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

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). sandharbnkamble

Below is the implementation of the above approach:

C++

// C++ program to print all subarrays
// in the array which has sum 0
#include <bits/stdc++.h>
using namespace std;

vector<pair<int, int> > findSubArrays(int arr[], int n)
{

    // Array to store all the start and end
    // indices of subarrays with 0 sum
    vector<pair<int, int> > output;
    for (int i = 0; i < n; i++) {
        int prefix = 0;
        for (int j = i; j < n; j++) {
            prefix += arr[j];
            if (prefix == 0)
                output.push_back({ i, j });
        }
    }

    return output;
}

// Function to print all subarrays with 0 sum
void print(vector<pair<int, int> > output)
{
    for (auto it = output.begin(); it != output.end(); it++)
        cout << "Subarray found from Index " << it->first
             << " to " << it->second << endl;
}

// Driver code
int main()
{

    // Given array
    int arr[] = { 6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    vector<pair<int, int> > output = findSubArrays(arr, n);

    // if we didn’t find any subarray with 0 sum,
    // then subarray doesn’t exists
    if (output.size() == 0) {
        cout << "No subarray exists";
    }
    else {
        print(output);
    }
    return 0;
}

// This article is contributed by Arpit Jain

Java

// 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
{
  static ArrayList<Pair> findSubArrays(int[] arr, int n)
  {
    // Array to store all the start and end
    // indices of subarrays with 0 sum
    ArrayList<Pair> out = new ArrayList<>();

    for (int i = 0; i < n; i++)
    {
      int prefix = 0;
      for(int j = i; j < n; j++){
        prefix += arr[j];
        if(prefix == 0)
          out.add(new Pair(i, j));
      }
    }
    return out;
  }

  // Function to print all subarrays with 0 sum
  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[])
  {

    // Given array
    int[] arr = {6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7};
    int n = arr.length;

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

    // if we didn’t find any subarray with 0 sum,
    // then subarray doesn’t exists
    if (out.size() == 0)
      System.out.println("No subarray exists");
    else
      print(out);
  }
}

// This code is contributed by sandharbnkamble.

Python3

# User defined pair class
class Pair :
    first = 0
    second = 0
    def __init__(self, a,  b) :
        self.first = a
        self.second = b
class GFG :
    @staticmethod
    def  findSubArrays( arr,  n) :

        # Array to store all the start and end
        # indices of subarrays with 0 sum
        out =  []
        i = 0
        while (i < n) :
            prefix = 0
            j = i
            while (j < n) :
                prefix += arr[j]
                if (prefix == 0) :
                    out.append(Pair(i, j))
                j += 1
            i += 1
        return out

    # Function to print all subarrays with 0 sum
    @staticmethod
    def print( out) :
        i = 0
        while (i < len(out)) :
            p = out[i]
            print("Subarray found from Index " + str(p.first) + " to " + str(p.second))
            i += 1

    # Driver code
    @staticmethod
    def main( args) :

        # Given array
        arr = [6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7]
        n = len(arr)

        # Function Call
        out = GFG.findSubArrays(arr, n)

        # if we didn't find any subarray with 0 sum,
        # then subarray doesn't exists
        if (len(out) == 0) :
            print("No subarray exists")
        else :
            GFG.print(out)


if __name__=="__main__":
    GFG.main([])

    # This code is contributed by aadityaburujwale.

C#

using System;
using System.Collections.Generic;

class GFG
{

  // Array to store all the start and end
  // indices of subarrays with 0 sum
  static List<Tuple<int, int>> findSubArrays(int[] arr, int n)
  {
    var output = new List<Tuple<int, int>>();
    for (int i = 0; i < n; i++)
    {
      int prefix = 0;
      for (int j = i; j < n; j++)
      {
        prefix += arr[j];
        if (prefix == 0)
          output.Add(Tuple.Create(i, j));
      }
    }

    return output;
  }

  // Function to print all subarrays with 0 sum
  static void print(List<Tuple<int, int>> output)
  {
    foreach (var subArray in output)
      Console.Write("Subarray found from Index " + subArray.Item1 + " to " + subArray.Item2+"\n");
  }

  // Driver code
  public static void Main()
  {
    // Given array
    int[] arr = { 6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7 };
    int n = arr.Length;

    // Function Call
    List<Tuple<int, int>> output = findSubArrays(arr, n);

    // if we didn’t find any subarray with 0 sum,
    // then subarray doesn’t exists
    if (output.Count == 0)
    {
      Console.WriteLine("No subarray exists");
    }
    else
    {
      print(output);
    }
  }
}

// This code is contributed by ratiagarwal.

Javascript

// Javascript program to print all subarrays
// in the array which has sum 0
function findSubArrays(arr, n)
{

    // Array to store all the start and end
    // indices of subarrays with 0 sum
    let out =[];
    for (let i = 0; i < n; i++) {
        let prefix = 0;
        for (let j = i; j < n; j++) {
            prefix += arr[j];
            if (prefix == 0)
                out.push([i, j]);
        }
    }

    return out;
}

// Function to print all subarrays with 0 sum
function print(out)
{
    for (let it of out)
        console.log("Subarray found from Index " + it[0]
             + " to " + it[1]);
}

// Driver code
// Given array
let arr = [ 6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7 ];
let n = arr.length ;

// Function Call
let out = findSubArrays(arr, n);

// if we didn’t find any subarray with 0 sum,
// then subarray doesn’t exists
if (out.length == 0) {
    console.log("No subarray exists");
}
else {
    print(out);
}

 // This code is contributed by poojaagarwal2.

Output

Subarray found from Index 0 to 10
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

Time Complexity: O(N^2) since we are using 2 loops.
Auxiliary Space: O(1), as constant extra space is required.

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 already exists in the hash table then it indicates that this sum 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] which means that the sum of the sub-array arr[i+1]…arr[j] must be 0.
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++

// C++ program to print all subarrays
// in the array which has sum 0
#include <bits/stdc++.h>
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;
}

Java

// 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

Python3

# Python3 program to print all subarrays
# in the array which has sum 0

# Function to get all subarrays
# in the array which has sum 0
def findSubArrays(arr,n):

    # create a python dict
    hashMap = {}

    # create a python list
    # equivalent to ArrayList
    out = []

    # tracker for sum of elements
    sum1 = 0
    for i in range(n):

        # increment sum by element of array
        sum1 += arr[i]

        # if sum is 0, we found a subarray starting
        # from index 0 and ending at index i
        if sum1 == 0:
            out.append((0, i))
        al = []

        # If sum already exists in the map
        # there exists at-least one subarray
        # ending at index i with 0 sum
        if sum1 in hashMap:

            # map[sum] stores starting index
            # of all subarrays
            al = hashMap.get(sum1)
            for it in range(len(al)):
                out.append((al[it] + 1, i))
        al.append(i)
        hashMap[sum1] = al
    return out

# Utility function to print
# all subarrays with sum 0
def printOutput(output):
    for i in output:
        print ("Subarray found from Index " +
                str(i[0]) + " to " + str(i[1]))

# Driver Code
if __name__ == '__main__':
    arr = [6, 3, -1, -3, 4, -2,
              2, 4, 6, -12, -7]
    n = len(arr)
    out = findSubArrays(arr, n)

    # if we did not find any subarray with 0 sum,
    # then subarray does not exists
    if (len(out) == 0):
        print ("No subarray exists")
    else:
        printOutput (out)

# This code is contributed by Vikas Chitturi

C#

// C# program to print all subarrays
// in the array which has sum 0
using System;
using System.Collections.Generic;

// User defined pair class
class Pair
{
    public int first, second;
    public Pair(int a, int b)
    {
        first = a;
        second = b;
    }
}

class GFG
{
    // Function to print all subarrays
    // in the array which has sum 0
    static List<Pair> findSubArrays(int[] arr, int n)
    {
        // create an empty map
        Dictionary<int, List<int>> map =
                   new Dictionary<int, List<int>>();

        // create an empty vector of pairs to store
        // subarray starting and ending index
        List<Pair> outt = new List<Pair>();

        // 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)
                outt.Add(new Pair(0, i));
            List<int> al = new List<int>();

            // 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[sum];
                for (int it = 0; it < al.Count; it++)
                {
                    outt.Add(new Pair(al[it] + 1, i));
                }
            }
            al.Add(i);
            if(map.ContainsKey(sum))
                map[sum] = al;
            else
                map.Add(sum, al);
        }
        return outt;
    }

    // Utility function to print all subarrays with sum 0
    static void print(List<Pair> outt)
    {
        for (int i = 0; i < outt.Count; i++)
        {
            Pair p = outt[i];
            Console.WriteLine("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;

        List<Pair> outt = findSubArrays(arr, n);

        // if we did not find any subarray with 0 sum,
        // then subarray does not exists
        if (outt.Count == 0)
            Console.WriteLine("No subarray exists");
        else
            print(outt);
    }
}

// This code is contributed by Rajput-Ji

Javascript

// JavaScript program to print all subarrays
// in the array which has sum 0

// Function to print all subarrays in the array which
// has sum 0
function findSubArrays(arr, n)
{
    // create an empty map
    let map = {};

    // create an empty vector of pairs to store
    // subarray starting and ending index
    let out = [];

    // Maintains sum of elements so far
    let sum = 0;

    for (var 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([0, i]);

        // If sum already exists in the map there exists
        // at-least one subarray ending at index i with
        // 0 sum
        if (map.hasOwnProperty(sum))
        {
            // map[sum] stores starting index of all subarrays
            let vc = map[sum];
            for (let it of vc)
                out.push([it + 1, i]);
        }
        else
            map[sum] = [];

        // Important - no else
        map[sum].push(i);
    }

    // return output vector
    return out;
}

// Utility function to print all subarrays with sum 0
function print(out)
{
    for (let it of out)
        console.log("Subarray found from Index " + it[0] + " to " + it[1]);
}


// Driver code
let arr = [6, 3, -1, -3, 4, -2, 2, 4, 6, -12, -7];
let n = arr.length;

let out = findSubArrays(arr, n);

// if we didn’t find any subarray with 0 sum,
// then subarray doesn’t exists
if (out.length == 0)
    console.log("No subarray exists");
else
    print(out);

// This code is contributed by phasing17.

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

Time Complexity: O(N)
Auxiliary Space: O(N)

Approach 3:  Finding subarrays with sum 0 using dynamic programming:

Algorithm:

1. Create an empty vector of pairs to store the starting and ending indices of all subarrays with a 0 sum.

2. Create an unordered map with the starting index of all subarrays that have the same sum.

3. Initialize  variable “sum =0″.

4.  Add a dummy index -1 to represent the empty subarray , i.e. ” sums[0].push_back(-1);”

5. Traverse the array from left to right:

• Add current element to the sum.
• If sum is already present in map, retrieve the vector of indices and add all subarrays beginning with those indices and ending with the current index to the output vector.
• In the map, add the current index to the vector of indices for the current total.

6. Return the output vector “out”.

Here’s the implementation of above algorithm:

C++

#include <bits/stdc++.h>
using namespace std;

vector<pair<int, int>> findSubArrays(int arr[], int n) {
    vector<pair<int, int>> out;
    unordered_map<int, vector<int>> sums; // map to store the starting index of all subarrays with the same sum
    int sum = 0;
    sums[0].push_back(-1); // add a dummy index -1 to represent the empty subarray
    for (int i = 0; i < n; i++) {
        sum += arr[i];
        if (sums.find(sum) != sums.end()) {
            vector<int> indices = sums[sum];
            for (int j = 0; j < indices.size(); j++) {
                out.push_back(make_pair(indices[j] + 1, i));
            }
        }
        sums[sum].push_back(i);
    }
    return out;
}

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;
    }
}

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;
}

// This article is contributed by Vaibhav Saroj

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

The dynamic programming approach to finding subarrays with sum 0  is contributed by Vaibhav Saroj.

Time Complexity: O(n)
Auxiliary Space: O(n)

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