Create partition_problem.cpp

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Lajat5
2021-11-09 21:16:42 +05:30
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/******************************************************************************
* @file
* @brief Implementation of the [Partition Problem](https://en.wikipedia.org/wiki/Partition_problem )
* @details
* The partition problem, or number partitioning, is the task of deciding whether
* a given multiset S of positive integers can be partitioned into two subsets S1
* and S2 such that the sum of the numbers in S1 equals the sum of the numbers in S2.
* Although the partition problem is NP-complete, there is a pseudo-polynomial time
* dynamic programming solution, and there are heuristics that solve the problem
* in many instances, either optimally or approximately. For this reason,
* it has been called "the easiest hard problem".
*
* The worst case time complexity of Jarviss Algorithm is O(n^2). Using
* Grahams scan algorithm, we can find Convex Hull in O(nLogn) time.
*
* ### Implementation
*
* Step 1
* Calculate sum of the array. If sum is odd, there can not be two subsets with
* equal sum, so return false.
*
* Step 2
* If sum of array elements is even, calculate sum/2 and find a subset of array
* with sum equal to sum/2.
*
* @author [Lajat Manekar](https://github.com/Lazeeez)
*
*******************************************************************************/
#include <iostream> /// for IO Operations
#include <vector> /// for std::vector
#include <numeric> /// for std::accumulate
/******************************************************************************
* @namespace dp
* @brief Dynamic programming algorithms
*******************************************************************************/
namespace dp {
/******************************************************************************
* @namespace partitionProblem
* @brief Partition problem algorithm
*******************************************************************************/
namespace partitionProblem {
/******************************************************************************
* @brief Returns true if arr can be partitioned in two subsets of equal sum,
* otherwise false
* @param arr vector containing elements
* @param size Size of the vector.
* @returns @param bool whether the vector can be partitioned or not.
*******************************************************************************/
bool findPartiion(std::vector<uint64_t> &arr, uint64_t size) {
uint64_t sum = std::accumulate(arr.begin() , arr.end() , 0); // Calculate sum of all elements
if (sum % 2 != 0) return false; //if sum is odd, it cannot be divided into two equal sum
bool part[sum / 2 + 1];
// Initialize the part array as 0
for (uint64_t it = 0; it <= sum / 2; ++it) {
part[it] = false;
}
// Fill the partition table in bottom up manner
for (uint64_t it = 0; it < size; ++it) {
// The element to be included in the sum cannot be greater than the sum
for (uint64_t it2 = sum / 2; it2 >= arr[it]; --it2) { // Check if sum - arr[i]
// ould be formed from a subset using elements before index i
if (part[it2 - arr[it]] == 1 || it2 == arr[it])
part[it2] = true;
}
}
return part[sum / 2];
}
}
}
/*******************************************************************************
* @brief Self-test implementations
* @returns void
*******************************************************************************/
void test() {
std::vector<uint64_t> arr = {{ 1, 3, 3, 2, 3, 2 }};
uint64_t n = arr.size();
bool expected_result = true;
bool derived_result = dp::partitionProblem::findPartiion(arr, n);
std::cout << derived_result << std::endl;
std::cout << "1st test: ";
if(expected_result == derived_result) {
std::cout << "Passed!" << std::endl;
}
else {
std::cout << "Failed!" << std::endl;
}
}
/*******************************************************************************
* @brief Main function
* @returns 0 on exit
*******************************************************************************/
int main()
{
test(); // run self-test implementations
return 0;
}