TheAlgorithms-C-Plus-Plus/backtracking/subarray_sum.cpp

/**
 * @file
 * @brief We are given with an array and a sum value. The algorithms find all
 * the subarrays of that array with sum equal to 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 refering to
 * the continuous subsets as subarrays everywhere. Subarrays can be created
 * using deletion operation at the end or the front of an array only. The parent
 * array is also counted in subarrays having 0 number of deletion operations.
 * @details Subset sum(only continuous subsets) problem
 * (https://en.wikipedia.org/wiki/Subset_sum_problem)
 * @author [Swastika Gupta](https://github.com/Swastyy)
 */

#include <cassert>        /// for assert
#include <iostream>       /// for IO operations
#include <unordered_map>  /// for unordered_map
#include <vector>         /// for std::vector

/**
 * @namespace backtracking
 * @brief subarray sum algorithm
 */
namespace backtracking {
/**
 * @namespace subarraySum
 * @brief Functions for the [Subset sum](https://en.wikipedia.org/wiki/Subset_sum_problem) implementation
 */
namespace subarraySum {
/**
 * @brief The main function implements count of 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(int sum, const std::vector<int> &in_arr) {
    int nelement = in_arr.size();
    int count_of_subset = 0;
    int current_sum = 0;
    std::unordered_map<int, int> sumarray;  // to store the subarrays count
                                            // frequency having some sum value

    for (int 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 subarraySum
}  // namespace backtracking

/**
 * @brief Self-test implementations
 * @returns void
 */
static void test() {
    // Test 1
    std::cout << "1st test ";
    std::vector<int> array1 = {-7, -3, -2, 5, 8};  // input array
    assert(backtracking::subarraySum::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;

    // Test 2
    std::cout << "2nd test ";
    std::vector<int> array2 = {1, 2, 3, 3};
    assert(backtracking::subarraySum::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;

    // Test 3
    std::cout << "3rd test ";
    std::vector<int> array3 = {1, 1, 1, 1};
    assert(backtracking::subarraySum::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;

    // Test 4
    std::cout << "4th test ";
    std::vector<int> array4 = {3, 3, 3, 3};
    assert(backtracking::subarraySum::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;

    // Test 5
    std::cout << "5th test ";
    std::vector<int> array5 = {};
    assert(backtracking::subarraySum::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();  // execute the test
    return 0;
}