# 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, -3], [-3]. All possible subarray of the given input array are [-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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 `// C++ program to Count the total number of  ` `// subarrays for a given array such that its  ` `// subarray should have non zero sum ` ` `  `#include ` `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 = arr; ` ` `  `    ``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 = -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); ` ` `  `    ``cout << CountSubarray(arr, N); ` `} `

## Python3

 `# 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 `

Output:

```15
```

Time Complexity: O(N)

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