How do you solve Find the Maximum Length of a Good Subsequence II on LeetCode?

Find the Maximum Length of a Good Subsequence II (LeetCode #3177) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to count unique elements and their frequencies is crucial for optimizing the solution.

What is the brute force solution for Find the Maximum Length of a Good Subsequence II?

The brute force approach for Find the Maximum Length of a Good Subsequence II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible subsequences of the array and checking each one to see if it qualifies as a good subsequence. 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 Find the Maximum Length of a Good Subsequence II?

Find the Maximum Length of a Good Subsequence II uses the following concepts: Array, Hash Table, Dynamic Programming. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find the Maximum Length of a Good Subsequence II?

Always clarify the constraints and requirements of the problem before diving into coding. Think about edge cases, such as when k is 0 or when all elements are the same. Practice explaining your thought process as you code, as this will help you articulate your reasoning during interviews.

What are common mistakes when solving Find the Maximum Length of a Good Subsequence II?

Failing to consider the frequency of elements and treating all elements as unique. Not properly initializing the DP table or misunderstanding its dimensions.

#3177

Find the Maximum Length of a Good Subsequence II

Hard

ArrayHash TableDynamic ProgrammingHash MapArray

LeetCode ↗

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 approach uses dynamic programming to efficiently calculate the maximum length of a good subsequence by tracking the counts of each unique number and how they can be combined while respecting the constraints of differing indices.

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Algorithm

4 steps

1Step 1: Count the frequency of each unique number in nums.
2Step 2: Store these frequencies in a list and sort it.
3Step 3: Use a dynamic programming table where dp[i][j] represents the maximum length of a good subsequence using the first i unique numbers with at most j differing indices.
4Step 4: Iterate through the unique numbers and update the dp table based on the frequency of the current number and the previous states.

solution.py19 lines

1# Full working Python code
2from collections import Counter
3
4def max_good_subsequence(nums, k):
5    count = Counter(nums)
6    freq = list(count.values())
7    n = len(freq)
8    dp = [[0] * (k + 1) for _ in range(n + 1)]
9
10    for i in range(1, n + 1):
11        for j in range(k + 1):
12            dp[i][j] = dp[i - 1][j] + freq[i - 1]  # take all of current number
13            if j > 0:
14                dp[i][j] = max(dp[i][j], dp[i - 1][j - 1] + freq[i - 1])  # take one less
15
16    return dp[n][k]
17
18# Example usage
19print(max_good_subsequence([1, 2, 1, 1, 3], 2))

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Complexity note: The time complexity is O(n) because we only pass through the array a few times: once to count frequencies and once to fill the dynamic programming table. The space complexity is O(n) due to the storage of frequency counts and the DP table.

1Understanding how to count unique elements and their frequencies is crucial for optimizing the solution.
2Dynamic programming can effectively manage state transitions based on constraints, leading to efficient solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.