Print all subarrays with 0 sum

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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

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

This article is contributed by Aditya Goel. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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