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

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

Below is a dry run of the above approach:

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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









Java














                                    filter_none




                                    edit

                                    close




                                    play_arrow




                                    link

                                    brightness_4

                                    code
                                




















// Java program to print all subarrays 


// in the array which has sum 0 


import java.io.*; 


import java.util.*; 


  


// User defined pair class 


class Pair  


{ 


    int first, second; 


    Pair(int a, int b)  


    { 


        first = a; 


        second = b; 


    } 


} 


  


public class GFG 


{ 


    // Function to print all subarrays in the array which  


    // has sum 0 


    static ArrayList<Pair> findSubArrays(int[] arr, int n) 


    { 


            // create an empty map  


            HashMap<Integer,ArrayList<Integer>> map = new HashMap<>(); 


  


            // create an empty vector of pairs to store  


            // subarray starting and ending index  


            ArrayList<Pair> out = new ArrayList<>(); 


  


            // 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.add(new Pair(0, i)); 


                ArrayList<Integer> al = new ArrayList<>(); 


                  


                // If sum already exists in the map there exists  


                // at-least one subarray ending at index i with  


                // 0 sum  


                if (map.containsKey(sum)) 


                { 


                    // map[sum] stores starting index of all subarrays 


                    al = map.get(sum); 


                    for (int it = 0; it < al.size(); it++) 


                    { 


                            out.add(new Pair(al.get(it) + 1, i));  


                    } 


                } 


                al.add(i); 


                map.put(sum, al); 


            } 


            return out; 


    }  


  


    // Utility function to print all subarrays with sum 0 


    static void print(ArrayList<Pair> out) 


    { 


            for (int i = 0; i < out.size(); i++) 


            { 


                Pair p = out.get(i); 


                System.out.println("Subarray found from Index "


                        + p.first + " to " + p.second);  


            } 


    } 


  


    // Driver code 


    public static void main(String args[]) 


    { 


            int[] arr = {6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7}; 


            int n = arr.length; 


              


            ArrayList<Pair> out = findSubArrays(arr, n); 


              


            // if we did not find any subarray with 0 sum,  


            // then subarray does not exists  


            if (out.size() == 0) 


                System.out.println("No subarray exists"); 


            else


                print(out); 


    } 


} 


  


// This code is contributed by rachana soma 



















                        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.


Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



          
          
          
          

            

    

        My Personal Notes
        arrow_drop_up
    

    

        
        

            
            
        

    


Recommended Posts:

Improved By :  rachana soma