Introduction

In this blog post, we will explore how to solve the LeetCode problem "2831" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You are given a 0-indexed integer array nums and an integer k. A subarray is called equal if all of its elements are equal. Note that the empty subarray is an equal subarray. Return the length of the longest possible equal subarray after deleting at most k elements from nums. A subarray is a contiguous, possibly empty sequence of elements within an array. Example 1: Input: nums = [1,3,2,3,1,3], k = 3 Output: 3 Explanation: It's optimal to delete the elements at index 2 and index 4. After deleting them, nums becomes equal to [1, 3, 3, 3]. The longest equal subarray starts at i = 1 and ends at j = 3 with length equal to 3. It can be proven that no longer equal subarrays can be created. Example 2: Input: nums = [1,1,2,2,1,1], k = 2 Output: 4 Explanation: It's optimal to delete the elements at index 2 and index 3. After deleting them, nums becomes equal to [1, 1, 1, 1]. The array itself is an equal subarray, so the answer is 4. It can be proven that no longer equal subarrays can be created. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= nums.length 0 <= k <= nums.length

Explanation

Here's the solution to find the longest equal subarray after deleting at most k elements:

• Sliding Window: Use a sliding window approach to iterate through all possible subarrays.
• Window Validity: For each window, check if the number of elements to delete (window size minus the frequency of the most frequent element) is less than or equal to k.

• Maximize Length: Maintain a running maximum of the valid window lengths.

• Time Complexity: O(n), Space Complexity: O(1)

Code

    def longestEqualSubarray(nums, k):
    """
    Finds the length of the longest possible equal subarray after deleting at most k elements.

    Args:
        nums: A list of integers.
        k: The maximum number of elements that can be deleted.

    Returns:
        The length of the longest equal subarray.
    """

    max_length = 0
    left = 0
    counts = {}  # Store element frequencies within the window

    for right in range(len(nums)):
        num = nums[right]
        counts[num] = counts.get(num, 0) + 1

        window_size = right - left + 1
        most_frequent = max(counts.values())
        elements_to_delete = window_size - most_frequent

        while elements_to_delete > k:
            left_num = nums[left]
            counts[left_num] -= 1
            if counts[left_num] == 0:
                del counts[left_num]

            left += 1
            window_size = right - left + 1
            if counts:
                most_frequent = max(counts.values())
            else:
                most_frequent = 0 # Handle empty window case
            elements_to_delete = window_size - most_frequent

        max_length = max(max_length, window_size)

    return max_length