added solution to subarray product less than k

Author: Ayon95

problems/sliding-window/medium/subarr-product-less-than-k/README.md

@@ -0,0 +1,61 @@
+# Subarray product less than k
+
+| #   | Difficulty | Tag(s)         | Link                                                                        |
+| --- | ---------- | -------------- | --------------------------------------------------------------------------- |
+| 21  | Medium     | Sliding window | [View problem](https://leetcode.com/problems/subarray-product-less-than-k/) |
+
+## Approaches
+
+- Brute force
+- Sliding window
+
+### Brute force
+
+- starting at each number of the array, analyze each subarray ending at the last number of the array
+- calculate the product for each subarray and check if it is less than k
+- if the product for a subarray is not less than k, then we can break out of the inner loop and move on to the next number in the array
+- O(n^2) time complexity
+- O(1) space complexity
+
+### Sliding window
+
+- we have to use a dynamically-resizing window since no fixed size is given for a valid subarray
+- keep expanding the window until the product of the numbers in the window becomes greater than or equal to k
+- then keep shrinking the window until the product becomes less than k again
+- if k is 0, then there won't be any valid subarray
+- O(n) time complexity
+- O(1) space complexity
+
+#### How many valid subarrays are there in a window?
+
+To calculate the number of valid subarrays in a window, it is important to realize two things:
+
+- if the product of numbers in a window is less than k, then the product of numbers in all shorter subarrays ending with the last number in the window will also be less than k
+- the total number of valid subarrays in a window is `endIndex - startIndex + 1`
+  - `endIndex - startIndex` gives the number of subarrays other than the subarray formed by the end number itself
+  - 1 is added to account for the subarray formed by the end number
+  - so the number of subarrays is equal to the size of the window
+
+```js
+/*
+- let's consider this array -> [1, 2, 3, 4], and k = 15
+- window [1], startIndex = 0, endIndex = 0
+    - the only valid subarray is formed by the number itself, [1]
+    - count = 0 - 0 + 1 = 1
+- window [1, 2], startIndex = 0, endIndex = 1
+    - there are 2 valid subarrays ending with 2 - [2] and [1, 2]
+    - count = 1 - 0 + 1 = 2
+- window [1, 2, 3], startIndex = 0, endIndex = 2
+    - there are 3 valid subarrays ending with 3 - [3], [2, 3], [1, 2, 3]
+    - count = 2 - 0 + 1 = 3
+- window [1, 2, 3, 4], startIndex = 0, endIndex = 3
+    - product of this window is 24 which is not less than 15
+    - so we have to shrink the window
+- window [2, 3, 4], startIndex = 1, endIndex = 3
+    - product is still not less than 15, so shrink the window again
+- window [3, 4], startIndex = 2, endIndex = 3
+    - there are 2 valid subarrays ending with 4 - [4] and [3, 4]
+    - count = 3 - 2 + 1 = 2
+- total number of valid subarrays = 1 + 2 + 3 + 2 = 8
+*/
+```

problems/sliding-window/medium/subarr-product-less-than-k/subarrProductLessThanK.js

@@ -0,0 +1,64 @@
+// Sliding window
+function numSubarrayProductLessThanK(nums, k) {
+	if (k <= 1) return 0;
+	if (nums.length === 1) return nums[0] < k ? 1 : 0;
+
+	let start = 0;
+	let count = 0;
+	let currentProduct = 1;
+
+	for (let i = 0; i < nums.length; i++) {
+		const num = nums[i];
+		currentProduct = currentProduct * num;
+
+		while (currentProduct >= k) {
+			// shrink the window
+			currentProduct = currentProduct / nums[start];
+			start++;
+		}
+
+		count += i - start + 1;
+	}
+
+	return count;
+}
+
+console.log(numSubarrayProductLessThanK([10, 5, 2, 6], 100));
+console.log(numSubarrayProductLessThanK([4, 12, 6, 7, 60], 60));
+console.log(numSubarrayProductLessThanK([4, 12, 60, 7, 6], 60));
+console.log(numSubarrayProductLessThanKV2([10, 2, 2, 5, 4, 4, 4, 3, 7, 7], 289));
+console.log(numSubarrayProductLessThanK([1, 2, 3], 0));
+console.log(numSubarrayProductLessThanK([1, 1, 1], 1));
+console.log(numSubarrayProductLessThanK([5], 10));
+console.log(numSubarrayProductLessThanK([4], 2));
+console.log(numSubarrayProductLessThanK([4, 2], 2));
+
+// Brute force
+function numSubarrayProductLessThanKV2(nums, k) {
+	if (k === 0) return 0;
+	if (nums.length === 1) return nums[0] < k ? 1 : 0;
+
+	let count = 0;
+
+	for (let i = 0; i < nums.length; i++) {
+		let product = 1;
+
+		for (let j = i; j < nums.length; j++) {
+			product = product * nums[j];
+			if (product < k) count++;
+			else break;
+		}
+	}
+
+	return count;
+}
+
+console.log('---------');
+console.log(numSubarrayProductLessThanKV2([10, 5, 2, 6], 100));
+console.log(numSubarrayProductLessThanKV2([4, 12, 6, 7, 60], 60));
+console.log(numSubarrayProductLessThanKV2([4, 12, 60, 7, 6], 60));
+console.log(numSubarrayProductLessThanKV2([10, 2, 2, 5, 4, 4, 4, 3, 7, 7], 289));
+console.log(numSubarrayProductLessThanKV2([1, 2, 3], 0));
+console.log(numSubarrayProductLessThanKV2([5], 10));
+console.log(numSubarrayProductLessThanKV2([4], 2));
+console.log(numSubarrayProductLessThanKV2([4, 2], 2));