add numSubarrayProductLessThanK in cpp

Author: akgmage

Sliding Window/subarray_product_less_than_k.cpp

@@ -0,0 +1,49 @@
+/**
+     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.
+ */
+
+#include <vector>
+
+class Solution {
+public:
+    int numSubarrayProductLessThanK(std::vector<int>& nums, int k) {
+        int startWindow = 0; // The left end of the sliding window
+        int product = 1;    // Initialize product to 1 to accumulate the product
+        int count = 0;      // Count of subarrays with a product less than k
+
+        // Iterate through the array using a sliding window approach
+        for (int endWindow = 0; endWindow < nums.size(); endWindow++) {
+            // 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 && 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++;
+            }
+
+            // 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;
+    }
+};