Codility Test (09. MaxCounters)

Codility Test (09. 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].

問題回答

class Solution {
    public int[] solution(int N, int[] A) {
        int[] result = new int[N];
        int min = 0;
        int max = 0;
        for (int i = 0; i < A.length; i++) {
            if(A[i] >= 1 && A[i] <= N) {
                if(result[A[i] - 1] < min) {
                    result[A[i] - 1] = min;
                }
                result[A[i] - 1]++;
                if(result[A[i] - 1] > max) {
                    max = result[A[i] - 1];
                }
            } else if(A[i] == N + 1) {
                min = max;
            }
        }
        
        for (int j = 0; j < result.length; j++) {
            if(result[j] < min) {
                result[j] = min;
            }
        }
        
        return result;
    }
}