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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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() }
.
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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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