Codility - Lesson 11 Sieve of Eratosthenes - 2. CountNonDivisible

Source Link:
https://codility.com/programmers/lessons/11-sieve_of_eratosthenes/count_non_divisible/

Question:

You are given a non-empty zero-indexed array A consisting of N integers.
For each number A[i] such that 0 ≤ i < N, we want to count the number of elements of the array that are not the divisors of A[i]. We say that these elements are non-divisors.

For example, consider integer N = 5 and array A such that:
A[0] = 3 A[1] = 1 A[2] = 2 A[3] = 3 A[4] = 6 
For the following elements:

• A[0] = 3, the non-divisors are: 2, 6,
• A[1] = 1, the non-divisors are: 3, 2, 3, 6,
• A[2] = 2, the non-divisors are: 3, 3, 6,
• A[3] = 3, the non-divisors are: 2, 6,
• A[4] = 6, there aren't any non-divisors.

Write a function:
class Solution { public int[] solution(int[] A); }
that, given a non-empty zero-indexed array A consisting of N integers, returns a sequence of integers representing the amount of non-divisors.

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] = 1 A[2] = 2 A[3] = 3 A[4] = 6 
the function should return [2, 4, 3, 2, 0], as explained above.

Assume that:

• N is an integer within the range [1..50,000];
• each element of array A is an integer within the range [1..2 * N].

Complexity:

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

Elements of input arrays can be modified.

My Solution:
Notes:
1. main idea:
// using "HashMap" to count 
2. map1(key, value)
// key: the elements, 
//value: count of elements
3. map2(key, value)
// key: the elements, 
// value: count of "number of non-divisors" of elements