# Between Two Sets - HackerRank Challenge | C++ Implementation

# Problem

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

- The elements of the first array are all factors of the integer being considered
- 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

# Solution

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).

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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() }`

.

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2

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`

.

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2

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.

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2

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.

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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++;
}

### C++ Implementation

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#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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