codility - MinAvgTwoSlice ( java, javascript)

codility / traning center [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 zero-indexed 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 zero-indexed array A consisting of M integers, returns a sequence of integers representing the values of the counters.

The sequence should be returned as:

• a structure Results (in C), or
• a vector of integers (in C++), or
• a record Results (in Pascal), or
• an array of integers (in any other programming language).

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.

Assume that:

• 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].

Complexity:

• expected worst-case time complexity is O(N+M);
• expected worst-case space complexity is O(N), beyond input storage (not counting the storage required for input arguments).

이 문제는 솔직히 말하면 아직도 왜 88%밖에 안나오는지 모르겠다.

report를 보면 시간 복잡도가 조건에 맞는 O(N+M)인데 마지막 문항이 time out이 나왔다.

이 문제를 풀면서 계속 실수를 했던 것은 

첫 번째 if에서 maxResult += max; max = 0; 를 안해줬던 것이었다.

계속해서 새로 배열을 만들었는데 아무래도 이 것 때문에 시간이 오래 걸린 것은 아닌가 싶다.

public int[] solution(int N, int[] A) {
    int max = 0;
    int maxResult = 0;
    int[] result = new int[N];
    for ( int index : A ) {
        if ( index == N + 1 ) {
            maxResult += max;
            result = new int[N];
            max = 0;
        } else {
            result[index - 1]++;
            int temp = result[index - 1];
            max = max > temp ? max : temp;
        }
    }
    for ( int i = 0; i < N; i ++ ) {
        result[i] += maxResult;
    }
    return result;
}
 

Test result report - Java

Test result report - Javascript