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

• Difficulty Level : Hard
• Last Updated : 21 Aug, 2022

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:
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:

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 ``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 mp = ``new` `HashMap();`` ` `    ``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

 ``

Output:

`5`

Time Complexity: O(N)

Auxiliary Space: O(N)

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