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# Number of subsequences with negative product

• Difficulty Level : Easy
• Last Updated : 19 Aug, 2022

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

Examples:

Input: arr[] = {1, -5, -6}
Output:
Explanation
{-5}, {-6}, {1, -5} and {1, -6} are the only possible subsequences

Input: arr[] = {2, 3, 1}
Output:
Explanation
There is no such possible subsequence with negative product

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

Efficient Approach:

• Count the number of positive and negative elements in the array
• An odd number of negative elements can be chosen for the subsequence to maintain the negative product. The number of different combinations of subsequences with an odd number of negative elements will be pow(2, count of negative elements – 1)
• Any number of positive elements can be chosen for the subsequence to maintain the negative product. The number of different combinations of subsequences with all the positive elements will be pow(2, count of positive elements)

## C++

 `// C++ implementation of the approach` `#include ` `using` `namespace` `std;`   `// Function to return the count of all` `// the subsequences with negative 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);` `    ``else` `        ``result = 0;`   `    ``return` `result;` `}`   `// Driver code` `int` `main()` `{` `    ``int` `arr[] = { 3, -4, -1, 6 };` `    ``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 negative 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``); ` `    ``else` `        ``result = ``0` `;`   `    ``return` `result; ` `} `   `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `arr[] = { ``3``,-``4``, -``1``, ``6` `}; ` `    ``int` `n = arr.length; `   `    ``System.out.print(cntSubSeq(arr, n)); ` `} ` `} `   `// This code is contributed by ANKITKUMAR34 `

## Python3

 `# Python 3 implementation of the approach ` `import` `math `   `# Function to return the count of all ` `# the subsequences with negative 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``)) ` `    ``else``:` `        ``result ``=` `0`   `    ``return` `result `   `# Driver code ` `arr ``=` `[ ``2``, ``-``3``, ``-``1``, ``4` `] ` `n ``=` `len` `(arr); `   `print` `(cntSubSeq(arr, n)) `

## C#

 `// C# implementation of the approach ` `using` `System;`   `class` `GFG ` `{ `   `// Function to return the count of all ` `// the subsequences with negative 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); ` `    ``else` `        ``result = 0 ;`   `    ``return` `result; ` `} `   `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `[]arr = { 3,-4, -1, 6 }; ` `    ``int` `n = arr.Length; `   `    ``Console.Write(cntSubSeq(arr, n)); ` `} ` `}`   `// This code is contributed by PrinciRaj1992`

## Javascript

 ``

Output:

`8`

Time Complexity: O(n)

Auxiliary Space: O(1)
Another Approach:
We can also count the number of subsequences with a negative product by subtracting total number of subsequences with positive subsequences from the total number of subsequences
To find the total number of subsequences with a positive product using the approach discussed in this article.

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