Count subarrays with non-zero sum in the given Array

Given an array arr[] of size N, the task is to count the total number of subarrays for the given array arr[] which have a non-zero sum.

Examples:

Input: arr[] = {-2, 2, -3}
Output: 4
Explanation:
The subarrays with non zero sum are: [-2], [2], [2, -3], [-3]. All possible subarray of the given input array are [-2], [2], [-3], [2, -2], [2, -3], [-2, 2, -3]. Out of these [2, -2] is not included in the count because 2+(-2) = 0 and [-2, 2, -3] is not selected because one the subarray [2, -2] of this array has a zero sum of elements.

Input: arr[] = {1, 3, -2, 4, -1}
Output: 15
Explanation:
There are 15 subarray for the given array {1, 3, -2, 4, -1} which has a non zero sum.

Approach:



The main idea to solve the above question is to use the Prefix Sum Array and Map Data Structure.

  • First build the Prefix sum array of the given array and use the below formula to check if the subarray has 0 sum of elements.
  • Sum of Subarray[L, R] = Prefix[R] – Prefix[L – 1]. So, If Sum of Subarray[L, R] = 0

    Then, Prefix[R] – Prefix[L – 1] = 0. Hence, Prefix[R] = Prefix[L – 1]

  • Now, iterate from 1 to N and keep a Hash table for storing the index of the previous occurrence of the element and a variable lets say last and initialize it to 0.
  • Check if Prefix[i] is already present in the Hash or not. If yes then, update last as last = max(last, hash[Prefix[i]] + 1). Otherwise, Add i – last to the answer and update the Hash table.

Below is the implementation of the above approach:

C++

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// C++ program to Count the total number of 
// subarrays for a given array such that its 
// subarray should have non zero sum
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to build the Prefix sum array
vector<int> PrefixSumArray(int arr[], int n)
{
    vector<int> prefix(n);
  
    // Store prefix of the first position
    prefix[0] = arr[0];
  
    for (int i = 1; i < n; i++)
        prefix[i] = prefix[i - 1] + arr[i];
  
    return prefix;
}
  
// Function to return the Count of
// the total number of subarrays
int CountSubarray(int arr[], int n)
{
    vector<int> Prefix(n);
  
    // Calculating the prefix array
    Prefix = PrefixSumArray(arr, n);
  
    int last = 0, ans = 0;
  
    map<int, int> Hash;
  
    Hash[0] = -1;
  
    for (int i = 0; i <= n; i++) {
        // Check if the element already exists
        if (Hash.count(Prefix[i]))
            last = max(last, Hash[Prefix[i]] + 1);
  
        ans += max(0, i - last);
  
        // Mark the element
        Hash[Prefix[i]] = i;
    }
  
    // Return the final answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 3, -2, 4, -1 };
  
    int N = sizeof(arr) / sizeof(arr[0]);
  
    cout << CountSubarray(arr, N);
}

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Python3

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# Python3 program to count the total number  
# of subarrays for a given array such that 
# its subarray should have non zero sum 
  
# Function to build the prefix sum array 
def PrefixSumArray(arr, n):
  
    prefix = [0] * (n + 1); 
  
    # Store prefix of the first position 
    prefix[0] = arr[0]; 
  
    for i in range(1, n):
        prefix[i] = prefix[i - 1] + arr[i]; 
          
    return prefix; 
  
# Function to return the count of 
# the total number of subarrays 
def CountSubarray(arr, n): 
  
    Prefix = [0] * (n + 1); 
  
    # Calculating the prefix array 
    Prefix = PrefixSumArray(arr, n); 
  
    last = 0; ans = 0
  
    Hash = {}; 
  
    Hash[0] = -1
  
    for i in range(n + 1):
          
        # Check if the element already exists 
        if (Prefix[i] in Hash):
            last = max(last, Hash[Prefix[i]] + 1); 
  
        ans += max(0, i - last); 
  
        # Mark the element 
        Hash[Prefix[i]] = i; 
  
    # Return the final answer 
    return ans; 
  
# Driver code 
if __name__ == "__main__"
  
    arr = [ 1, 3, -2, 4, -1 ]; 
    N = len(arr); 
  
    print(CountSubarray(arr, N)); 
      
# This code is contributed by AnkitRai01

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Output:

15

Time Complexity: O(N)

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