Número de subarreglos con producto positivo

Dado un arreglo arr[] de N enteros, la tarea es encontrar el conteo de subarreglos con producto positivo.
Ejemplos: 
 

Entrada: arr[] = {-1, 2, -2} 
Salida: 2 
Subarreglos con producto positivo son {2} y {-1, 2, -2}.
Entrada: arr[] = {5, -4, -3, 2, -5} 
Salida: 7 
 

Enfoque: El enfoque para encontrar los subarreglos con producto negativo se ha discutido en este artículo. Si cntNeg es el conteo de subarreglos de productos negativos y total es el conteo de todos los subarreglos posibles del arreglo dado, entonces el conteo de subarreglos de productos positivos será cntPos = total – cntNeg .
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);
}
 
// Function to return the count of
// subarrays with positive product
int posProdSubArr(int arr[], int n)
{
 
    // Total subarrays possible
    int total = (n * (n + 1)) / 2;
 
    // Count to subarrays with negative product
    int cntNeg = negProdSubArr(arr, n);
 
    // Return the count of subarrays
    // with positive product
    return (total - cntNeg);
}
 
// Driver code
int main()
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << posProdSubArr(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);
}
 
// Function to return the count of
// subarrays with positive product
static int posProdSubArr(int arr[], int n)
{
 
    // Total subarrays possible
    int total = (n * (n + 1)) / 2;
 
    // Count to subarrays with negative product
    int cntNeg = negProdSubArr(arr, n);
 
    // Return the count of subarrays
    // with positive product
    return (total - cntNeg);
}
 
// Driver code
public static void main (String[] args)
{
    int arr[] = { 5, -4, -3, 2, -5 };
    int n = arr.length;
 
    System.out.println(posProdSubArr(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)
 
# Function to return the count of
# subarrays with positive product
def posProdSubArr(arr, n):
 
    total = (n * (n + 1)) / 2;
 
    # Count to subarrays with negative product
    cntNeg = negProdSubArr(arr, n);
 
    # Return the count of subarrays
    # with positive product
    return (total - cntNeg);
 
# Driver code
arr = [5, -4, -3, 2, -5]
n = len(arr)
print(posProdSubArr(arr, n))
 
# This code is contributed by Mehul Bhutalia

// 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);
}
 
// Function to return the count of
// subarrays with positive product
static int posProdSubArr(int []arr, int n)
{
 
    // Total subarrays possible
    int total = (n * (n + 1)) / 2;
 
    // Count to subarrays with negative product
    int cntNeg = negProdSubArr(arr, n);
 
    // Return the count of subarrays
    // with positive product
    return (total - cntNeg);
}
 
// Driver code
public static void Main (String[] args)
{
    int []arr = { 5, -4, -3, 2, -5 };
    int n = arr.Length;
 
    Console.WriteLine(posProdSubArr(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

<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);
}
 
// Function to return the count of
// subarrays with positive product
function posProdSubArr(arr, n)
{
 
    // Total subarrays possible
    let total = parseInt((n * (n + 1)) / 2);
 
    // Count to subarrays with negative product
    let cntNeg = negProdSubArr(arr, n);
 
    // Return the count of subarrays
    // with positive product
    return (total - cntNeg);
}
 
// Driver code
    let arr = [ 5, -4, -3, 2, -5 ];
    let n = arr.length;
 
    document.write(posProdSubArr(arr, n));
 
// This code is contributed by rishavmahato348.
</script>

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