Count of subarray that does not contain any subarray with sum 0

Given an array arr, the task is to find the total number of subarrays of the given array which do not contain any subarray whose sum of elements is equal to zero. All the array elements may not be distinct.
Examples:

Input: arr = {2, 4, -6} 
Output: 5 
Explanation:
There are 5 subarrays which do not contain any subarray whose elements sum is equal to zero: [2], [4], [-6], [2, 4], [4, -6]
Input: arr = {10, -10, 10} 
Output: 3 

Approach:

1. Firstly store all elements of array as sum of its previous element. 
   
2. Now take two pointers, increase second pointer and store the value in a map while a same element not encounter. 
   
3. If an element encounter which is already exist in map, this means there exist a subarray between two pointers whose elements sum is equal to 0. 
   
4. Now increase first pointer and remove the element from map while the two same elements exists. 
   
5. Store the answer in a variable and finally return it.

Below is the implementation of above approach:

C++

// C++ program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero

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

// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
void numberOfSubarrays(int arr[], int n)
{
    vector<int> v(n + 1);
    v[0] = 0;

    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }

    map<int, int> mp;

    int begin = 0, end = 0, answer = 0;

    mp[0] = 1;

    while (begin < n) {

        while (end < n
               && mp.find(v[end + 1])
                      == mp.end()) {
            end++;
            mp[v[end]] = 1;
        }

        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;

        // Erase beginning element from map
        mp.erase(v[begin]);

        // Increase begin
        begin++;
    }

    // Print the result
    cout << answer << endl;
}

// Driver Code
int main()
{

    int arr[] = { 2, 4, -6 };
    int size = sizeof(arr) / sizeof(arr[0]);

    numberOfSubarrays(arr, size);

    return 0;
}

Java

// Java program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
import java.util.*;

class GFG{

// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
static void numberOfSubarrays(int arr[], int n)
{
    int []v = new int[n + 1];
    v[0] = 0;

    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }

    HashMap<Integer,Integer> mp = new HashMap<Integer,Integer>();

    int begin = 0, end = 0, answer = 0;

    mp.put(0, 1);

    while (begin < n) {

        while (end < n
               && !mp.containsKey(v[end + 1])) {
            end++;
            mp.put(v[end],  1);
        }

        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;

        // Erase beginning element from map
        mp.remove(v[begin]);

        // Increase begin
        begin++;
    }

    // Print the result
    System.out.print(answer +"\n");
}

// Driver Code
public static void main(String[] args)
{

    int arr[] = { 2, 4, -6 };
    int size = arr.length;

    numberOfSubarrays(arr, size);
}
}

// This code is contributed by sapnasingh4991

Python3

# Python 3 program to Count the no of subarray
# which do not contain any subarray
# whose sum of elements is equal to zero

# Function that print the number of
# subarrays which do not contain any subarray
# whose elements sum is equal to 0
def numberOfSubarrays(arr, n):

    v = [0]*(n + 1)

    # Storing each element as sum
    # of its previous element
    for i in range( n):
        v[i + 1] = v[i] + arr[i]

    mp = {}

    begin, end, answer = 0 , 0 , 0

    mp[0] = 1

    while (begin < n):

        while (end < n
            and (v[end + 1]) not in mp):
            end += 1
            mp[v[end]] = 1

        # Check if another same element found
        # this means a subarray exist between
        # end and begin whose sum
        # of elements is equal to 0
        answer = answer + end - begin

        # Erase beginning element from map
        del mp[v[begin]]

        # Increase begin
        begin += 1

    # Print the result
    print(answer)

# Driver Code
if __name__ == "__main__":

    arr = [ 2, 4, -6 ]
    size = len(arr)
    numberOfSubarrays(arr, size)

# This code is contributed by chitranayal

C#

// C# program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero
using System;
using System.Collections.Generic;

class GFG{

// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
static void numberOfSubarrays(int []arr, int n)
{
    int []v = new int[n + 1];
    v[0] = 0;

    // Storing each element as sum
    // of its previous element
    for (int i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }

    Dictionary<int,int> mp = new Dictionary<int,int>();

    int begin = 0, end = 0, answer = 0;

    mp.Add(0, 1);

    while (begin < n) {

        while (end < n
               && !mp.ContainsKey(v[end + 1])) {
            end++;
            mp.Add(v[end],  1);
        }

        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;

        // Erase beginning element from map
        mp.Remove(v[begin]);

        // Increase begin
        begin++;
    }

    // Print the result
    Console.Write(answer +"\n");
}

// Driver Code
public static void Main(String[] args)
{

    int []arr = { 2, 4, -6 };
    int size = arr.Length;

    numberOfSubarrays(arr, size);
}
}

// This code is contributed by Rajput-Ji

Javascript

<script>

// JavaScript program to Count the no of subarray
// which do not contain any subarray
// whose sum of elements is equal to zero

// Function that print the number of
// subarrays which do not contain any subarray
// whose elements sum is equal to 0
function numberOfSubarrays(arr, n)
{
    let v = new Array(n + 1);
    v[0] = 0;

    // Storing each element as sum
    // of its previous element
    for (let i = 0; i < n; i++) {
        v[i + 1] = v[i] + arr[i];
    }

    let mp = new Map();

    let begin = 0, end = 0, answer = 0;

    mp.set(0, 1);

    while (begin < n) {

        while (end < n && !mp.has(v[end + 1])) {
            end++;
            mp.set(v[end], 1);
        }

        // Check if another same element found
        // this means a subarray exist between
        // end and begin whose sum
        // of elements is equal to 0
        answer = answer + end - begin;

        // Erase beginning element from map
        mp.clear();

        // Increase begin
        begin++;
    }

    // Print the result
    document.write(answer + "<br>");
}

// Driver Code

let arr = [ 2, 4, -6 ];
let size = arr.length;

numberOfSubarrays(arr, size);

// This code is contributed by _saurabh_jaiswal

</script>

Output:

5

Time Complexity: O(N)

Auxiliary Space: O(N)