Print all subarrays with 0 sum

Here given code implementation process.

//C++ Program 
//Print all subarrays with 0 sum
//compile code like this
//g++ -std=c++11 test.cpp
//here test.cpp is file name
#include <iostream>
#include <map>
#include <vector>

using namespace std;
//This is finding the all subarrays of zero sum in array
void zero_sum(int collection[], int size)
{
	//Define a map which is is an integer and that is capable to collect 
	//Group of integer value [like a vector of integer]
	map < int, vector < int > > mp;
	//This is collect information of resultant location
	vector < pair < int, int > > result;
	vector < int > ::iterator it;
	int current_sum = 0;
	int i = 0;
	//loop are collecting the information of zero sum subarray
	for (i = 0; i < size; ++i)
	{
		//Sum the array elements  which is exist on ith location
		current_sum += collection[i];
		if (current_sum == 0)
		{
			//base case when index 0 to i sum is zero
			result.push_back(make_pair(0, i));
		}
		if (mp.find(current_sum) != mp.end())
		{
			//When current sums already exist in map collection
			for (it = mp[current_sum].begin(); it != mp[current_sum].end(); it++)
			{
				//Add new pair into resultant vector
				//(*it) is a previous index which are same sum, 
				//add new pair (*it +1) to current element location [i]
				result.push_back(make_pair( *it + 1, i));
			}
		}
		//Add new sum and location into map
		mp[current_sum].push_back(i);
	}
	cout << " Zero sum subarray is None : ";
	if (result.size() == 0)
	{
		cout << "None" << endl;
	}
	else
	{
		//Display resultant index
		for (auto out_value = result.begin(); out_value != result.end(); out_value++)
		{
			cout << "\n [";
			for (int i = out_value->first; i <= out_value->second; ++i)
			{
				cout << "(" << collection[i] << ")";
				if (i < out_value->second)
				{
					cout << " + ";
				}
			}
			cout << "]";
		}
	}
}
int main()
{
	//Define collection of integer elements
	//This is an input array
	int collection[] = {
		1 , 4 , -5 , 3 , 2 , -5 , 7 , -2
	};
	//Get the size of array
	int size = sizeof(collection) / sizeof(collection[0]);
	//Find the sum of resulting to zero subarray
	zero_sum(collection, size);
	return 0;
}

Output

 Zero sum subarray is None :
 [(1) + (4) + (-5)]
 [(-5) + (3) + (2)]
 [(1) + (4) + (-5) + (3) + (2) + (-5)]
 [(3) + (2) + (-5)]
 [(-5) + (3) + (2) + (-5) + (7) + (-2)]
 [(-5) + (7) + (-2)]


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