2958. Length of Longest Subarray With at Most K Frequency - Explanation

class Solution:
    def maxSubarrayLength(self, nums: List[int], k: int) -> int:
        n, res = len(nums), 0

        for i in range(n):
            count = defaultdict(int)
            for j in range(i, n):
                count[nums[j]] += 1
                if count[nums[j]] > k:
                    break
                res = max(res, j - i + 1)

        return res

class Solution:
    def maxSubarrayLength(self, nums: List[int], k: int) -> int:
        res = 0
        count = defaultdict(int)
        l = 0
        for r in range(len(nums)):
            count[nums[r]] += 1
            while count[nums[r]] > k:
                count[nums[l]] -= 1
                l += 1
            res = max(res, r - l + 1)
        return res

class Solution:
    def maxSubarrayLength(self, nums: List[int], k: int) -> int:
        count = defaultdict(int)
        l = res = 0
        cnt = 0  # count of numbers with freq > k
        for r in range(len(nums)):
            count[nums[r]] += 1
            cnt += count[nums[r]] > k
            if cnt > 0:
                cnt -= count[nums[l]] > k
                count[nums[l]] -= 1
                l += 1
        return len(nums) - l