What is the brute force solution for Count Subarrays Where Max Element Appears at Least K Times?

The brute force approach for Count Subarrays Where Max Element Appears at Least K Times 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 the maximum element appears at least k times. 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 Where Max Element Appears at Least K Times?

Count Subarrays Where Max Element Appears at Least K Times uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Count Subarrays Where Max Element Appears at Least K Times?

Always consider edge cases, such as when k is larger than the number of occurrences of any element. Explain your thought process clearly while coding, as it shows your understanding. Practice implementing sliding window techniques, as they are common in array problems.

What are common mistakes when solving Count Subarrays Where Max Element Appears at Least K Times?

Not resetting the frequency count when a new maximum is found. Failing to correctly count valid subarrays when the left pointer moves.

#2962

Count Subarrays Where Max Element Appears at Least K Times

Medium

ArraySliding WindowSliding WindowTwo Pointers

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Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to efficiently count valid subarrays. We maintain a window that expands and contracts based on the occurrences of the maximum element.

⚙️

Algorithm

4 steps

1Step 1: Initialize pointers for the sliding window and a counter for occurrences of the current maximum element.
2Step 2: Expand the right pointer to include new elements and update the maximum and its frequency.
3Step 3: If the maximum appears at least k times, count the valid subarrays ending at the right pointer.
4Step 4: Move the left pointer to shrink the window while maintaining the condition.

solution.py18 lines

1def countSubarrays(nums, k):
2    count = 0
3    left = 0
4    max_elem = -1
5    freq = 0
6    n = len(nums)
7    for right in range(n):
8        if nums[right] > max_elem:
9            max_elem = nums[right]
10            freq = 1
11        elif nums[right] == max_elem:
12            freq += 1
13        while freq >= k:
14            count += n - right
15            if nums[left] == max_elem:
16                freq -= 1
17            left += 1
18    return count

ℹ

Complexity note: The time complexity is O(n) because each element is processed at most twice (once when expanding and once when contracting the window). The space complexity is O(1) since we only use a few extra variables.

1The maximum element must be tracked efficiently to avoid redundant calculations.
2Using a sliding window allows us to dynamically adjust the subarray and count valid configurations.

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