How do you solve Count Subarrays With Cost Less Than or Equal to K on LeetCode?

Count Subarrays With Cost Less Than or Equal to K (LeetCode #3835) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using deques allows efficient tracking of max/min values.

What is the time complexity of Count Subarrays With Cost Less Than or Equal to K?

The optimal solution for Count Subarrays With Cost Less Than or Equal to K runs in O(n) time and uses O(n) space. The sliding window approach processes each element at most twice, leading to O(n) time complexity.

What is the brute force solution for Count Subarrays With Cost Less Than or Equal to K?

The brute force approach for Count Subarrays With Cost Less Than or Equal to K is Brute Force, with O(n²) time complexity and O(1) space complexity. We can generate all possible subarrays and calculate their costs. This method 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 Cost Less Than or Equal to K?

Count Subarrays With Cost Less Than or Equal to K uses the following concepts: Array, Queue, Monotonic Queue. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Subarrays With Cost Less Than or Equal to K?

Explain your thought process clearly. Discuss edge cases, like empty arrays. Practice coding the sliding window technique.

What are common mistakes when solving Count Subarrays With Cost Less Than or Equal to K?

Not maintaining the deques correctly. Forgetting to adjust the left pointer when cost exceeds k.

#3835

Count Subarrays With Cost Less Than or Equal to K

Medium

ArrayQueueMonotonic QueueHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

Using a sliding window and monotonic deques allows us to efficiently track the max and min values in the current subarray, reducing the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize two deques to maintain indices of max and min values.
2Step 2: Use a sliding window approach to expand the right end until the cost exceeds k.
3Step 3: Count valid subarrays by adjusting the left end of the window.

solution.py18 lines

1from collections import deque
2
3def countSubarrays(nums, k):
4    n = len(nums)
5    left = 0
6    count = 0
7    max_deque, min_deque = deque(), deque()
8    for right in range(n):
9        while max_deque and nums[max_deque[-1]] <= nums[right]: max_deque.pop()
10        while min_deque and nums[min_deque[-1]] >= nums[right]: min_deque.pop()
11        max_deque.append(right)
12        min_deque.append(right)
13        while (nums[max_deque[0]] - nums[min_deque[0]]) * (right - left + 1) > k:
14            left += 1
15            if max_deque[0] < left: max_deque.popleft()
16            if min_deque[0] < left: min_deque.popleft()
17        count += right - left + 1
18    return count

ℹ

Complexity note: The sliding window approach processes each element at most twice, leading to O(n) time complexity.

1Using deques allows efficient tracking of max/min values.
2Sliding window helps maintain valid subarray bounds.

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