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
}
}