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2958. Length of Longest Subarray With at Most K Frequency

This is a very common subarray technique, remember since it’s a subarray it has to be a sliding window. If it were a $subsequence$, it becomes a DP or greedy solution.

We just keep track of current elements and their increasing/decreasing frequencies in a dict/table. At every iteration, if the frequency of the current element ($freq[nums[right]]$) goes beyond k, we start shrinking the window from the left.

Code:

Python

class Solution:
    def maxSubarrayLength(self, nums: List[int], k: int) -> int:
        res = 0
        freq = dict()
 
        left = 0
 
        for right, n in enumerate(nums):
            if n in freq.keys():
                freq[n] += 1
                if freq[n] > k:
                    while freq[nums[right]] > k:
                        freq[nums[left]] -= 1
                        left += 1
            else:
                freq[n] = 1
 
            res = max(right - left + 1, res)
 
        return res

Big O Analysis

• Runtime The runtime here is $O(N)$ since we are visiting each element once.
• Memory We use a frequency table so the memory order is $O(N)$

— A

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