Recuento de subarreglo que no contiene ningún subarreglo con suma 0

Dada una array arr , la tarea es encontrar el número total de subarreglos de la array dada que no contienen ningún subarreglo cuya suma de elementos sea igual a cero. Todos los elementos de la array son distintos.
Ejemplos: 
 

Entrada: arr = {2, 4, -6} 
Salida: 5 
Explicación: 
Hay 5 subarreglos que no contienen ningún subarreglo cuya suma de elementos sea igual a cero: [2], [4], [-6], [2 , 4], [4, -6]
Entrada: array = {10, -10, 10} 
Salida: 3 
 

Acercarse: 
 

1. En primer lugar, almacene todos los elementos de la array como suma de su elemento anterior. 
    
2. Ahora tome dos punteros, aumente el segundo puntero y almacene el valor en un mapa mientras no se encuentra un mismo elemento. 
    
3. Si se encuentra un elemento que ya existe en el mapa, esto significa que existe un subarreglo entre dos punteros cuya suma de elementos es igual a 0. 
    
4. Ahora aumente el primer puntero y elimine el elemento del mapa mientras existan los dos mismos elementos. 
    
5. Almacene la respuesta en una variable y finalmente devuélvala.

A continuación se muestra la implementación del enfoque anterior: 
 

// 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];
    }
 
    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>

5

Complejidad de tiempo: O(N)

Espacio Auxiliar: O(N)