Recuento de elementos que es la suma de un subarreglo del Array dado

Dada una array arr[] , la tarea es contar los elementos de una array de modo que exista una subarreglo cuya suma sea igual a este elemento.
Nota: La longitud del subarreglo debe ser mayor que 1. 

Ejemplos: 

Entrada: arr[] = {1, 2, 3, 4, 5, 6, 7} 
Salida: 4 
Explicación: 
Hay 4 de esos elementos en el arreglo – 
arr[2] = 3 => arr[0-1] => 1 + 2 = 3 
arr[4] = 5 => arr[1-2] => 2 + 3 = 5 
arr[5] = 6 => arr[0-2] => 1 + 2 + 3 = 6 
arr [6] = 7 => arr[2-3] => 3 + 4 = 7
Entrada: arr[] = {1, 2, 3, 3} 
Salida: 2 
Hay 2 de esos elementos en la array: 
arr[2] = 3 => array[0-1] => 1 + 2 = 3 
array[3] = 3 => array[0-1] => 1 + 2 = 3  

Enfoque: La idea es almacenar la frecuencia de los elementos de la array en un mapa hash . Luego, itere sobre cada subarreglo posible y verifique que su suma esté presente en el mapa hash. En caso afirmativo, aumente el recuento de dichos elementos por su frecuencia.

A continuación se muestra la implementación del enfoque anterior:  

// C++ implementation to count the
// elements such that their exist
// a subarray whose sum is equal to
// this element
 
#include <bits/stdc++.h>
 
using namespace std;
 
// Function to count the elements
// such that their exist a subarray
// whose sum is equal to this element
int countElement(int arr[], int n)
{
    map<int, int> freq;
    int ans = 0;
 
    // Loop to count the frequency
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
 
    // Loop to iterate over every possible
    // subarray of the array
    for (int i = 0; i < n - 1; i++) {
        int tmpsum = arr[i];
        for (int j = i + 1; j < n; j++) {
            tmpsum += arr[j];
            if (freq.find(tmpsum) != freq.end()) {
                ans += freq[tmpsum];
            }
        }
    }
 
    return ans;
}
 
// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << countElement(arr, n) << endl;
 
    return 0;
}

// Java implementation to count the
// elements such that their exist
// a subarray whose sum is equal to
// this element
import java.util.*;
 
class GFG {
 
// Function to count the elements
// such that their exist a subarray
// whose sum is equal to this element
static int countElement(int arr[], int n)
{
    int freq[] = new int[n + 1];
    int ans = 0;
     
    // Loop to count the frequency
    for(int i = 0; i < n; i++)
    {
       freq[arr[i]]++;
    }
 
    // Loop to iterate over every possible
    // subarray of the array
    for(int i = 0; i < n - 1; i++)
    {
       int tmpsum = arr[i];
 
       for(int j = i + 1; j < n; j++)
       {
          tmpsum += arr[j];
 
          if (tmpsum <= n)
          {
              ans += freq[tmpsum];
              freq[tmpsum] = 0;
          }
       }
    }
     
    return ans;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.length;
    System.out.println(countElement(arr, n));
}
}
 
// This code is contributed by rutvik_56

# Python3 implementation to count the
# elements such that their exist
# a subarray whose sum is equal to
# this element
 
# Function to count element such
# that their exist a subarray whose
# sum is equal to this element
def countElement(arr, n):
    freq = {}
    ans = 0
     
    # Loop to compute frequency
    # of the given elements
    for i in range(n):
        freq[arr[i]] = \
            freq.get(arr[i], 0) + 1
     
    # Loop to iterate over every
    # possible subarray of array
    for i in range(n-1):
        tmpsum = arr[i]
        for j in range(i + 1, n):
            tmpsum += arr[j]
            if tmpsum in freq:
                ans += freq[tmpsum]
    return ans
          
  
# Driver Code
if __name__ == "__main__":
    arr =[1, 2, 3, 4, 5, 6, 7]
    n = len(arr)
    print(countElement(arr, n))

// C# implementation to count the
// elements such that their exist
// a subarray whose sum is equal to
// this element
using System;
 
class GFG {
 
// Function to count the elements
// such that their exist a subarray
// whose sum is equal to this element
static int countElement(int[] arr, int n)
{
    int[] freq = new int[n + 1];
    int ans = 0;
     
    // Loop to count the frequency
    for(int i = 0; i < n; i++)
    {
       freq[arr[i]]++;
    }
 
    // Loop to iterate over every possible
    // subarray of the array
    for(int i = 0; i < n - 1; i++)
    {
       int tmpsum = arr[i];
 
       for(int j = i + 1; j < n; j++)
       {
          tmpsum += arr[j];
           
          if (tmpsum <= n)
          {
              ans += freq[tmpsum];
              freq[tmpsum] = 0;
          }   
       }
    }
    return ans;
}
 
// Driver Code
public static void Main()
{
    int[] arr = { 1, 2, 3, 4, 5, 6, 7 };
    int n = arr.Length;
     
    Console.WriteLine(countElement(arr, n));
}
}
 
// This code is contributed by AbhiThakur

<script>
 
// Javascript implementation to count the
// elements such that their exist
// a subarray whose sum is equal to
// this element
 
// Function to count the elements
// such that their exist a subarray
// whose sum is equal to this element
function countElement(arr, n)
{
    let freq = Array.from({length: n+1}, (_, i) => 0);
    let ans = 0;
       
    // Loop to count the frequency
    for(let i = 0; i < n; i++)
    {
       freq[arr[i]]++;
    }
   
    // Loop to iterate over every possible
    // subarray of the array
    for(let i = 0; i < n - 1; i++)
    {
       let tmpsum = arr[i];
   
       for(let j = i + 1; j < n; j++)
       {
          tmpsum += arr[j];
   
          if (tmpsum <= n)
          {
              ans += freq[tmpsum];
              freq[tmpsum] = 0;
          }
       }
    }
       
    return ans;
}
 
// Driver Code
     
    let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
    let n = arr.length;
    document.write(countElement(arr, n));
            
</script>

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Complejidad de tiempo: O(N ² )  
Complejidad de espacio: O(N)