Number of subsequences with positive product

Given an array arr[] of N integers, the task is to find the count of all the subsequences of the array that have positive product.

Example:

Input: arr[] = {2, -3, -1}
Output: 3
{2}, {-3, -1} and {2, -3, -1} are the only possible subsequences.

Input: arr[] = {2, 3, -1, 4, 5}
Output: 15

Naive approach: Generate all the subsequences of the array and compute the product of all the subsequences. If the product is positive then increment the count by 1.



Efficient approach:

  1. Count the number of positive and negative elements in the array.
  2. Any number of positive elements can be chosen for the subsequence to maintain the positive product. The number of different combinations of subsequences with all the positive elements will be pow(2, count of positive elements).
  3. An even number of negative elements can be chosen for the subsequence to maintain the positive product. The number of different combinations of subsequences with an even number of negative elements will be pow(2, count of negative elements – 1).
  4. After that, remove 1 from the results for the empty subsequence.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the count of all
// the subsequences with positive product
int cntSubSeq(int arr[], int n)
{
    // To store the count of positive
    // elements in the array
    int pos_count = 0;
  
    // To store the count of negative
    // elements in the array
    int neg_count = 0;
  
    int result;
  
    for (int i = 0; i < n; i++) {
  
        // If the current element
        // is positive
        if (arr[i] > 0)
            pos_count++;
  
        // If the current element
        // is negative
        if (arr[i] < 0)
            neg_count++;
    }
  
    // For all the positive
    // elements of the array
    result = pow(2, pos_count);
  
    // For all the negative
    // elements of the array
    if (neg_count > 0)
        result *= pow(2, neg_count - 1);
  
    // For the empty subsequence
    result -= 1;
  
    return result;
}
  
// Driver code
int main()
{
    int arr[] = { 2, -3, -1, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << cntSubSeq(arr, n);
  
    return 0;
}

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Java

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// Java implementation of the approach
import java.util.*;
  
class GFG
{
  
// Function to return the count of all
// the subsequences with positive product
static int cntSubSeq(int arr[], int n)
{
    // To store the count of positive
    // elements in the array
    int pos_count = 0;
  
    // To store the count of negative
    // elements in the array
    int neg_count = 0;
  
    int result;
  
    for (int i = 0; i < n; i++)
    {
  
        // If the current element
        // is positive
        if (arr[i] > 0)
            pos_count++;
  
        // If the current element
        // is negative
        if (arr[i] < 0)
            neg_count++;
    }
  
    // For all the positive
    // elements of the array
    result = (int) Math.pow(2, pos_count);
  
    // For all the negative
    // elements of the array
    if (neg_count > 0)
        result *= Math.pow(2, neg_count - 1);
  
    // For the empty subsequence
    result -= 1;
  
    return result;
}
  
// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, -3, -1, 4 };
    int n = arr.length;
  
    System.out.print(cntSubSeq(arr, n));
  
}
}
  
// This code is contributed by 29AjayKumar

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Python3

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# Python 3 implementation of the approach 
import math
  
# Function to return the count of all 
# the subsequences with positive product 
def cntSubSeq(arr, n):
  
    # To store the count of positive 
    # elements in the array 
    pos_count = 0
  
    # To store the count of negative 
    # elements in the array 
    neg_count = 0
  
    for i in range (n):
  
        # If the current element 
        # is positive 
        if (arr[i] > 0) :
            pos_count += 1
  
        # If the current element 
        # is negative 
        if (arr[i] < 0): 
            neg_count += 1
  
    # For all the positive 
    # elements of the array 
    result = int(math.pow(2, pos_count)) 
  
    # For all the negative 
    # elements of the array 
    if (neg_count > 0):
        result *= int(math.pow(2, neg_count - 1)) 
  
    # For the empty subsequence 
    result -= 1
  
    return result
  
# Driver code 
arr = [ 2, -3, -1, 4
n = len (arr); 
  
print (cntSubSeq(arr, n)) 
  
# This code is contributed by ANKITKUMAR34

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
  
    // Function to return the count of all
    // the subsequences with positive product
    static int cntSubSeq(int []arr, int n)
    {
        // To store the count of positive
        // elements in the array
        int pos_count = 0;
      
        // To store the count of negative
        // elements in the array
        int neg_count = 0;
      
        int result;
      
        for (int i = 0; i < n; i++)
        {
      
            // If the current element
            // is positive
            if (arr[i] > 0)
                pos_count++;
      
            // If the current element
            // is negative
            if (arr[i] < 0)
                neg_count++;
        }
      
        // For all the positive
        // elements of the array
        result = (int) Math.Pow(2, pos_count);
      
        // For all the negative
        // elements of the array
        if (neg_count > 0)
            result *= (int)Math.Pow(2, neg_count - 1);
      
        // For the empty subsequence
        result -= 1;
      
        return result;
    }
      
    // Driver code
    public static void Main()
    {
        int []arr = { 2, -3, -1, 4 };
        int n = arr.Length;
      
        Console.Write(cntSubSeq(arr, n));
      
    }
}
  
// This code is contributed by AnkitRai01

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Output:

7

Time Complexity: O(n)

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