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;
}
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
}
}
# 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
}
}
// 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-
5
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 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
# 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
<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>
Time Complexity: O(N)
Auxiliary Space: O(N)
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