Given an array, print all subarrays in the array which has sum 0.

Examples:

Input:arr = [6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7]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

Related posts: Find if there is a subarray with 0 sum

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.

Do following for each element in the array

- Maintain sum of elements encountered so far in a variable (say sum).
- If current sum is 0, we found a subarray starting from index 0 and ending at index current index
- Check if current sum exists in the hash table or not.
- If current sum exists in the hash table, that means we have subarray(s) present with 0 sum that ends at current index.
- Insert current sum into the hash table

Below is C++ implementation of above idea –

// C++ program to print all subarrays // in the array which has sum 0 #include <iostream> #include <unordered_map> #include <vector> using namespace std; // Function to print all subarrays in the array which // has sum 0 vector< pair<int, int> > findSubArrays(int arr[], int n) { // create an empty map unordered_map<int, vector<int> > map; // create an empty vector of pairs to store // subarray starting and ending index 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 sum already exists in the map there exists // at-least one subarray ending at index i with // 0 sum 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 we didn’t find any subarray with 0 sum, // then subarray doesn’t exists 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

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