Between Two Sets - HackerRank Challenge | C++ Implementation

Publish date: 2019-11-29
Tags: Cpp, Algorithm, Competitive-Programming, Hackerrank, Data-Structure

You will be given two arrays of integers and asked to determine all integers that satisfy the following two conditions:

  1. The elements of the first array are all factors of the integer being considered
  2. The integer being considered is a factor of all elements of the second array

These numbers are referred to as being between the two arrays. You must determine how many such numbers exist.

For example, given the arrays a =[2, 6] and b = [24, 36], there are two numbers between them: 6 and 12. 6 % 2 = 0, 6 % 6 = 0, 24 % 6 = 0 and 36 % 6 = 0 for the first value. Similarly, 12 % 2 = 0, 12 % 6 = 0 and 24 % 12 = 0, 36 % 12 = 0.

Read the full problem here: Between Two Sets

Between Two Sets Hackerrank

First we sort both the arrays in increasing order because values can be entered in any order. Let the first array is factors and second array is multiples (See conditions 1 and 2).

std::sort(factors.begin(), factors.end());
std::sort(multiples.begin(), multiples.end());

The range of integer being considered is from last element of factors to first element of multiples. Let that integer is num. So num = { factors.back(), ... , multiplies.front() }.

unsigned short num = factors.back();
unsigned short times_multiplied = 2;

Here num also contain integers that are not multiples of 2 and 6 for eg. 7, 11, 13, 15 ... To make program efficient num should contain multiples of factors.back() i. e. in the above example multiples of 6. So num = {6, 12, 18, 24 } and the numbers which satisfy the above two conditions are 6 and 12.

num = factors.back() * times_multiplied;
times_multiplied++;

Bool variable is_multiple tells if a number is multiple of the num and is_factor tells if the num is the factor of a number.

bool is_multiple = true;
bool is_factor = true;

If is_multiple is true for all the elements of factors and is_factor is true for all the elements of multiplies for a num then that num belongs to between the two arrays and our count variable is incremented by 1.

for (unsigned int i = 0; i < factors.size(); ++i)
{
    if (num % factors[i] != 0)
    {
        is_multiple = false;
        break;
    }
}

for (unsigned int j = 0; j < multiples.size(); ++j)
{
    if (is_multiple && multiples[j] % num != 0)
    {
        is_factor = false;
        break;
    }
}

if (is_multiple && is_factor)
{
    count++;
}

#include <iostream>
#include <vector>
#include <algorithm>

unsigned int count_between_two_sets(std::vector<unsigned short>& factors, std::vector<unsigned short>& multiples)
{
    std::sort(factors.begin(), factors.end());
    std::sort(multiples.begin(), multiples.end());

    unsigned int count = 0;
    unsigned short times_multiplied = 2;
    unsigned short num = factors.back();

    while ( num <= multiples.front())
    {
    	bool is_multiple = true;
        bool is_factor = true;
        
        for (unsigned int i = 0; i < factors.size(); ++i)
        {
            if (num % factors[i] != 0)
            {
                is_multiple = false;
                break;
            }
        }

        for (unsigned int j = 0; j < multiples.size(); ++j)
        {
            if (is_multiple && multiples[j] % num != 0)
            {
                is_factor = false;
                break;
            }
        }

        if (is_multiple && is_factor)
        {
            count++;
        }
        num = factors.back() * times_multiplied;
        times_multiplied++;
    }
    return count;
}

int main()
{
    unsigned short num_factors, num_multiples;
    std::cin >> num_factors >> num_multiples;
    std::vector<unsigned short> factors(num_factors);
    std::vector<unsigned short> multiples(num_multiples);

    for (unsigned short i = 0; i < num_factors; ++i)
    {
        std::cin >> factors[i];
    }
    for (unsigned short i = 0; i < num_multiples; ++i)
    {
        std::cin >> multiples[i];
    }
  
    std::cout << count_between_two_sets(factors, multiples) << "\n";
}

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Tags: Cpp, Algorithm, Competitive-Programming, Hackerrank, Data-Structure