Subarray Product Less Than K

Author: Shikha-code36

README.md

@@ -37,4 +37,5 @@ Check the notes for the explaination - [Notes](https://stingy-shallot-4ea.notion
     - [x] [Squares of a Sorted Array](Two-Pointers/977-Squares-of-a-Sorted-Array.py)
     - [x] [Backspace String Compare](Two-Pointers/844-Backspace-String-Compare.py)
     - [x] [Find the Duplicate Number](Two-Pointers/287-Find-the-Duplicate-Number.py)
+    - [x] [Subarray Product Less Than K](Two-Pointers/713-Subarray-Product-Less-Than-K.py)
 

Two-Pointers/713-Subarray-Product-Less-Than-K.py

@@ -0,0 +1,49 @@
+"Leetcode- https://leetcode.com/problems/subarray-product-less-than-k/ "
+'''
+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.
+'''
+
+# Solution-1 TLE
+def numSubarrayProductLessThanK(self, nums, k):
+    count = 0
+    ans = 1
+    for i in range(0, len(nums)):
+        if nums[i] < k:
+            count += 1
+        ans = nums[i]
+        for j in range(i+1, len(nums)):
+            ans = ans*nums[j]
+
+            if ans < k:
+                count += 1
+            else:
+                break
+    return count
+
+# T:O(N^2)
+# S:O(1)
+
+# Solution-2
+def numSubarrayProductLessThanK(self, nums, k):
+    if k <= 1:
+        return 0
+    prod = 1
+    cur = left = 0
+    for right, val in enumerate(nums):
+        prod *= val
+        while prod >= k:
+            prod /= nums[left]
+            left += 1
+        cur += right - left + 1
+    return cur
+
+# T:O(N)
+# S:O(1)