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:

	package CountNonDivisible;
	import java.util.*;
	class Solution {
	public int[] solution(int[] A) {
	// main idea:
	// using "HashMap" to count
	// map1(key, value)
	HashMap<Integer, Integer> map1 = new HashMap<>();
	// key: the elements, value, count of elements
	for(int i=0; i< A.length; i++){
	if(map1.containsKey(A[i]) == false){
	map1.put(A[i], 1); // add new element
	}
	else{
	map1.put(A[i], map1.get(A[i])+1 ); // count++
	}
	}
	// map2(key, value)
	HashMap<Integer, Integer> map2 = new HashMap<>();
	// key: the elements, value, count of "number of non-divisors" of elements
	for( int n : map1.keySet() ){
	int numDivisors =0;
	// find divisors from 1 to sqrt(n)
	int sqrtN = (int)Math.sqrt(n);
	for(int i=1; i<=sqrtN; i++ ){
	if( n % i == 0){ // means: i could be a divisor
	int anotherDivisor = n/i;
	if(map1.containsKey(i) == true ){
	numDivisors = numDivisors + map1.get(i);
	}
	if(anotherDivisor != i){ // avoid double count (be careful)
	if(map1.containsKey(anotherDivisor) == true){
	numDivisors = numDivisors + map1.get(anotherDivisor);
	}
	}
	}
	}
	int numNonDivisors = A.length - numDivisors;
	map2.put(n, numNonDivisors);
	}
	// results: number of non-divisors
	int[] results = new int[A.length];
	for (int i = 0; i < A.length; i++) {
	results[i] = map2.get(A[i]);
	}
	return results;
	}
	}

view raw Codility_Lession_11-2_CountNonDivisible.java hosted with ❤ by GitHub

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