
            if(ans[i] < defaultNum){
                ans[i] = defaultNum;
            }
        }
        
        return ans;
    }
}