Given an integer array nums and two integers left and right, return the number of contiguous non-empty subarrays such that the value of the maximum array element in that subarray is in the range [left, right].
The test cases are generated so that the answer will fit in a 32-bit integer.
Example 1:
Input: nums = [2,1,4,3], left = 2, right = 3 Output: 3 Explanation: There are three subarrays that meet the requirements: [2], [2, 1], [3].
Example 2:
Input: nums = [2,9,2,5,6], left = 2, right = 8 Output: 7
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1090 <= left <= right <= 109To 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.
The function countSubarrays counts the number of subarrays where the maximum element is less than or equal to a given bound. By computing this for both the 'right' and 'left-1' bounds and subtracting the two results, we can find the count where the maximum element lies between 'left' and 'right'.
C++
Java
Python
C#
JavaScript
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.
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.
The sliding window approach in Python for the specified problem uses a queue-like logic to maintain ranges that fall within the boundaries. Counter increments track the number of valid sequences built during iterations.
JavaScript
Time Complexity: O(n), scanning and resolving limits in a single pass.
Space Complexity: O(1) maintaining concise storage need.
| Approach | Complexity |
|---|---|
| Consider Maximum Allowed Value | Time Complexity: O(n), where n is the length of the array because we pass through the array twice both in O(n) time. |
| Sliding Window Technique | Time Complexity: O(n), scanning and resolving limits in a single pass. |
Maximum Subarray - Amazon Coding Interview Question - Leetcode 53 - Python • NeetCode • 605,117 views views
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