[Codility]Lesson11/SieveOfEratosthenes/CountNonDivisible

Mijeong Ryu·2023년 5월 27일
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Codility

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문제

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].

코드1

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");
import java.util.*;
class Solution {
public int[] solution(int[] A) {
int[] arr = new int[A.length];
ArrayList<Integer> list = new ArrayList<>();
for(int i=0; i<A.length; i++){
int cnt =0;
list = new ArrayList<>();
for(int j=1; j*j<=A[i]; j++){
if(j*j==A[i]) {
}
else if(A[i]%j==0){
}
}
for(int z=0; z<A.length; z++){
if(!list.contains(A[z])){
cnt++;
}
}
arr[i] = cnt;

}
return arr;
}