/**
* @file
* @brief [Subset-sum](https://en.wikipedia.org/wiki/Subset_sum_problem) (only
* continuous subsets) problem
* @details We are given an array and a sum value. The algorithms find all
* the subarrays of that array with sum equal to the given sum and return such
* subarrays count. This approach will have \f$O(n)\f$ time complexity and
* \f$O(n)\f$ space complexity. NOTE: In this problem, we are only referring to
* the continuous subsets as subarrays everywhere. Subarrays can be created
* using deletion operation at the end of the front of an array only. The parent
* array is also counted in subarrays having 0 number of deletion operations.
*
* @author [Swastika Gupta](https://github.com/Swastyy)
*/
#include <cassert> /// for assert
#include <cstdint>
#include <iostream> /// for IO operations
#include <unordered_map> /// for unordered_map
#include <vector> /// for std::vector
/**
* @namespace backtracking
* @brief Backtracking algorithms
*/
namespace backtracking {
/**
* @namespace subarray_sum
* @brief Functions for the [Subset
* sum](https://en.wikipedia.org/wiki/Subset_sum_problem) implementation
*/
namespace subarray_sum {
/**
* @brief The main function that implements the count of the subarrays
* @param sum is the required sum of any subarrays
* @param in_arr is the input array
* @returns count of the number of subsets with required sum
*/
uint64_t subarray_sum(int64_t sum, const std::vector<int64_t> &in_arr) {
int64_t nelement = in_arr.size();
int64_t count_of_subset = 0;
int64_t current_sum = 0;
std::unordered_map<int64_t, int64_t>
sumarray; // to store the subarrays count
// frequency having some sum value
for (int64_t i = 0; i < nelement; i++) {
current_sum += in_arr[i];
if (current_sum == sum) {
count_of_subset++;
}
// If in case current_sum is greater than the required sum
if (sumarray.find(current_sum - sum) != sumarray.end()) {
count_of_subset += (sumarray[current_sum - sum]);
}
sumarray[current_sum]++;
}
return count_of_subset;
}
} // namespace subarray_sum
} // namespace backtracking
/**
* @brief Self-test implementations
* @returns void
*/
static void test() {
// 1st test
std::cout << "1st test ";
std::vector<int64_t> array1 = {-7, -3, -2, 5, 8}; // input array
assert(
backtracking::subarray_sum::subarray_sum(0, array1) ==
1); // first argument in subarray_sum function is the required sum and
// second is the input array, answer is the subarray {(-3,-2,5)}
std::cout << "passed" << std::endl;
// 2nd test
std::cout << "2nd test ";
std::vector<int64_t> array2 = {1, 2, 3, 3};
assert(backtracking::subarray_sum::subarray_sum(6, array2) ==
2); // here we are expecting 2 subsets which sum up to 6 i.e.
// {(1,2,3),(3,3)}
std::cout << "passed" << std::endl;
// 3rd test
std::cout << "3rd test ";
std::vector<int64_t> array3 = {1, 1, 1, 1};
assert(backtracking::subarray_sum::subarray_sum(1, array3) ==
4); // here we are expecting 4 subsets which sum up to 1 i.e.
// {(1),(1),(1),(1)}
std::cout << "passed" << std::endl;
// 4rd test
std::cout << "4th test ";
std::vector<int64_t> array4 = {3, 3, 3, 3};
assert(backtracking::subarray_sum::subarray_sum(6, array4) ==
3); // here we are expecting 3 subsets which sum up to 6 i.e.
// {(3,3),(3,3),(3,3)}
std::cout << "passed" << std::endl;
// 5th test
std::cout << "5th test ";
std::vector<int64_t> array5 = {};
assert(backtracking::subarray_sum::subarray_sum(6, array5) ==
0); // here we are expecting 0 subsets which sum up to 6 i.e. we
// cannot select anything from an empty array
std::cout << "passed" << std::endl;
}
/**
* @brief Main function
* @returns 0 on exit
*/
int main() {
test(); // run self-test implementations
return 0;
}