Skip to content
Related Articles

Related Articles

Improve Article

Number of subsequences with positive product

  • Difficulty Level : Hard
  • Last Updated : 02 Jun, 2021
Geek Week

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

Example: 

Input: arr[] = {2, -3, -1} 
Output:
{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++




// 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;
}

Java




// 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

Python3




# 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

C#




// 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

Javascript




<script>
// Javascript implementation of the approach
 
// Function to return the count of all
// the subsequences with positive product
function cntSubSeq(arr, n) {
    // To store the count of positive
    // elements in the array
    let pos_count = 0;
 
    // To store the count of negative
    // elements in the array
    let neg_count = 0;
 
    let result;
 
    for (let 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 = 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
 
let arr = [2, -3, -1, 4];
let n = arr.length;
 
document.write(cntSubSeq(arr, n));
</script>
Output: 
7

 

Time Complexity: O(n)
 

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :