CountNonDivisible - Codility training task

Question

I'm training on codility now. Some tasks I can solve by myself, but with some tasks have problems.
Difficulty of this task is <**>. It's medium, but I stalled.

Problem:

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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[6] = 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 numbers 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.

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I have written some solutions. But my solutions bulky and still have O(n^2) complexity.
Can you help me with some ideas or algorithms how to do it optimally? It's not an interview task or something else. I'm just training and try to solve all tasks.
You can find this task here: http://codility.com/demo/train/ Lesson 9, first task in lesson.

Answer 1

I thought I'll share my solution in C++ which gets 100 score. I think it's pretty straightforward.

https://codility.com/demo/results/demoQFK5R5-YGD/

1. First it counts the occurrences of each number in the array.

2. Then for each array element i it finds the number of its divisors in a range from 1 to sqrt(i), including the divisors which are the result of the division.

3. Finally it subtracts a total number of divisors for given element from a total number of elements in the array.

   vector<int> solution(vector<int> &A) {
   
       int N = A.size();
       vector<int> counts (*std::max_element(A.begin(), A.end()) + 1,0);
   
       // Calculate occurences of each number in the array
       for (int i = 0; i < N; ++i)
       {
           counts[A[i]] += 1;
       }
   
       std::vector<int> answer(N,0);
   
       // For each element of the array
       for (int i = 0; i < N; ++i)
       {
           // Calulate how many of its divisors are in the array
           int divisors = 0;
   
           for (int j = 1; j * j <= A[i]; ++j)
           {
               if (A[i] % j == 0)
               {
                   divisors += counts[j];
                   if (A[i] / j != j)
                   {
                       divisors += counts[A[i] / j];
                   }
               }
           }
   
           // Subtract the number of divisors from the number of elements in the array
           answer[i] = N - divisors;
       }
   
       return answer;
   }
   

Answer 2

A solution attempt: (EDITED, see below)

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

// Solution for Lesson 9, "CountNonDivisible"
// of http://codility.com/demo/train/
public class Solution
{
    public static void main(String[] args)
    {
        int A[] = new int[5];
        A[0] = 3;
        A[1] = 1;
        A[2] = 2;
        A[3] = 3;
        A[4] = 6;

        Solution s = new Solution();
        int B[] = s.solution(A);
        System.out.println("Input  : "+Arrays.toString(A));
        System.out.println("Result : "+Arrays.toString(B));
    }

    public int[] solution(int[] A)
    {
        Set<Integer> setA = asSet(A);
        List<Set<Integer>> divisors = computeDivisors(A.length * 2);
        int occurrences[] = computeOccurrences(A);
        int nonDivisors[] = new int[A.length];
        for (int i=0; i<A.length; i++)
        {
            int value = A[i];
            Set<Integer> d = divisors.get(value);
            int totalOccurances = 0;
            for (Integer divisor : d)
            {
                if (setA.contains(divisor))
                {
                    totalOccurances += occurrences[divisor];
                }
            }
            nonDivisors[i] = A.length-totalOccurances;
        }
        return nonDivisors;
    }


    /**
     * Returns a set containing all elements of the given array
     * 
     * Space: O(N)
     * Time: O(N)
     * 
     * @param A The input array
     * @return The set
     */
    private static Set<Integer> asSet(int A[])
    {
        Set<Integer> result = new HashSet<Integer>();
        for (int value : A)
        {
            result.add(value);
        }
        return result;
    }


    /**
     * Computes a list that contains for each i in [0...maxValue+1] a set
     * with all divisors of i. This is basically an "Eratosthenes Sieve". 
     * But in addition to setting the entries of a list to 'false' 
     * (indicating that the respective numbers are non-prime), this 
     * methods inserts the divisors into the corresponding set.
     *  
     * Space: O(N) (?)
     * Time: O(N*logN) (?)
     * 
     * @param maxValue The maximum value
     * @return The list 
     */
    private static List<Set<Integer>> computeDivisors(int maxValue)
    {
        List<Boolean> prime = new ArrayList<Boolean>();
        prime.addAll(Collections.nCopies(maxValue+1, Boolean.TRUE));
        List<Set<Integer>> divisors = new ArrayList<Set<Integer>>();
        for (int i = 0; i < maxValue + 1; i++)
        {
            Set<Integer> d = new HashSet<Integer>();
            d.add(1);
            d.add(i);
            divisors.add(d);
        }
        for (int i = 2; i <= maxValue; i++)
        {
            int next = i + i;
            while (next <= maxValue)
            {
                divisors.get(next).addAll(divisors.get(i));
                prime.set(next, Boolean.FALSE);
                next += i;
            }
        }
        return divisors;
    }

    /**
     * Computes an array of length 2*A.length+1, where each entry i contains
     * the number of occurrences of value i in array A
     * 
     * Space: O(N)
     * Time: O(N)
     * 
     * @param A The input array
     * @return The occurrences array
     */
    private static int[] computeOccurrences(int A[])
    {
        int occurances[] = new int[A.length * 2 + 1];
        for (int i=0; i<A.length; i++)
        {
            int value = A[i];
            occurances[value]++;
        }
        return occurances;
    }
}

The maximum value for the numbers that occur in the array was defined to be 2*arrayLength. For each number that MAY occur in the array, it computes

• The set of divisors of this number (using the Erathostenes Sieve)
• How often the number actually occurs in the array

Given this information, one can walk through the array. For each value that is found in the array, one can look up the set of divisors, and compute the total number occurences of all the divisors. The result is then simply the array length, minus this total number of occurances of divisors.

Since it uses only the Sieve of Erathostenes for the computation (and only walks through the set of divisors for each number, which should be logN as well), it should have a worst-case time complexity of O(N*logN). But I'm not entirely sure whether the storage complexity really can be considered to be strictly O(N), because for each of the N numbers, it has to store the set of divisors. Maybe this can somehow be avoided, by combining some of the methods, but in any case, the storage is at least in O(N*logN) as well.

EDIT: The exceptions came from the array for the occurrences storing only values up to A.length*2-1, this was fixed now. Additionally, the set of divisors was not computed properly, this should also be fixed now. Apart from that, analysis results like

got      [8, 8, 9, 10, 6, 8, .. 
expected [8, 8, 9, 10, 6, 8, ..

are not really helpful. Maybe this is part of the "game", but I'm not playing this game right now. The basic idea should be clear, and I assume that it's now working properly until someone shows be a counterexample ;-P It still does not reach the O(N) storage complexity, but I haven't thought about a possible way to achieve this thoroughly...

Answer 3

This solution gives 100 score. https://codility.com/demo/results/demo63KVRG-Q63/

public int[] solution(int[] A) {
    int[][] D = new int[A.length*2 + 1][2];

    for (int i = 0; i < A.length; i++) {
        D[A[i]][0]++;
        D[A[i]][1] = -1;
    }

    for (int i = 0; i < A.length; i++) {
        if (D[A[i]][1] == -1) {
            D[A[i]][1] = 0;
            for (int j = 1; j <= Math.sqrt(A[i]) ; j++) {
                if (A[i] % j == 0 && A[i] / j != j) {
                    D[A[i]][1] += D[j][0];
                    D[A[i]][1] += D[A[i]/j][0];
                } else if (A[i] % j == 0 && A[i] / j == j) {
                    D[A[i]][1] += D[j][0];
                }
            }
        }
    }
    for (int i = 0; i < A.length; i++) {
        A[i] = A.length - D[A[i]][1];
    }
    return A;
}

Thank you all for your help.

Answer 4

Here is my 100 score Python solution. Hope it is helpful to others.

def solution(A):
    ''' Solution for the CountNonDivisible by codility
        Author: Sheng Yu - codesays.com
    '''
    from math import sqrt

    A_max = max(A)
    A_len = len(A)

    # Compute the frequency of occurrence of each
    # element in array A
    count = {}
    for element in A:
        count[element] = count.get(element,0)+1

    # Compute the divisors for each element in A
    divisors = {}
    for element in A:
        # Every nature number has a divisor 1.
        divisors[element] = [1]
    # In this for loop, we only find out all the
    # divisors less than sqrt(A_max), with brute
    # force method.
    for divisor in xrange(2, int(sqrt(A_max))+1):
        multiple = divisor
        while multiple  <= A_max:
            if multiple in divisors and not divisor in divisors[multiple]:
                divisors[multiple].append(divisor)
            multiple += divisor
    # In this loop, we compute all the divisors
    # greater than sqrt(A_max), filter out some
    # useless ones, and combine them.
    for element in divisors:
        temp = [element/div for div in divisors[element]]
        # Filter out the duplicate divisors
        temp = [item for item in temp if item not in divisors[element]]
        divisors[element].extend(temp)

    # The result of each number should be, the array length minus
    # the total number of occurances of its divisors.
    result = []
    for element in A:
        result.append(A_len-sum([count.get(div,0) for div in divisors[element]]))

    return result