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