In a product-based subarray problem, a negative number can flip the sign, turning a big minimum into a potential maximum. So, we track both the maximum and minimum products at each step. This helps handle negative numbers effectively.

Start by setting the answer, current max product, and current min product to the first number.
For each number in the array starting from the second:

If the number is negative, swap current max and min.
If the product of the current number and previous max is larger than the number itself, update current max to that product; otherwise, set it to the number.
Similarly, if the product of the current number and previous min is smaller, update current min; otherwise, set it to the number.
If the current max is greater than the answer so far, update the answer.

Return the answer at the end.

Code

#include <bits/stdc++.h>
using namespace std;

class Solution {
public:
    // This function returns the maximum product
    // of any contiguous subarray using optimized approach
    int maxProduct(vector<int>& nums) {
        // Initialize answer, max and min product as first element
        int res = nums[0];
        int maxProd = nums[0];
        int minProd = nums[0];

        // Traverse from second element
        for (int i = 1; i < nums.size(); i++) {
            // Store current number
            int curr = nums[i];

            // If current number is negative, swap max and min
            if (curr < 0) sw