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To solve this problem, we iterate through the array and use two helper functions to find the number of subarrays with maximum elements less than or equal to `right` and strictly less than `left`. The result is the difference between these two values.
Time Complexity: O(n), where n is the length of the array because we pass through the array twice both in O(n) time.
Space Complexity: O(1) as we only use a fixed amount of extra space.
1def countSubarrays(nums, bound):
2 count = result = current = 0
3 for num in nums:
4 if num <= bound:
5 current += 1
6 else:
7 current = 0
8 result += current
9 return result
10
11
12def numSubarrayBoundedMax(nums, left, right):
13 return countSubarrays(nums, right) - countSubarrays(nums, left - 1)
14
15
16nums = [2, 1, 4, 3]
17left = 2
18right = 3
19print(numSubarrayBoundedMax(nums, left, right))Using Python, the solution is composed of countSubarrays function to determine subarray counts in bounds. This solution follows a common pattern across implementations with linear complexity.
This approach utilizes a sliding window to dynamically check and adjust the range within the array where subarrays satisfy the maximum constraints between `left` and `right`. We scan and adjust pointers to pinpoint valid ranges continuously.
Time Complexity: O(n), scanning and resolving limits in a single pass.
Space Complexity: O(1) maintaining concise storage need.
1function numSubarrayBoundedMax(nums, left, right) {
2
This JavaScript implementation executes a sliding-window-like mechanics through a loop that assesses alignments of elements within set bounds. This efficiently measures valid subarray formations by extending arrays.