Number of Subarrays with positive product

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Given an array arr[] of N integers, the task is to find the count of subarrays with positive product.
Examples:

Input: arr[] = {-1, 2, -2}
Output: 2
Subarrays with positive product are {2} and {-1, 2, -2}.
Input: arr[] = {5, -4, -3, 2, -5}
Output: 7

Approach: The approach to find the subarrays with negative product has been discussed in this article. If cntNeg is the count of negative product subarrays and total is the count of all possible subarrays of the given array then the count of positive product subarrays will be cntPos = total – cntNeg.
Below is the implementation of the above approach:

C++

// 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

// 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

# 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#

// 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

Javascript

<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>

Output:

Time Complexity: O(n)

Auxiliary Space: O(1), since no extra space has been taken.

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