Leetcode-Problems/152. Maximum Product Subarray at master · alexkoby/Leetcode-Problems · GitHub

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class Solution {
    public int maxProduct(int[] nums) {
        int numberConsecutiveNegatives = 0;
        boolean repeat = false;
        int globalMax = nums[0];
        int interMaxProduct = 1;
        if(nums.length == 1){
            return nums[0];
        }
        int negativeNumber = 0;
        for(int i = 0; i < nums.length; i++){
            //Multiply by the next number
            interMaxProduct *= nums[i];

            //If bigger than globalMax, set globalMax = interMax
            if(interMaxProduct > globalMax){
                globalMax = interMaxProduct;
            }

            //If the number you multiplied by is 0, even if next number is negative, the subarray will have a 0 in it --> sum = 0
            //Want to start over with the next value so you don't continuously multiply by 0
            if(nums[i] == 0){
                negativeNumber = 0;
                interMaxProduct = 1;
                if(numberConsecutiveNegatives % 2 == 1 && numberConsecutiveNegatives > 2){
                    repeat = true;
                }
                numberConsecutiveNegatives = 0;
            }

            //If you just multiplied by a negative number, see if there is a substring that's attachable that was negative (- * - = +)
            //Else, keep track of this negative substring and start over and see if you will hit another negative
            if(interMaxProduct < 0){
                numberConsecutiveNegatives++;
                if(negativeNumber != 0){
                    interMaxProduct *= negativeNumber;
                    negativeNumber = 0;

                    if(interMaxProduct > globalMax){
                        globalMax = interMaxProduct;
                    }

                }
                else{
                    negativeNumber = interMaxProduct;
                    interMaxProduct = 1;
                }

            }
        }
        if(numberConsecutiveNegatives % 2 == 1 && numberConsecutiveNegatives > 2){
            repeat = true;
        }

        if(repeat){
            //copy over numbers in reverse
            int [] newArray = new int[nums.length];
            for(int i = 0, length = nums.length; i < length; i++){
                newArray[i] = nums[length - i - 1];
            }
            return Math.max(globalMax, maxProductHelper(newArray));
        }
        else{
            return globalMax;
        }
    }

    public int maxProductHelper(int [] nums){
        int globalMax = nums[0];
        int interMaxProduct = 1;
        if(nums.length == 1){
            return nums[0];
        }
        int negativeNumber = 0;
        for(int i = 0; i < nums.length; i++){
            //Multiply by the next number
            interMaxProduct *= nums[i];

            //If bigger than globalMax, set globalMax = interMax
            if(interMaxProduct > globalMax){
                globalMax = interMaxProduct;
            }

            //If the number you multiplied by is 0, even if next number is negative, the subarray will have a 0 in it --> sum = 0
            //Want to start over with the next value so you don't continuously multiply by 0
            if(nums[i] == 0){
                negativeNumber = 0;
                interMaxProduct = 1;
            }

            //If you just multiplied by a negative number, see if there is a substring that's attachable that was negative (- * - = +)
            //Else, keep track of this negative substring and start over and see if you will hit another negative
            if(interMaxProduct < 0){
                if(negativeNumber != 0){
                    interMaxProduct *= negativeNumber;
                    negativeNumber = 0;

                    if(interMaxProduct > globalMax){
                        globalMax = interMaxProduct;
                    }
                }
                else{
                    negativeNumber = interMaxProduct;
                    interMaxProduct = 1;
                }
            }
        }
        return globalMax;
    }
}