# 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

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; ` `} ` |

*chevron_right*

*filter_none*

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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