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Length of longest subarray with positive product
• Difficulty Level : Medium
• Last Updated : 08 Jun, 2021

Given an array arr[] consisting of N integers, the task is to print the length of the longest subarray with a positive product.

Examples:

Input: arr[] ={0, 1, -2, -3, -4}
Output: 3
Explanation:
The longest subarray with positive products is: {1, -2, -3}. Therefore, the required length is 3.

Input: arr[]={-1, -2, 0, 1, 2}
Output: 2
Explanation:
The longest subarray with positive products are: {-1, -2}, {1, 2}. Therefore, the required length is 2.

Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays and check if its product is positive or not. Among all such subarrays, print the length of the longest subarray obtained.
Time Complexity: (N3)
Auxiliary Space: O(1)
Efficient Approach: The problem can be solved using Dynamic Programming. The idea here is to maintain the count of positive elements and negative elements such that their product is positive. Follow the steps below to solve the problem:

1. Initialize the variable, say res, to store the length of the longest subarray with the positive product.
2. Initialize two variables, Pos and Neg, to store the length of the current subarray with the positive and negative products respectively.
3. Iterate over the array.
4. If arr[i] = 0: Reset the value of Pos and Neg.
5. If arr[i] > 0: Increment Pos by 1. If at least one element is present in the subarray with the negative product, then increment Neg by 1.
6. If arr[i] < 0: Swap Pos and Neg and increment the Neg by 1. If at least one element is present in the subarray with the positive product, then increment Pos also.
7. Update res=max(res, Pos).

## C++

 `// C++ program to implement``// the above approach` `#include ``using` `namespace` `std;` `// Function to find the length of``// longest subarray whose product``// is positive``int` `maxLenSub(``int` `arr[], ``int` `N)``{``    ``// Stores the length of current``    ``// subarray with positive product``    ``int` `Pos = 0;` `    ``// Stores the length of current``    ``// subarray with negative product``    ``int` `Neg = 0;` `    ``// Stores the length of the longest``    ``// subarray with positive product``    ``int` `res = 0;` `    ``for` `(``int` `i = 0; i < N; i++) {` `        ``if` `(arr[i] == 0) {` `            ``// Reset the value``            ``Pos = Neg = 0;``        ``}` `        ``// If current element is positive``        ``else` `if` `(arr[i] > 0) {` `            ``// Increment the length of``            ``// subarray with positive product``            ``Pos += 1;` `            ``// If at least one element is``            ``// present in the subarray with``            ``// negative product``            ``if` `(Neg != 0) {` `                ``Neg += 1;``            ``}` `            ``// Update res``            ``res = max(res, Pos);``        ``}` `        ``// If current element is negative``        ``else` `{` `            ``swap(Pos, Neg);` `            ``// Increment the length of subarray``            ``// with negative product``            ``Neg += 1;` `            ``// If at least one element is present``            ``// in the subarray with positive product``            ``if` `(Pos != 0) {` `                ``Pos += 1;``            ``}` `            ``// Update res``            ``res = max(res, Pos);``        ``}``    ``}``    ``return` `res;``}` `// Driver Code``int` `main()``{``    ``int` `arr[] = { -1, -2, -3, 0, 1 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);``    ``cout << maxLenSub(arr, N);``}`

## Python3

 `# Python3 program to implement``# the above approach` `# Function to find the length of``# longest subarray whose product``# is positive``def` `maxLenSub(arr, N):``    ` `    ``# Stores the length of current``    ``# subarray with positive product``    ``Pos ``=` `0` `    ``# Stores the length of current``    ``# subarray with negative product``    ``Neg ``=` `0` `    ``# Stores the length of the longest``    ``# subarray with positive product``    ``res ``=` `0` `    ``for` `i ``in` `range``(N):``        ``if` `(arr[i] ``=``=` `0``):` `            ``# Reset the value``            ``Pos ``=` `Neg ``=` `0` `        ``# If current element is positive``        ``elif` `(arr[i] > ``0``):` `            ``# Increment the length of``            ``# subarray with positive product``            ``Pos ``+``=` `1` `            ``# If at least one element is``            ``# present in the subarray with``            ``# negative product``            ``if` `(Neg !``=` `0``):``                ``Neg ``+``=` `1` `            ``# Update res``            ``res ``=` `max``(res, Pos)` `        ``# If current element is negative``        ``else``:``            ``Pos, Neg ``=` `Neg, Pos` `            ``# Increment the length of subarray``            ``# with negative product``            ``Neg ``+``=` `1` `            ``# If at least one element is present``            ``# in the subarray with positive product``            ``if` `(Pos !``=` `0``):``                ``Pos ``+``=` `1` `            ``# Update res``            ``res ``=` `max``(res, Pos)``            ` `    ``return` `res` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``arr ``=` `[ ``-``1``, ``-``2``, ``-``3``, ``0``, ``1` `]``    ``N ``=` `len``(arr)``    ` `    ``print``(maxLenSub(arr, N))` `# This code is contributed by mohit kumar 29`

## Java

 `// Java program to implement``// the above approach``import` `java.util.*;``class` `GFG{` `// Function to find the length of``// longest subarray whose product``// is positive``static` `int` `maxLenSub(``int` `arr[], ``int` `N)``{``    ``// Stores the length of current``    ``// subarray with positive product``    ``int` `Pos = ``0``;` `    ``// Stores the length of current``    ``// subarray with negative product``    ``int` `Neg = ``0``;` `    ``// Stores the length of the longest``    ``// subarray with positive product``    ``int` `res = ``0``;` `    ``for` `(``int` `i = ``0``; i < N; i++)``    ``{``        ``if` `(arr[i] == ``0``)``        ``{``            ``// Reset the value``            ``Pos = Neg = ``0``;``        ``}` `        ``// If current element is positive``        ``else` `if` `(arr[i] > ``0``)``        ``{``            ``// Increment the length of``            ``// subarray with positive product``            ``Pos += ``1``;` `            ``// If at least one element is``            ``// present in the subarray with``            ``// negative product``            ``if` `(Neg != ``0``)``            ``{``                ``Neg += ``1``;``            ``}` `            ``// Update res``            ``res = Math.max(res, Pos);``        ``}` `        ``// If current element is negative``        ``else``        ``{``            ``Pos = Pos + Neg;``            ``Neg = Pos - Neg;``            ``Neg = Pos - Neg;` `            ``// Increment the length of subarray``            ``// with negative product``            ``Neg += ``1``;` `            ``// If at least one element is present``            ``// in the subarray with positive product``            ``if` `(Pos != ``0``)``            ``{``                ``Pos += ``1``;``            ``}` `            ``// Update res``            ``res = Math.max(res, Pos);``        ``}``    ``}``    ``return` `res;``}` `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `arr[] = {-``1``, -``2``, -``3``, ``0``, ``1``};``    ``int` `N = arr.length;``    ``System.out.print(maxLenSub(arr, N));``}``}` `// This code is contributed by Rajput-Ji`

## C#

 `// C# program to implement``// the above approach``using` `System;``class` `GFG{` `// Function to find the length of``// longest subarray whose product``// is positive``static` `int` `maxLenSub(``int``[] arr, ``int` `N)``{``    ``// Stores the length of current``    ``// subarray with positive product``    ``int` `Pos = 0;` `    ``// Stores the length of current``    ``// subarray with negative product``    ``int` `Neg = 0;` `    ``// Stores the length of the longest``    ``// subarray with positive product``    ``int` `res = 0;` `    ``for` `(``int` `i = 0; i < N; i++)``    ``{``        ``if` `(arr[i] == 0)``        ``{``            ``// Reset the value``            ``Pos = Neg = 0;``        ``}` `        ``// If current element is positive``        ``else` `if` `(arr[i] > 0)``        ``{``            ``// Increment the length of``            ``// subarray with positive product``            ``Pos += 1;` `            ``// If at least one element is``            ``// present in the subarray with``            ``// negative product``            ``if` `(Neg != 0)``            ``{``                ``Neg += 1;``            ``}` `            ``// Update res``            ``res = Math.Max(res, Pos);``        ``}` `        ``// If current element is negative``        ``else``        ``{``            ``Pos = Pos + Neg;``            ``Neg = Pos - Neg;``            ``Neg = Pos - Neg;` `            ``// Increment the length of subarray``            ``// with negative product``            ``Neg += 1;` `            ``// If at least one element is present``            ``// in the subarray with positive product``            ``if` `(Pos != 0)``            ``{``                ``Pos += 1;``            ``}` `            ``// Update res``            ``res = Math.Max(res, Pos);``        ``}``    ``}``    ``return` `res;``}` `// Driver Code``public` `static` `void` `Main()``{``    ``int``[] arr = {-1, -2, -3, 0, 1};``    ``int` `N = arr.Length;``    ``Console.Write(maxLenSub(arr, N));``}``}` `// This code is contributed by Chitranayal`

## Javascript

 ``
Output:
`2`

Time Complexity: O(N)
Auxiliary Space: O(1)

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