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

Last Updated : 14 Oct, 2023

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

Naive Approach-

The idea is to find all subarrays and then find those subarrays whose any of the subarrays does not have a sum equal to zero.

Steps to implement-

Declare a variable count with value 0 to store the final answer
Run two for loops to find all subarray
For each subarray find its all subarray by running two another for loops
If it's every subarray has a non-zero sum then increment the count
In the last print the value of the count

Code-

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) {     //To store final answer     int count=0;         //Find all subarray    for(int i=0;i<N;i++){       for(int j=i;j<N;j++){                      //Boolean variable to tell whether its any subarray           //have sum is equal to zero or not           bool val=true;                      //Find all subarray of this subarray           for(int m=i;m<=j;m++){               //To store sum of all elements of subarray               int sum=0;               for(int n=m;n<=j;n++){                   sum+=arr[n];                   if(sum==0){                       val=false;                        break;                   }               }                              if(val==false){break;}           }           if(val==true){count++;}      }     }    //Print final answer    cout<<count<<endl; }  // Driver Code int main() {     int arr[] = { 2, 4, -6 };     int size = sizeof(arr) / sizeof(arr[0]);     numberOfSubarrays(arr, size);     return 0; }

Java

public class SubarraySum {      // Function that returns the number of subarrays which do not contain     // any subarray whose elements sum is equal to 0     public static int numberOfSubarrays(int[] arr, int N) {         // To store the final answer         int count = 0;          // Find all subarrays         for (int i = 0; i < N; i++) {             for (int j = i; j < N; j++) {                  // Boolean variable to tell whether any subarray has sum equal to zero or not                 boolean val = true;                  // Find all subarrays of this subarray                 for (int m = i; m <= j; m++) {                     // To store sum of all elements of subarray                     int sum = 0;                     for (int n = m; n <= j; n++) {                         sum += arr[n];                         if (sum == 0) {                             val = false;                             break;                         }                     }                      if (!val) {                         break;                     }                 }                 if (val) {                     count++;                 }             }         }          return count;     }      public static void main(String[] args) {         int[] arr = { 2, 4, -6 };         int size = arr.length;         int result = numberOfSubarrays(arr, size);         System.out.println(result);          // This Code Is Contributed By Shubham Tiwari     } }

Python

# Function that returns the number of # subarrays which do not contain any subarray # whose elements sum is equal to 0 def numberOfSubarrays(arr):     N = len(arr)     # To store the final answer     count = 0          # Find all subarrays     for i in range(N):         for j in range(i, N):                          # Boolean variable to tell whether any subarray             # has a sum equal to zero or not             val = True                          # Find all subarrays of this subarray             for m in range(i, j+1):                 s = 0                 for n in range(m, j+1):                     s += arr[n]                     if s == 0:                         val = False                         break                                  if not val:                     break             if val:                 count += 1                      # Return the final answer     return count  # Driver Code arr = [2, 4, -6] print(numberOfSubarrays(arr)) #This Code Is Contributed By Shubham Tiwari

using System;  public class GFG {     // Function that print the number of     // subarrays which do not contain any subarray     // whose elements sum is equal to 0     public static void NumberOfSubarrays(int[] arr, int N) {         //To store final answer         int count=0;                 //Find all subarray        for(int i=0; i<N; i++){           for(int j=i; j<N; j++){                              //Boolean variable to tell whether its any subarray               //have sum is equal to zero or not               bool val=true;                              //Find all subarray of this subarray               for(int m=i; m<=j; m++){                   //To store sum of all elements of subarray                   int sum=0;                   for(int n=m; n<=j; n++){                       sum += arr[n];                       if(sum==0){                           val = false;                           break;                       }                   }                                      if(val==false) { break; }               }               if(val==true) { count++; }          }         }        //Print final answer        Console.WriteLine(count);     }      // 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 Shubham Tiwari     } }

JavaScript

// Function that prints the number of // subarrays which do not contain any subarray // whose elements sum is equal to 0 function numberOfSubarrays(arr) {     // To store final answer     let count = 0;      // Find all subarrays     for (let i = 0; i < arr.length; i++) {         for (let j = i; j < arr.length; j++) {              // Boolean variable to tell whether any subarray             // has a sum equal to zero or not             let val = true;              // Find all subarrays of this subarray             for (let m = i; m <= j; m++) {                 // To store the sum of all elements of subarray                 let sum = 0;                 for (let n = m; n <= j; n++) {                     sum += arr[n];                     if (sum === 0) {                         val = false;                         break;                     }                 }                  if (val === false) {                     break;                 }             }             if (val === true) {                 count++;             }         }     }     // Print the final answer     console.log(count); }  // Driver Code const arr = [2, 4, -6]; numberOfSubarrays(arr);  // This code is contributed by rambabuguphka

Output-

Time Complexity: O(N⁴), because two loops to find all subarray and two more loops to find all subarray of a particular subarray
Auxiliary Space: O(1), because no extra space has been used

Approach:

Firstly store all elements of array as sum of its previous element.
Now take two pointers, increase second pointer and store the value in a map while a same element not encounter.
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.
Now increase first pointer and remove the element from map while the two same elements exists.
Store the answer in a variable and finally return it.

Below is the implementation of the 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];     }      unordered_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# 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

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

Count subarrays having sum modulo K same as the length of the subarray

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

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