Given an array **arr[]** of **N** integers, the task is to find the length of the longest subarray whose product is greater than or equals to 0.

**Examples:**

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

Output:6

Explanation:

The longest subarray with product ≥ 0 = {1, 1, -2, 3, 2, -1} and {-1, 1, 1, -2, 3, 2}.

Length of each = 6.

Input:arr[] = {-1, -2, -3, -4}

Output:4

Explanation:

The longest subarray with product ≥ 0 = {-1, -2, -3, -4}.

Length = 4.

**Approach:**

- Check whether the product of all the elements in the given array is greater than or equals zero or not.
- If Yes then, the length of the longest subarray with a product greater than or equals to zero is the
**length of the array**. - If the above statement is not true, then the array contains an odd number of negative elements. In this case, to find the longest subarray do the following:
- For each negative element occurs in the array, the subarray to left and right of the current element gives the product which is greater than or equals to 0. Therefore the length of required longest subarray will be:
L = max(L, max(i, N - i - 1))

- Keep updating the length of the subarray for each negative element found in the array.
- The value of
**L**is the length of longest subarray with product greater than equals to 0.

- For each negative element occurs in the array, the subarray to left and right of the current element gives the product which is greater than or equals to 0. Therefore the length of required longest subarray will be:

Below is the implementation of the above approach:

## C++

`// C++ implementation of the above approach ` ` ` `#include <bits/stdc++.h> ` `using` `namespace` `std; ` ` ` `// Function that count the length ` `// of longest subarray with product ` `// greater than or equals to zero ` `int` `maxLength(` `int` `arr[], ` `int` `N) ` `{ ` ` ` `int` `product = 1, len = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` `product *= arr[i]; ` ` ` `} ` ` ` ` ` `// If product is greater than ` ` ` `// zero, return array size ` ` ` `if` `(product >= 0) { ` ` ` `return` `N; ` ` ` `} ` ` ` ` ` `// Traverse the array and if ` ` ` `// any negative element found ` ` ` `// then update the length of ` ` ` `// longest subarray with the ` ` ` `// length of left and right subarray ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` `if` `(arr[i] < 0) { ` ` ` `len = max(len, ` ` ` `max(N - i - 1, i)); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `len; ` `} ` ` ` `// Driver Code ` `int` `main() ` `{ ` ` ` `int` `arr[] = { -1, 1, 1, -2, 3, 2, -1 }; ` ` ` `int` `N = ` `sizeof` `(arr) / ` `sizeof` `(arr[0]); ` ` ` ` ` `cout << maxLength(arr, N) << endl; ` ` ` ` ` `return` `0; ` `} ` |

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## Java

`// Java implementation of the above approach ` `import` `java.util.*; ` ` ` `public` `class` `GFG{ ` `// Function that count the length ` `// of longest subarray with product ` `// greater than or equals to zero ` ` ` `static` `int` `maxLength(` `int` `arr[], ` `int` `N) ` ` ` `{ ` ` ` `int` `product = ` `1` `, len = ` `0` `; ` ` ` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) { ` ` ` `product *= arr[i]; ` ` ` `} ` ` ` ` ` `// If product is greater than ` ` ` `// zero, return array size ` ` ` `if` `(product >= ` `0` `) { ` ` ` `return` `N; ` ` ` `} ` ` ` ` ` `// Traverse the array and if ` ` ` `// any negative element found ` ` ` `// then update the length of ` ` ` `// longest subarray with the ` ` ` `// length of left and right subarray ` ` ` `for` `(` `int` `i = ` `0` `; i < N; i++) { ` ` ` `if` `(arr[i] < ` `0` `) { ` ` ` `len = Math.max(len, Math.max(N - i - ` `1` `, i)); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `len; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `main(String args[]) ` ` ` `{ ` ` ` `int` `arr[] = { -` `1` `, ` `1` `, ` `1` `, -` `2` `, ` `3` `, ` `2` `, -` `1` `}; ` ` ` `int` `N = arr.length; ` ` ` `System.out.println(maxLength(arr, N)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by AbhiThakur ` |

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## Python3

`# Python3 implementation of the above approach ` ` ` `# Function that count the Length ` `# of longest subarray with product ` `# greater than or equals to zero ` `def` `maxLength(arr, N): ` ` ` `product ` `=` `1` ` ` `Len` `=` `0` ` ` ` ` `for` `i ` `in` `arr: ` ` ` `product ` `*` `=` `i ` ` ` ` ` `# If product is greater than ` ` ` `# zero, return array size ` ` ` `if` `(product >` `=` `0` `): ` ` ` `return` `N ` ` ` ` ` `# Traverse the array and if ` ` ` `# any negative element found ` ` ` `# then update the Length of ` ` ` `# longest subarray with the ` ` ` `# Length of left and right subarray ` ` ` `for` `i ` `in` `range` `(N): ` ` ` `if` `(arr[i] < ` `0` `): ` ` ` `Len` `=` `max` `(` `Len` `,` `max` `(N ` `-` `i ` `-` `1` `, i)) ` ` ` ` ` `return` `Len` ` ` `# Driver Code ` `if` `__name__ ` `=` `=` `'__main__'` `: ` ` ` `arr ` `=` `[` `-` `1` `, ` `1` `, ` `1` `, ` `-` `2` `, ` `3` `, ` `2` `, ` `-` `1` `] ` ` ` `N ` `=` `len` `(arr) ` ` ` ` ` `print` `(maxLength(arr, N)) ` ` ` `# This code is contributed by mohit kumar 29 ` |

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## C#

`// C# implementation of the above approach ` `using` `System; ` ` ` `class` `GFG{ ` `// Function that count the length ` `// of longest subarray with product ` `// greater than or equals to zero ` ` ` `static` `int` `maxLength(` `int` `[]arr, ` `int` `N) ` ` ` `{ ` ` ` `int` `product = 1, len = 0; ` ` ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` `product *= arr[i]; ` ` ` `} ` ` ` ` ` `// If product is greater than ` ` ` `// zero, return array size ` ` ` `if` `(product >= 0) { ` ` ` `return` `N; ` ` ` `} ` ` ` ` ` `// Traverse the array and if ` ` ` `// any negative element found ` ` ` `// then update the length of ` ` ` `// longest subarray with the ` ` ` `// length of left and right subarray ` ` ` `for` `(` `int` `i = 0; i < N; i++) { ` ` ` `if` `(arr[i] < 0) { ` ` ` `len = Math.Max(len, Math.Max(N - i - 1, i)); ` ` ` `} ` ` ` `} ` ` ` ` ` `return` `len; ` ` ` `} ` ` ` ` ` `// Driver Code ` ` ` `public` `static` `void` `Main() ` ` ` `{ ` ` ` `int` `[]arr = { -1, 1, 1, -2, 3, 2, -1 }; ` ` ` `int` `N = arr.Length; ` ` ` `Console.WriteLine(maxLength(arr, N)); ` ` ` ` ` `} ` `} ` ` ` `// This code is contributed by abhaysingh290895 ` |

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**Output:**

6

* Time Complexity: O(N)*, where N is the length of the array.

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