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# Number of subsequences with positive product

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

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:

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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 ``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);` `    ``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

 ``
Output:
`7`

Time Complexity: O(n)

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