负积的子阵列数

给定N个整数的数组arr [] ，任务是查找具有负积的子数组的数量。

例子：

Input: arr[] = {-1, 2, -2}
Output: 4
Subarray with negative product are {-1}, {-2}, {-1, 2} and {2, -2}.

Input: arr[] = {5, -4, -3, 2, -5}
Output: 8

方法：

• 将正数组元素替换为1 ，将负数组元素替换为-1 。
• 创建一个前缀乘积数组pre [] ，其中pre [i]存储从索引arr [0]到arr [i]的所有元素的乘积。
• 现在，可以注意到，只有当pre [i] * pre [j]为负时，子阵列arr [i…j]才为负。
• 因此，具有负乘积的子数组的总计数将是前缀乘积数组中正和负元素计数的乘积。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
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

// 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

# 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#

// 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.

输出：

8