Length of longest subarray with negative product

Given an array arr[] of N elements. The task is to find the length of the longest subarray such that the product of the subarray is negative. If there is no such subarray available, print -1.
Examples:

Input: N = 6, arr[] = {-1, 2, 3, 2, 1, -4}
Output:
Explanation:
In the example the subarray
in range [1, 5] has product -12 which is negative,
so the length is 5.
Input: N = 4, arr[] = {1, 2, 3, 2}
Output: -1

Approach:

• First, check if the total product of the array is negative. If the total product of the array is negative then the answer will be N.
• If the total product of the array is not negative, means it is positive. So, the idea is to find a negative element from the array such that excluding that element and comparing the length of both parts of the array we can obtain the max length of the subarray with the negative product.
• It is obvious that the-

subarray with negative product will exist in the range [1, x) or (x, N], where 1 <= x <= N, and arr[x] is negative.

Below is the implementation of the above approach-

 `// C++ implementation of the above approach` `#include ` `using` `namespace` `std;`   `// Function to find length of the` `// longest subarray such that product` `// of the subarray is negative` `int` `maxLength(``int` `a[], ``int` `n)` `{` `    ``int` `product = 1, len = -1;`   `    ``// Check if product of complete` `    ``// array is negative` `    ``for` `(``int` `i = 0; i < n; i++)` `        ``product *= a[i];`   `    ``// Total product is already` `    ``// negative` `    ``if` `(product < 0)` `        ``return` `n;`   `    ``// Find an index i such the a[i]` `    ``// is negative and compare length` `    ``// of both halfs excluding a[i] to` `    ``// find max length subarray` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(a[i] < 0)` `            ``len = max(len,` `                      ``max(n - i - 1, i));` `    ``}`   `    ``return` `len;` `}`   `// Driver Code` `int` `main()` `{` `    ``int` `arr[] = { 1, 2, -3, 2, 5, -6 };` `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);`   `    ``cout << maxLength(arr, N)` `         ``<< ``"\n"``;`   `    ``int` `arr1[] = { 1, 2, 3, 4 };` `    ``N = ``sizeof``(arr1) / ``sizeof``(arr1[0]);`   `    ``cout << maxLength(arr1, N)` `         ``<< ``"\n"``;`   `    ``return` `0;` `}`

 `// Java implementation of the above approach ` `import` `java.util.Arrays;`   `class` `GFG{ ` `    `  `// Function to find length of the ` `// longest subarray such that product ` `// of the subarray is negative ` `static` `int` `maxLength(``int` `a[], ``int` `n) ` `{ ` `    ``int` `product = ``1``, len = -``1``; `   `    ``// Check if product of complete ` `    ``// array is negative ` `    ``for``(``int` `i = ``0``; i < n; i++) ` `        ``product *= a[i]; `   `    ``// Total product is already ` `    ``// negative ` `    ``if` `(product < ``0``) ` `        ``return` `n; `   `    ``// Find an index i such the a[i] ` `    ``// is negative and compare length ` `    ``// of both halfs excluding a[i] to ` `    ``// find max length subarray ` `    ``for``(``int` `i = ``0``; i < n; i++)` `    ``{ ` `        ``if` `(a[i] < ``0``) ` `            ``len = Math.max(len, ` `                  ``Math.max(n - i - ``1``, i)); ` `    ``} ` `    ``return` `len; ` `} ` `    `  `// Driver code ` `public` `static` `void` `main (String[] args) ` `{ ` `    `  `    ``// Given array arr[]` `    ``int` `arr[] = ``new` `int``[]{ ``1``, ``2``, -``3``,` `                           ``2``, ``5``, -``6` `};` `    ``int` `N = arr.length;` `    `  `    ``System.out.println(maxLength(arr, N));` `    `  `    ``// Given array arr[]` `    ``int` `arr1[] = ``new` `int``[]{ ``1``, ``2``, ``3``, ``4` `};` `    ``N = arr1.length;`   `    ``System.out.println(maxLength(arr1, N));` `} ` `} `   `// This code is contributed by Pratima Pandey `

 `# Python3 implementation of the above approach` `# Function to find length of the` `# longest subarray such that product` `# of the subarray is negative` `def` `maxLength(a, n):` `    ``product ``=` `1` `    ``length ``=` `-``1`   `    ``# Check if product of complete` `    ``# array is negative` `    ``for` `i ``in` `range` `(n):` `        ``product ``*``=` `a[i]`   `    ``# Total product is already` `    ``# negative` `    ``if` `(product < ``0``):` `        ``return` `n`   `    ``# Find an index i such the a[i]` `    ``# is negative and compare length` `    ``# of both halfs excluding a[i] to` `    ``# find max length subarray` `    ``for` `i ``in` `range` `(n):` `        ``if` `(a[i] < ``0``):` `            ``length ``=` `max``(length,` `                         ``max``(n ``-` `i ``-` `1``, i))`   `    ``return` `length`   `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:  ` `    ``arr ``=` `[``1``, ``2``, ``-``3``, ``2``, ``5``, ``-``6``]` `    ``N ``=` `len``(arr)` `    ``print` `(maxLength(arr, N))` `    ``arr1 ``=` `[``1``, ``2``, ``3``, ``4``]` `    ``N ``=` `len``(arr1)` `    ``print` `(maxLength(arr1, N))` `         `  `# This code is contributed by Chitranayal`

 `// C# implementation of the above approach ` `using` `System;`   `class` `GFG{ ` `    `  `// Function to find length of the ` `// longest subarray such that product ` `// of the subarray is negative ` `static` `int` `maxLength(``int` `[]a, ``int` `n) ` `{ ` `    ``int` `product = 1, len = -1; `   `    ``// Check if product of complete ` `    ``// array is negative ` `    ``for``(``int` `i = 0; i < n; i++) ` `        ``product *= a[i]; `   `    ``// Total product is already ` `    ``// negative ` `    ``if` `(product < 0) ` `        ``return` `n; `   `    ``// Find an index i such the a[i] ` `    ``// is negative and compare length ` `    ``// of both halfs excluding a[i] to ` `    ``// find max length subarray ` `    ``for``(``int` `i = 0; i < n; i++)` `    ``{ ` `        ``if` `(a[i] < 0) ` `            ``len = Math.Max(len, ` `                  ``Math.Max(n - i - 1, i)); ` `    ``} ` `    ``return` `len; ` `} ` `    `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    `  `    ``// Given array []arr` `    ``int` `[]arr = ``new` `int``[]{ 1, 2, -3,` `                           ``2, 5, -6 };` `    ``int` `N = arr.Length;` `    `  `    ``Console.WriteLine(maxLength(arr, N));` `    `  `    ``// Given array []arr` `    ``int` `[]arr1 = ``new` `int``[]{ 1, 2, 3, 4 };` `    ``N = arr1.Length;`   `    ``Console.WriteLine(maxLength(arr1, N));` `} ` `} `   `// This code is contributed by Amit Katiyar`

Output:
```5
-1

```

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

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