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Find the Maximum Length of a Good Subsequence II

Difficulty: Hard

Problem Description

You are given an integer array nums and a non-negative integer k. A sequence of integers seq is called good if there are at most k indices i in the range [0, seq.length - 2] such that seq[i] != seq[i + 1]. Return the maximum possible length of a good subsequence of nums.

Key Insights

  • A "good" subsequence allows for at most k transitions between different values.
  • The problem can be rephrased as counting the number of occurrences of each unique number in nums.
  • The maximum length of a good subsequence can be determined by selecting the most frequent numbers while respecting the allowed k transitions.

Space and Time Complexity

Time Complexity: O(n + m), where n is the length of nums and m is the number of unique elements in nums.
Space Complexity: O(m), where m is the number of unique elements in nums.

Solution

To solve this problem, we can use a frequency map (or hash table) to count the occurrences of each number in the array. We then sort the unique numbers based on their frequencies. We can select the most frequent numbers up to k + 1 different values. The length of the good subsequence will be the sum of the counts of these selected numbers.

Code Solutions

def maxLengthGoodSubsequence(nums, k):
    from collections import Counter

    # Count the frequency of each number
    frequency = Counter(nums)
    
    # Sort the counts in descending order
    sorted_counts = sorted(frequency.values(), reverse=True)
    
    # We can take at most k + 1 different numbers
    max_length = sum(sorted_counts[:k + 1])
    
    return max_length
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