Introduction

In this blog post, we will explore how to solve the LeetCode problem "2958" 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 an integer array nums and an integer k. The frequency of an element x is the number of times it occurs in an array. An array is called good if the frequency of each element in this array is less than or equal to k. Return the length of the longest good subarray of nums. A subarray is a contiguous non-empty sequence of elements within an array. Example 1: Input: nums = [1,2,3,1,2,3,1,2], k = 2 Output: 6 Explanation: The longest possible good subarray is [1,2,3,1,2,3] since the values 1, 2, and 3 occur at most twice in this subarray. Note that the subarrays [2,3,1,2,3,1] and [3,1,2,3,1,2] are also good. It can be shown that there are no good subarrays with length more than 6. Example 2: Input: nums = [1,2,1,2,1,2,1,2], k = 1 Output: 2 Explanation: The longest possible good subarray is [1,2] since the values 1 and 2 occur at most once in this subarray. Note that the subarray [2,1] is also good. It can be shown that there are no good subarrays with length more than 2. Example 3: Input: nums = [5,5,5,5,5,5,5], k = 4 Output: 4 Explanation: The longest possible good subarray is [5,5,5,5] since the value 5 occurs 4 times in this subarray. It can be shown that there are no good subarrays with length more than 4. Constraints: 1 <= nums.length <= 105 1 <= nums[i] <= 109 1 <= k <= nums.length

Explanation

Here's a breakdown of the solution:

• Sliding Window: Use a sliding window to iterate through all possible subarrays.
• Frequency Tracking: Maintain a dictionary (or hash map) to store the frequency of each element within the current window.

• Good Subarray Check: Expand the window until an element's frequency exceeds k. Then, shrink the window from the left until all element frequencies are within the limit. Track the maximum length of a valid "good" subarray encountered.

• Runtime Complexity: O(n), where n is the length of the input array nums.

• Storage Complexity: O(n) in the worst case, to store the frequency of each element if all elements are distinct.

Code

    def longest_good_subarray(nums: list[int], k: int) -> int:
    """
    Finds the length of the longest good subarray of nums.
    A subarray is called good if the frequency of each element in this array is less than or equal to k.
    """
    freq = {}
    left = 0
    max_len = 0
    for right in range(len(nums)):
        num = nums[right]
        freq[num] = freq.get(num, 0) + 1
        while freq[num] > k:
            freq[nums[left]] -= 1
            if freq[nums[left]] == 0:
                del freq[nums[left]]
            left += 1
        max_len = max(max_len, right - left + 1)
    return max_len