How do you solve Count Subarrays With Fixed Bounds on LeetCode?

Count Subarrays With Fixed Bounds (LeetCode #2444) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The problem can be efficiently solved using the sliding window technique by tracking the last seen indices of minK and maxK.

What is the brute force solution for Count Subarrays With Fixed Bounds?

The brute force approach for Count Subarrays With Fixed Bounds is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking every possible subarray to see if it meets the criteria of having minK as the minimum and maxK as the maximum. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Subarrays With Fixed Bounds?

Count Subarrays With Fixed Bounds uses the following concepts: Array, Queue, Sliding Window, Monotonic Queue. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Count Subarrays With Fixed Bounds?

Always clarify the problem constraints and edge cases before diving into coding. Think about how to reduce the number of iterations or checks needed to solve the problem. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Subarrays With Fixed Bounds?

Not considering elements outside the bounds of minK and maxK, which can invalidate a subarray. Failing to correctly update the last seen indices of minK and maxK during iteration.

#2444

Count Subarrays With Fixed Bounds

Hard

ArrayQueueSliding WindowMonotonic QueueSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently count the valid subarrays without generating all possible subarrays. We keep track of the last positions of minK and maxK to determine valid subarrays quickly.

⚙️

Algorithm

3 steps

1Step 1: Initialize variables to track the last positions of minK and maxK, and a count of valid subarrays.
2Step 2: Iterate through the array, updating the last seen positions of minK and maxK.
3Step 3: For each element, calculate the number of valid subarrays that can end at the current position based on the last seen positions.

solution.py14 lines

1# Full working Python code
2
3def countSubarrays(nums, minK, maxK):
4    count = 0
5    lastMinK = lastMaxK = leftBound = -1
6    for i, num in enumerate(nums):
7        if num < minK or num > maxK:
8            leftBound = i
9        if num == minK:
10            lastMinK = i
11        if num == maxK:
12            lastMaxK = i
13        count += max(0, min(lastMinK, lastMaxK) - leftBound)
14    return count

ℹ

Complexity note: This complexity is achieved because we only make a single pass through the array, updating indices and counts without needing nested loops.

1The problem can be efficiently solved using the sliding window technique by tracking the last seen indices of minK and maxK.
2Understanding the boundaries of valid subarrays is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.