Número de subarreglos con producto negativo

Dada una array arr[] de N enteros, la tarea es encontrar el recuento de subarreglos con producto negativo.
Ejemplos: 
 

Entrada: array[] = {-1, 2, -2} 
Salida: 4 
subarreglo con producto negativo son {-1}, {-2}, {-1, 2} y {2, -2}.
Entrada: arr[] = {5, -4, -3, 2, -5} 
Salida: 8 
 

Acercarse: 
 

• Reemplace los elementos de array positivos con 1 y los elementos de array negativos con -1 .
• Cree una array de productos de prefijo pre[] donde pre[i] almacena el producto de todos los elementos desde el índice arr[0] hasta arr[i] .
• Ahora, se puede notar que el subarreglo arr[i…j] tiene un producto negativo solo si pre[i] * pre[j] es negativo.
• Por lo tanto, el conteo total de subarreglos con producto negativo será el producto del conteo de elementos positivos y negativos en el arreglo del producto de prefijos.

A continuación se muestra la implementación del enfoque anterior: 
 

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// subarrays with negative product
int negProdSubArr(int arr[], int n)
{
    int positive = 1, negative = 0;
    for (int i = 0; i < n; i++) {
 
        // Replace current element with 1
        // if it is positive else replace
        // it with -1 instead
        if (arr[i] > 0)
            arr[i] = 1;
        else
            arr[i] = -1;
 
        // Take product with previous element
        // to form the prefix product
        if (i > 0)
            arr[i] *= arr[i - 1];
 
        // Count positive and negative elements
        // in the prefix product array
        if (arr[i] == 1)
            positive++;
        else
            negative++;
    }
 
    // Return the required count of subarrays
    return (positive * negative);
}
 
// Driver code
int main()
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << negProdSubArr(arr, n);
 
    return (0);
}

// Java implementation of the approach
class GFG
{
     
    // Function to return the count of
    // subarrays with negative product
    static int negProdSubArr(int arr[], int n)
    {
        int positive = 1, negative = 0;
        for (int i = 0; i < n; i++)
        {
     
            // Replace current element with 1
            // if it is positive else replace
            // it with -1 instead
            if (arr[i] > 0)
                arr[i] = 1;
            else
                arr[i] = -1;
     
            // Take product with previous element
            // to form the prefix product
            if (i > 0)
                arr[i] *= arr[i - 1];
     
            // Count positive and negative elements
            // in the prefix product array
            if (arr[i] == 1)
                positive++;
            else
                negative++;
        }
     
        // Return the required count of subarrays
        return (positive * negative);
    }
     
    // Driver code
    public static void main (String[] args)
    {
        int arr[] = { 5, -4, -3, 2, -5 };
        int n = arr.length;
     
        System.out.println(negProdSubArr(arr, n));
    }
}
 
// This code is contributed by AnkitRai01

# Python3 implementation of the approach
 
# Function to return the count of
# subarrays with negative product
def negProdSubArr(arr, n):
    positive = 1
    negative = 0
    for i in range(n):
 
        # Replace current element with 1
        # if it is positive else replace
        # it with -1 instead
        if (arr[i] > 0):
            arr[i] = 1
        else:
            arr[i] = -1
 
        # Take product with previous element
        # to form the prefix product
        if (i > 0):
            arr[i] *= arr[i - 1]
 
        # Count positive and negative elements
        # in the prefix product array
        if (arr[i] == 1):
            positive += 1
        else:
            negative += 1
 
    # Return the required count of subarrays
    return (positive * negative)
 
# Driver code
arr = [5, -4, -3, 2, -5]
n = len(arr)
 
print(negProdSubArr(arr, n))
 
# This code is contributed by Mohit Kumar

// C# implementation of the approach
using System;
 
class GFG
{
         
    // Function to return the count of
    // subarrays with negative product
    static int negProdSubArr(int []arr, int n)
    {
        int positive = 1, negative = 0;
        for (int i = 0; i < n; i++)
        {
     
            // Replace current element with 1
            // if it is positive else replace
            // it with -1 instead
            if (arr[i] > 0)
                arr[i] = 1;
            else
                arr[i] = -1;
     
            // Take product with previous element
            // to form the prefix product
            if (i > 0)
                arr[i] *= arr[i - 1];
     
            // Count positive and negative elements
            // in the prefix product array
            if (arr[i] == 1)
                positive++;
            else
                negative++;
        }
     
        // Return the required count of subarrays
        return (positive * negative);
    }
     
    // Driver code
    static public void Main ()
    {
        int []arr = { 5, -4, -3, 2, -5 };
        int n = arr.Length;
     
        Console.Write(negProdSubArr(arr, n));
    }
}
 
// This code is contributed by Sachin.

<script>
 
// Javascript implementation of the approach
 
// Function to return the count of
// subarrays with negative product
function negProdSubArr(arr, n)
{
    let positive = 1, negative = 0;
    for (let i = 0; i < n; i++) {
 
        // Replace current element with 1
        // if it is positive else replace
        // it with -1 instead
        if (arr[i] > 0)
            arr[i] = 1;
        else
            arr[i] = -1;
 
        // Take product with previous element
        // to form the prefix product
        if (i > 0)
            arr[i] *= arr[i - 1];
 
        // Count positive and negative elements
        // in the prefix product array
        if (arr[i] == 1)
            positive++;
        else
            negative++;
    }
 
    // Return the required count of subarrays
    return (positive * negative);
}
 
// Driver code
    let arr = [ 5, -4, -3, 2, -5 ];
    let n = arr.length;
 
    document.write(negProdSubArr(arr, n));
 
</script>

8

Complejidad de tiempo: O(n)

Espacio Auxiliar: O(1)