#2962

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

Count Subarrays Where Max Element Appears at Least K Times

Find the number of subarrays where the maximum element appears at least k times using a sliding window approach.

Problem Statement

Given an integer array nums and a positive integer k, return the number of subarrays where the maximum element appears at least k times in that subarray.

A subarray is a contiguous sequence of elements within an array, and a valid subarray is one where the maximum element appears at least k times.

Examples

Example 1

Input: nums = [1,3,2,3,3], k = 2

Output: 6

The subarrays that contain the element 3 at least 2 times are: [1,3,2,3], [1,3,2,3,3], [3,2,3], [3,2,3,3], [2,3,3] and [3,3].

Example 2

Input: nums = [1,4,2,1], k = 3

Output: 0

No subarray contains the element 4 at least 3 times.

Constraints

1 <= nums.length <= 105
1 <= nums[i] <= 106
1 <= k <= 105

Solution Approach

Sliding Window with Running State Updates

The sliding window technique is used to dynamically adjust the window size while keeping track of the frequency of the maximum element. As the window slides, we incrementally check if the maximum element appears at least k times within the window. This results in an efficient solution by processing each element only once.

Track the Frequency of Maximum Element

Within the sliding window, maintain the count of the maximum element. This helps quickly identify when the current subarray satisfies the condition of having the maximum element appear at least k times, avoiding redundant checks and unnecessary recalculations.

Adjust the Window Dynamically

As you move the window across the array, adjust the window’s start point dynamically. This ensures that the subarrays that do not meet the condition are excluded, while valid subarrays are counted efficiently. The dynamic adjustment ensures that no unnecessary subarrays are checked multiple times.

Complexity Analysis

Metric	Value
Time	O(N)
Space	O(N)

The time complexity is O(N) because we traverse the array once, and the space complexity is O(N) due to the storage of frequency data for the sliding window.

What Interviewers Usually Probe

Look for the candidate’s ability to apply the sliding window pattern effectively.
Evaluate if the candidate is keeping track of the state of the maximum element efficiently.
Check if the candidate understands the importance of dynamic window adjustment to avoid redundant work.

Common Pitfalls or Variants

Common pitfalls

Failing to dynamically adjust the window size, which may lead to unnecessary calculations.
Not properly keeping track of the frequency of the maximum element, resulting in missed subarrays.
Overcomplicating the solution by not leveraging the sliding window technique, which could lead to slower performance.

Follow-up variants

Instead of finding subarrays with the maximum element appearing at least k times, find subarrays where the maximum element appears exactly k times.
Change the problem to count subarrays where the sum of the elements equals a certain value, using a sliding window approach.
Consider solving the problem with a different constraint, such as finding subarrays where the maximum element appears at most k times.

FAQ

What is the optimal approach for the Count Subarrays Where Max Element Appears at Least K Times problem?

The optimal approach is to use a sliding window with dynamic updates to track the frequency of the maximum element and adjust the window size accordingly.

How does sliding window help in solving this problem?

Sliding window allows for efficient traversal of the array by maintaining a dynamic window and updating its state, avoiding redundant checks.

What are common mistakes when solving this problem?

Common mistakes include failing to adjust the window size dynamically or not properly tracking the frequency of the maximum element within the window.

Can this problem be solved without using sliding window?

While it is possible to use brute force, sliding window is the most efficient approach as it reduces the time complexity to O(N).

What is the time and space complexity of this problem?

The time complexity is O(N), and the space complexity is O(N) due to the frequency tracking in the sliding window.

terminal

Solution

Solution 1: Two Pointers

Let's denote the maximum value in the array as $mx$.

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class Solution:
    def countSubarrays(self, nums: List[int], k: int) -> int:
        mx = max(nums)
        n = len(nums)
        ans = cnt = j = 0
        for x in nums:
            while j < n and cnt < k:
                cnt += nums[j] == mx
                j += 1
            if cnt < k:
                break
            ans += n - j + 1
            cnt -= x == mx
        return ans

Continue Practicing

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