Merge pull request akgmage#1780 from akgmage/dev

add numSubarrayProductLessThanK in python

Author: akgmage

Sliding Window/subarray_product_less_than_k.py

@@ -0,0 +1,43 @@
+'''
+   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.
+
+  The provided code defines a Java class Solution with a method numSubarrayProductLessThanK that counts the number
+  of subarrays in an input array nums whose product is less than a given threshold k. It uses a sliding window
+  approach to efficiently compute this count.
+
+  Time Complexity:
+
+  The code iterates through the nums array once, using two pointers (startWindow and endWindow) to define the
+  sliding window. This results in a time complexity of O(N), where N is the length of the input array nums.
+  Space Complexity:
+
+  The code uses a constant amount of additional space to store integer variables (startWindow, product, and count).
+  Therefore, the space complexity is O(1), which means it is independent of the size of the input array.
+'''
+from typing import List
+
+class Solution:
+    def numSubarrayProductLessThanK(self, nums: List[int], k: int) -> int:
+        startWindow = 0  # The left end of the sliding window
+        product = 1     # Initialize product to 1 to accumulate the product
+        count = 0       # Count of subarrays with a product less than k
+
+        # Iterate through the list using a sliding window approach
+        for endWindow in range(len(nums)):
+            # Multiply the current element to the product
+            product *= nums[endWindow]
+
+            # Shrink the window by moving the start pointer as long as the product is greater than or equal to k
+            while startWindow <= endWindow and product >= k:
+                # Divide the product by the element at the start of the window
+                product /= nums[startWindow]
+                # Move the start of the window to the right
+                startWindow += 1
+
+            # Update the count with the number of valid subarrays within the current window
+            count += endWindow - startWindow + 1
+
+        # Return the count, which represents the number of subarrays with a product less than k
+        return count