algoadvance

You are given a string s consisting of lowercase English letters. An adjacent group of characters is a group of consecutive characters that are all the same in the string. For example, in the string “abaaaacc”, the groups are ‘a’, ‘b’, ‘aaaa’, ‘cc’. The function longestUnequalGroupsSubsequence(s) should return the longest subsequence where no two adjacent groups are equal.

Clarifying Questions

  1. Can the input string be empty?
    • No, an input string s with at least one character will be provided.
  2. What should the function return?
    • The length of the longest subsequence of groups where no two adjacent groups in the sequence are equal.
  3. Are there any constraints on the length of the string?
    • The string’s length will be between 1 and (10^5), inclusive.
  4. What characters does the string contain?
    • The string contains only lowercase English letters (‘a’ to ‘z’).

Strategy

  1. Identify Groups: First, identify all the groups of consecutive identical characters.
  2. Track Adjacent Groups: Traverse the identified groups and track the longest subsequence where no two adjacent groups are the same.
  3. Return Result: Count the length of this longest subsequence and return it.

Code

def longestUnequalGroupsSubsequence(s: str) -> int:
    if not s:
        return 0

    n = len(s)
    prev_char = s[0]
    groups = []
    count = 1

    for i in range(1, n):
        if s[i] == prev_char:
            count += 1
        else:
            groups.append(prev_char * count)
            prev_char = s[i]
            count = 1
    
    # Add the last group
    groups.append(prev_char * count)

    # Initialize the longest subsequence count
    longest_length = 1
    current_length = 1

    for i in range(1, len(groups)):
        if groups[i] != groups[i - 1]:
            current_length += 1
            longest_length = max(longest_length, current_length)
        else:
            current_length = 1

    return longest_length

# Example usage:
print(longestUnequalGroupsSubsequence("abaaaacc"))  # Expected output: 4

Time Complexity

  1. Identify Groups: Takes O(n) time to traverse through the string once to identify all groups.
  2. Track Adjacent Groups: Takes O(m) time where m is the number of groups, which in the worst case can be O(n).

Therefore, the overall time complexity is O(n), where n is the length of the string, which is efficient for the input size constraints.

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