What is the brute force solution for Length of Longest Subarray With at Most K Frequency?

The brute force approach for Length of Longest Subarray With at Most K Frequency is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach checks all possible subarrays to see if they meet the frequency condition. This is straightforward but inefficient, especially for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Length of Longest Subarray With at Most K Frequency?

Length of Longest Subarray With at Most K Frequency uses the following concepts: Array, Hash Table, Sliding Window. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Length of Longest Subarray With at Most K Frequency?

Always consider edge cases, such as all elements being the same or k being larger than the number of unique elements. Explain your thought process clearly while coding; it helps the interviewer understand your approach. Practice writing clean and efficient code, as readability is important in interviews.

What are common mistakes when solving Length of Longest Subarray With at Most K Frequency?

Not resetting the frequency count properly when moving the left pointer. Failing to account for edge cases where all elements are the same.

#2958

Length of Longest Subarray With at Most K Frequency

Medium

ArrayHash TableSliding WindowHash MapArray

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Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal solution uses the sliding window technique to efficiently find the longest good subarray in a single pass. This reduces the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Initialize two pointers (left and right) to represent the current window of the subarray.
2Step 2: Use a hashmap to track the frequency of elements in the current window.
3Step 3: Expand the right pointer to include new elements and update their frequency.
4Step 4: If any frequency exceeds k, increment the left pointer to shrink the window until all frequencies are valid.
5Step 5: Update the maximum length of the good subarray during the process.

solution.py18 lines

1# Full working Python code
2
3def longest_good_subarray(nums, k):
4    left = 0
5    max_length = 0
6    freq = {}
7    for right in range(len(nums)):
8        freq[nums[right]] = freq.get(nums[right], 0) + 1
9        while freq[nums[right]] > k:
10            freq[nums[left]] -= 1
11            if freq[nums[left]] == 0:
12                del freq[nums[left]]
13            left += 1
14        max_length = max(max_length, right - left + 1)
15    return max_length
16
17# Example usage
18print(longest_good_subarray([1,2,3,1,2,3,1,2], 2))  # Output: 6

ℹ

Complexity note: The time complexity is O(n) because we traverse the array with two pointers, each moving at most n times. The space complexity is O(n) due to the hashmap storing frequencies of elements.

1Using a sliding window allows us to efficiently manage the current subarray without needing to recheck previous elements.
2Hashmaps are useful for counting frequencies dynamically as we expand and contract our window.

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