subarray-product-less-than-k.py

# Time:  O(n)
# Space: O(1)

# Your are given an array of positive integers nums.
#
# Count and print the number of (contiguous) subarrays where the product of all the elements
# in the subarray is 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.
#
# Note:
# 0 < nums.length <= 50000.
# 0 < nums[i] < 1000.
# 0 <= k < 10^6.

# Sliding window solution.
class Solution(object):
    def numSubarrayProductLessThanK(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        if k <= 1: return 0
        result, start, prod = 0, 0, 1
        for i, num in enumerate(nums):
            prod *= num
            while prod >= k:
                prod /= nums[start]
                start += 1
            result += i-start+1
        return result