Codility 32번 CountNonDivisible

CountNonDivisible

 

You are given an 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 an array A consisting of N integers, returns a sequence of integers representing the amount of non-divisors.

Result array should be returned as an array of integers.

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.

Write an efficient algorithm for the following assumptions:

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

 

class Solution {
    public int[] solution(int[] A) {
        int[] cnt = new int[A.length*2 + 1];
        int[] p = new int[A.length];
        for(int i=0; i<A.length; i++){
            cnt[A[i]]++;
        }
        for(int i=0; i<A.length; i++){
            for(int j=1; j<=Math.sqrt(A[i]); j++){
                if(A[i]/j == j){
                    p[i] += cnt[j];
                }else{
                    p[i] += (cnt[j] + cnt[A[i]/j]);
                }
            }
        }
        int[] ans = new int[A.length];
        for(int i=0; i<ans.length; i++){
            ans[i] = A.length - p[i];
        }
        
        return ans;
        // write your code in Java SE 8
    }
}

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