Maximum Product Subarray LT152

Given an integer array nums, find the contiguous subarray within an array (containing at least one number) which has the largest product.

Example 1:

Input: [2,3,-2,4]
Output: 6
Explanation: [2,3] has the largest product 6.

Example 2:

Input: [-2,0,-1]
Output: 0
Explanation: The result cannot be 2, because [-2,-1] is not a subarray.

Idea 1. How can we extend a solution from nums[0..i] to nums[0..i+1]? what shall we do for a newly added element nums[i+1]. Similar to Maximum Subarray LT53, assume we know the largest product for array nums[0..i],

    if nums[i+1] >= 0, maxProduct(i+1) = Math.max(maxProduct(i) * nums[i+1], nums[i+1])

    if nums[i+1] < 0, to get the max product ending at (i+1), we also need to know the min element ending at i which could be negative, maxProduct(i+1) = Math.max(minProduct(i) * nums[i+1], nums[i+1]).

    For any element at i+1, maxProduct(i+1) = Math.max( nums[i+1], minProduct(i) * nums[i+1], maxProduct(i) * nums[i+1] )

Time complexity: O(n)

Space complexity: O(1)

class Solution {
    public int maxProduct(int[] nums) {
        int minHere = 1, maxHere = 1, result = Integer.MIN_VALUE;
        
        for(int num: nums) {     
            int res1 = minHere * num;
            int res2 = maxHere * num;
            if(res1 < res2) {
                minHere = Math.min(res1, num);
                maxHere = Math.max(res2, num);
            }
            else {
                minHere = Math.min(res2, num);
                maxHere = Math.max(res1, num);
            }
            result = Math.max(result, maxHere);
        }
        return result;
    }
}

class Solution {
    public int maxProduct(int[] nums) {
        int minHere = 1, maxHere = 1, result = Integer.MIN_VALUE;
        
        for(int num: nums) {     
            int res1 = minHere * num;
            int res2 = maxHere * num;
            
            minHere = Math.min(Math.min(res1, res2), num);
            maxHere = Math.max(Math.max(res1, res2), num);
            
            result = Math.max(result, maxHere);
        }
        return result;
    }
}