You are given an integer array nums and two integers minK and maxK.
A fixed-bound subarray of nums is a subarray that satisfies the following conditions:
minK.maxK.Return the number of fixed-bound subarrays.
A subarray is a contiguous part of an array.
Example 1:
Input: nums = [1,3,5,2,7,5], minK = 1, maxK = 5 Output: 2 Explanation: The fixed-bound subarrays are [1,3,5] and [1,3,5,2].
Example 2:
Input: nums = [1,1,1,1], minK = 1, maxK = 1 Output: 10 Explanation: Every subarray of nums is a fixed-bound subarray. There are 10 possible subarrays.
Constraints:
2 <= nums.length <= 1051 <= nums[i], minK, maxK <= 106This approach uses a sliding window technique with two pointers to efficiently count the subarrays. As we iterate through the array, we maintain a range that tracks the valid subarrays containing both minK and maxK. When a valid subarray is found, we slide the window to find other subarrays that meet the criteria.
In this solution, we iterate over the array with a single loop. For each element, we determine if it's part of a valid subarray by checking if both minK and maxK have been encountered after the last invalid boundary. We maintain variables to track the most recent positions of minK and maxK, and a pointer to the last invalid position. The count of valid subarrays increases by the number of such subarrays ending at each valid position.
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Time Complexity: O(n), where n is the length of nums, as we iterate through the array once.
Space Complexity: O(1), as we are using a fixed amount of space.
This approach uses a brute force method to count fixed-bound subarrays by examining all possible subarrays in the provided array. For each subarray, check if the minimum and maximum elements match minK and maxK respectively.
This brute force implementation in C iterates over all possible subarrays of the given array and checks if each subarray has a minimum value of minK and a maximum value of maxK. For each valid subarray, it increases the count.
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Time Complexity: O(n2) due to the nested loops for subarray examination.
Space Complexity: O(1) as it uses a fixed number of variables.
| Approach | Complexity |
|---|---|
| Sliding Window with Two Pointers | Time Complexity: O(n), where n is the length of nums, as we iterate through the array once. |
| Brute Force | Time Complexity: O(n2) due to the nested loops for subarray examination. |
Count Subarrays With Fixed Bounds | Made Simple | Microsoft | Leetcode - 2444 | codestorywithMIK • codestorywithMIK • 16,736 views views
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