Naive Approach (Brute Force)

The most straightforward approach is to generate all possible subarrays and check if each one is "good" (i.e., the frequency of each element is less than or equal to k).

1. Generate all subarrays: Iterate through all possible start and end indices to define subarrays.
2. Check if a subarray is good: For each subarray, count the frequency of each element. If any element's frequency exceeds k, the subarray is not good.
3. Track the maximum length: Keep track of the length of the longest good subarray found so far.

Code (Python):

def longest_good_subarray_brute_force(nums, k):
    max_len = 0
    for i in range(len(nums)):
        for j in range(i, len(nums)):
            subarray = nums[i:j+1]
            counts = {}
            is_good = True
            for num in subarray:
                counts[num] = counts.get(num, 0) + 1
                if counts[num] > k:
                    is_good = False
                    break
            if is_good:
                max_len = max(max_len, len(subarray))
    return max_len

Time Complexity: O(n³). Generating all subarrays takes O(n²) time, and checking each subarray takes O(n) time.

Space Complexity: O(n). In the worst case, we might store all n elements of the input array in the counts dictionary.

Optimal Approach (Sliding Window)

We can use a sliding window approach to solve this problem more efficiently. The idea is to maintain a window and expand it to the right as long as the subarray within the window is "good". If the subarray becomes "bad" (i.e., an element's frequency exceeds k), we shrink the window from the left until it becomes "good" again.

1. Initialize a sliding window: Start with an empty window (left = 0, right = 0).
2. Expand the window: Move the right pointer one step at a time. Update the frequency count of the element just added to the window.
3. Check if the window is good: If any element's frequency exceeds k, shrink the window from the left until all frequencies are within the limit.
4. Update the maximum length: After each adjustment, update the maximum length of a good subarray.

Code (Python):

def longest_good_subarray(nums, k):
    max_len = 0
    left = 0
    counts = {}

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

        while counts[nums[right]] > k:
            counts[nums[left]] -= 1
            left += 1

        max_len = max(max_len, right - left + 1)

    return max_len

Time Complexity: O(n). Each element is visited at most twice (once by the right pointer and once by the left pointer).

Space Complexity: O(n). In the worst case, the counts dictionary might store the frequency of all n elements if they are distinct.

Edge Cases

• Empty array: If the input array is empty, the length of the longest good subarray is 0.
• k = 0: If k is 0, then no element can appear more than 0 times. Thus, the longest "good" array can only consist of unique elements if the first element of the array is kept; every other element must be removed from the array. However, according to the problem constraints, k >= 1, so this case is not possible.
• k >= n: If k is greater than or equal to the length of the array, the entire array is a good subarray.
• All elements are the same: If all elements in the array are the same, the longest good subarray is of length k.