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 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++

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// C++ implementation of the above approach
#include <bits/stdc++.h>
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;
}

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Java

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// 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 

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C#

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// 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

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Output: 

5
-1

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

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