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.
1#include <stdio.h>
2
3int numSubarrayProductLessThanK(int* nums, int numsSize, int k) {
4 if (k <= 1) return 0;
5 int prod = 1, count = 0, left = 0;
6 for (int right = 0; right < numsSize; right++) {
7 prod *= nums[right];
8 while (prod >= k) prod /= nums[left++];
9 count += right - left + 1;
10 }
11 return count;
12}
The solution leverages a sliding window (or two-pointer) technique. The product of elements within the window is maintained, and pointers left
and right
are used to denote the current subarray. If the product exceeds k
, the left pointer is moved to shrink the window until the product is less than k
. The count of subarrays ending at each index is added to the total count.
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.
1def numSubarrayProductLessThanK_bruteforce(nums, k):
2 if k <= 1: return 0
3 count = 0
4 for start in range(len(nums)):
5 prod = 1
6 for end in range(start, len(nums)):
7 prod *= nums[end]
8 if prod >= k:
9 break
10 count += 1
11 return count
This Python function iterates through each start position and expands the subarray to examine all subarray products. If a product reaches k
, it breaks out of the loop to improve efficiency.