The sliding window approach allows us to efficiently find subarrays where the product is less than k. By maintaining a window (two pointers), we can dynamically adjust the size of the current subarray based on the product of its elements. As the window size grows, the product is updated; if it exceeds k, we reduce the window from the left until the product is again less than k. This method minimizes redundant calculations.
Time Complexity: O(n) - Each element is processed at most twice.
Space Complexity: O(1) - Only a constant amount of extra space is used.
1var numSubarrayProductLessThanK = function(nums, k) {
2 if (k <= 1) return 0;
3 let prod = 1, count = 0, left = 0;
4 for (let right = 0; right < nums.length; right++) {
5 prod *= nums[right];
6 while (prod >= k) prod /= nums[left++];
7 count += right - left + 1;
8 }
9 return count;
10};The JavaScript solution follows the same logic as the previous approaches. Utilizing a sliding window technique, we manage and update a running product to find the number of valid subarrays efficiently. Adjusting the window minimizes unnecessary recalculations.
The brute force approach involves checking every possible subarray within nums and calculating their products. Each subarray is counted if its product is less than k. Though less efficient, it guarantees correctness by examining all potential subarrays.
Time Complexity: O(n^2) - Considers all possible subarrays.
Space Complexity: O(1) - Constant extra space used.
1var numSubarrayProductLessThanK_bruteforce = function(nums, k) {
2 if (k <= 1) return 0;
3 let count = 0;
4 for (let start = 0; start < nums.length; start++) {
5 let prod = 1;
6 for (let end = start; end < nums.length; end++) {
7 prod *= nums[end];
8 if (prod >= k) break;
9 count++;
10 }
11 }
12 return count;
13};The JavaScript brute force function involves checking each potential subarray product, stopping once those products are no longer below k. This complete search checks each element pairwise throughout the array.