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 let’s 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 <bits/stdc++.h>using namespace std;// Function to build the Prefix sum arrayvector<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 subarraysint 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 codeint main(){ int arr[] = { 1, 3, -2, 4, -1 }; int N = sizeof(arr) / sizeof(arr[0]); cout << CountSubarray(arr, N);} |
Java
// Java program to count the total number of // subarrays for a given array such that its // subarray should have non zero sumimport java.util.*;class GFG{// Function to build the Prefix sum arraystatic int[] PrefixSumArray(int arr[], int n){ int []prefix = new int[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 subarraysstatic int CountSubarray(int arr[], int n){ int []Prefix = new int[n]; // Calculating the prefix array Prefix = PrefixSumArray(arr, n); int last = 0, ans = 0; HashMap<Integer, Integer> Hash = new HashMap<Integer, Integer>(); Hash.put(0, -1); for(int i = 0; i <= n; i++) { // Check if the element already exists if (i < n && Hash.containsKey(Prefix[i])) last = Math.max(last, Hash.get(Prefix[i]) + 1); ans += Math.max(0, i - last); // Mark the element if (i < n) Hash.put(Prefix[i], i); } // Return the final answer return ans;}// Driver codepublic static void main(String[] args){ int arr[] = { 1, 3, -2, 4, -1 }; int N = arr.length; System.out.print(CountSubarray(arr, N));}}// This code is contributed by amal kumar choubey |
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 |
C#
// C# program to count the total number of // subarrays for a given array such that its // subarray should have non zero sumusing System;using System.Collections.Generic;class GFG{// Function to build the Prefix sum arraystatic int[] PrefixSumArray(int []arr, int n){ int []prefix = new int[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 subarraysstatic int CountSubarray(int []arr, int n){ int []Prefix = new int[n]; // Calculating the prefix array Prefix = PrefixSumArray(arr, n); int last = 0, ans = 0; Dictionary<int, int> Hash = new Dictionary<int, int>(); Hash.Add(0, -1); for(int i = 0; i <= n; i++) { // Check if the element already exists if (i < n && Hash.ContainsKey(Prefix[i])) last = Math.Max(last, Hash[Prefix[i]] + 1); ans += Math.Max(0, i - last); // Mark the element if (i < n) Hash.Add(Prefix[i], i); } // Return the readonly answer return ans;}// Driver codepublic static void Main(String[] args){ int []arr = {1, 3, -2, 4, -1}; int N = arr.Length; Console.Write(CountSubarray(arr, N));}}// This code is contributed by shikhasingrajput |
Javascript
<script>// Javascript 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 arrayfunction PrefixSumArray(arr, n){ let prefix = Array.from({length: n}, (_, i) => 0); // Store prefix of the first position prefix[0] = arr[0]; for(let i = 1; i < n; i++) prefix[i] = prefix[i - 1] + arr[i]; return prefix;} // Function to return the Count of// the total number of subarraysfunction CountSubarray(arr, n){ let Prefix = Array.from({length: n}, (_, i) => 0); // Calculating the prefix array Prefix = PrefixSumArray(arr, n); let last = 0, ans = 0; let Hash = new Map(); Hash.set(0, -1); for(let i = 0; i <= n; i++) { // Check if the element already exists if (i < n && Hash.has(Prefix[i])) last = Math.max(last, Hash.get(Prefix[i]) + 1); ans += Math.max(0, i - last); // Mark the element if (i < n) Hash.set(Prefix[i], i); } // Return the final answer return ans;}// Driver code let arr = [ 1, 3, -2, 4, -1 ]; let N = arr.length; document.write(CountSubarray(arr, N)); // This code is contributed by code_hunt.</script> |
Output:
15
Time Complexity: O(N)
Auxiliary Space: O(N)
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