subarray product less than K medium

DP/SubarrayProductLessThanK.py

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+"""
+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.
+
+思路 Dp，处理下 1 即可。 不考虑 0，nums[i] 不会为 0。
+
+beat 19%
+
+测试地址:
+https://leetcode.com/problems/subarray-product-less-than-k/description/
+
+可剪枝优化。
+"""
+class Solution(object):
+    def numSubarrayProductLessThanK(self, nums, k):
+        """
+        :type nums: List[int]
+        :type k: int
+        :rtype: int
+        """
+
+        dp = []
+        result = 0
+        start = 0
+        for i in range(len(nums)):
+            if nums[i] < k:
+                result += 1
+                dp = [nums[i]]
+                start = i
+                break
+
+        for i in range(start+1, len(nums)):
+            if nums[i] == 1 and nums[i] < k:
+                dp.append(1)
+                result += len(dp)
+                continue
+
+            new = []
+
+            if nums[i] < k:
+                result += 1
+                new.append(nums[i])
+
+
+            for j in dp:
+                if j * nums[i] < k:
+                    result += 1
+                    new.append(j * nums[i])
+
+            dp = new
+        return result
+