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

Find the Maximum Length of a Good Subsequence I (LeetCode #3176) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(n) space complexity. Key insight: Understanding the constraints on differences is crucial for optimizing the solution.

What is the time complexity of Find the Maximum Length of a Good Subsequence I?

The optimal solution for Find the Maximum Length of a Good Subsequence I runs in O(n log n) time and uses O(n) space. The time complexity is O(n log n) due to sorting the unique elements, while the space complexity is O(n) for storing the frequency map and DP array.

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

The brute force approach for Find the Maximum Length of a Good Subsequence I 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 meets the criteria of being a good subsequence. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Find the Maximum Length of a Good Subsequence I?

Find the Maximum Length of a Good Subsequence I 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 I?

Always clarify the problem constraints and edge cases before diving into coding. Discuss your thought process and potential approaches with the interviewer. Don’t forget to optimize your solution; interviewers appreciate candidates who can improve their initial ideas.

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

Failing to account for the maximum allowed differences when calculating subsequences. Overlooking the need to sort unique elements before processing.

#3176

Find the Maximum Length of a Good Subsequence I

Medium

ArrayHash TableDynamic ProgrammingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n)Space O(n)

The optimal solution uses dynamic programming to efficiently calculate the maximum length of a good subsequence by leveraging the properties of the subsequences and counting differences in a structured way.

⚙️

Algorithm

3 steps

1Step 1: Create a frequency map to count occurrences of each number in nums.
2Step 2: Sort the unique numbers and initialize a DP array to store maximum lengths.
3Step 3: Iterate through the unique numbers and update the DP array based on the allowed differences (k).

solution.py15 lines

1# Full working Python code
2from collections import Counter
3
4def max_good_subsequence(nums, k):
5    freq = Counter(nums)
6    unique_nums = list(freq.keys())
7    unique_nums.sort()
8    dp = [0] * (len(unique_nums) + 1)
9
10    for i in range(1, len(unique_nums) + 1):
11        dp[i] = dp[i - 1] + freq[unique_nums[i - 1]]
12        if i > k:
13            dp[i] -= freq[unique_nums[i - k - 1]]
14
15    return dp[len(unique_nums)]

ℹ

Complexity note: The time complexity is O(n log n) due to sorting the unique elements, while the space complexity is O(n) for storing the frequency map and DP array.

1Understanding the constraints on differences is crucial for optimizing the solution.
2Using frequency maps can significantly reduce the complexity of counting occurrences.

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