713. Subarray Product Less Than K

Medium

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Given an array of integers nums and an integer k, return the number of contiguous subarrays where the product of all the elements in the subarray is strictly less than k.

Example 1:

Input: nums = [10,5,2,6], k = 100
Output: 8
Explanation: The 8 subarrays that have product less than 100 are:
[10], [5], [2], [6], [10, 5], [5, 2], [2, 6], [5, 2, 6]
Note that [10, 5, 2] is not included as the product of 100 is not strictly less than k.

Example 2:

Input: nums = [1,2,3], k = 0
Output: 0

Constraints:

• 1 <= nums.length <= 3 * 104

• 1 <= nums[i] <= 1000

• 0 <= k <= 106

class Solution:
    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
        product = 1
        start = 0
        index = 0
        count = 0
        while index < len(nums):
            product *= nums[index]
            print('Product', product)
            # If Product goes out of scope then move pointer from left until product is in range
            while product >= k and start < len(nums):
                product = (product//nums[start])
                start += 1
            # If Product is less than k then its elements will be less than K :D
            if product < k:
                count += (index - start + 1)
            index += 1
        return count
        
        
    def numSubarrayProductLessThanKOLD(self, nums: List[int], k: int) -> int:
        # Got TTL With this : O(n^2)
        count = 0
        for index in range(len(nums)):
            product = nums[index]
            if product < k:
                count += 1
            for j in range(index+1, len(nums)):
                product *= nums[j]
                if product < k:
                    count += 1
                else:
                    break
        return count