Maximum product subarray

Find the contiguous subarray within an array (containing at least one number) which has the largest product.

For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.

 

Note that, we need to consider two cases:
(1) negative numbers
          Store the minimum product to handle the case that new element < 0. Because if current element < 0, the product of two negative numbers (new element, and minimum product before the new element) become positive.
(2) zeros
        When meets zero, current max and min product become 0, new search is needed from the next element.

Therefore,  we can write down to function to store both + and - products:

max_product = max{A[i]*min_product (when A[i]<0),  A[i]*max_product (when A[i]>0),  A[i] (when 0 occurs before A[i]) }.

min_product = min{A[i]*min_product,  A[i]*max_product,  A[i] }.

 

Because current sequence might start from any element, to get the final result, we also need to store the max product after each iteration "res = max(res, maxp);".

 

class Solution {
public:
    int maxProduct(vector<int>& nums) {
        int currentMin = nums[0];
        int currentMax = nums[0];
        int result = nums[0];
        
        for(int i = 1; i < nums.size(); i++){
            int tempMax = currentMax;
            int tempMin = currentMin;
            currentMax = max(nums[i], max(tempMin * nums[i], tempMax * nums[i]));
            currentMin = min(nums[i], min(tempMin * nums[i], tempMax * nums[i]));
            
            result = max(result, currentMax);
        }
        return result;
    }
};