# 2444 Count Subarrays With Fixed Bounds

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:

• The minimum value in the subarray is equal to `minK`.
• The maximum value in the subarray is equal to `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 <= 105`
• `1 <= nums[i], minK, maxK <= 106`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 `````` ``````class Solution: def countSubarrays(self, nums: List[int], minK: int, maxK: int) -> int: # min_position, max_position: the MOST RECENT positions of minK and maxK. # left_bound: the MOST RECENT value outside the range [minK, maxK]. answer = 0 min_position = max_position = left_bound = -1 # Iterate over nums, for each number at index i: for i, number in enumerate(nums): # If the number is outside the range [minK, maxK], update the most recent left_bound. if number < minK or number > maxK: left_bound = i # If the number is minK or maxK, update the most recent position. if number == minK: min_position = i if number == maxK: max_position = i # The number of valid subarrays equals the number of elements between left_bound and # the smaller of the two most recent positions. answer += max(0, min(min_position, max_position) - left_bound) return answer``````