Print All Subarrays With 0 Sum in C++
Print all subarrays with 0 sum
Here, on this page, we will discuss the program to Print all subarrays with 0 sum in C++ Programming language.
A simple solution is to consider all subarrays one by one and check if sum of every subarray is equal to 0 or not. The complexity of this solution would be O(n^2).A better approach is to use Hashing.
Algorithm :
- Keep the sum of the elements encountered thus far in a variable (say sum).
If the current sum is zero, we discovered a subarray beginning at index 0 and ending at index current index. - Check to see if the current sum is in the hash table.
- If the current sum already exists in the hash table, it means that it was the sum of some sub-array elements arr[0]…arr[i], and now the same sum is obtained for the current sub-array arr[0]…arr[j], implying that the sum of the sub-array arr[i+1]…arr[j] must be 0.
- Insert the current total into the hash table.
Run
#include <bits/stdc++.h>
using namespace std;
vector< pair<int, int> > findSubArrays(int arr[], int n)
{
// create an empty map
unordered_map<int, vector<int> > map;
vector <pair<int, int>> out;
// Maintains sum of elements so far
int sum = 0;
for (int i = 0; i < n; i++)
{
// add current element to sum
sum += arr[i];
// if sum is 0, we found a subarray starting
// from index 0 and ending at index i
if (sum == 0)
out.push_back(make_pair(0, i));
if (map.find(sum) != map.end())
{
// map[sum] stores starting index of all subarrays
vector<int> vc = map[sum];
for (auto it = vc.begin(); it != vc.end(); it++)
out.push_back(make_pair(*it + 1, i));
}
// Important - no else
map[sum].push_back(i);
}
// return output vector
return out;
}
// Utility function to print all subarrays with sum 0
void print(vector<pair<int, int>> out)
{
for (auto it = out.begin(); it != out.end(); it++)
cout << "Subarray found from Index " <<
it->first << " to " << it->second << endl;
}
// Driver code
int main()
{
int arr[] = {6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7};
int n = sizeof(arr)/sizeof(arr[0]);
vector<pair<int, int> > out = findSubArrays(arr, n);
if (out.size() == 0)
cout << "No subarray exists";
else
print(out);
return 0;
}Output
Subarray found from Index 2 to 4
Subarray found from Index 2 to 6
Subarray found from Index 5 to 6
Subarray found from Index 6 to 9
Subarray found from Index 0 to 10
Note:-
Time Complexity: O(N)
Auxiliary Space: O(N)

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