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Given a list of strings, you need to find the longest uncommon subsequence among them. The longest uncommon subsequence is defined as the longest subsequence of one of these strings that does not occur as a subsequence in any other strings.

A subsequence is a sequence that can be derived from another sequence by deleting some or no elements without changing the order of the remaining elements.

Clarifying Questions

  1. Can there be duplicate strings in the input list?
    • Yes, there can be duplicates, but they should be considered as individual strings.
  2. What is the maximum length of each string or the total number of strings?
    • Often the lengths and number of strings would be within reasonable limits for algorithmic problems (say length 100 and up to 200 strings per constraints).
  3. Do we need to output the subsequence itself or just the length?
    • Only the length of the longest uncommon subsequence is needed.
  4. Are special characters, spaces, or lowercase/uppercase characters considered?
    • Usually, the problem will mention specifically, otherwise, assume typical alphanumeric.

Strategy

  1. Sort the Strings by Length: Start by sorting the strings primarily by length in descending order. This allows us to check longer strings first, as we are interested in the longest uncommon subsequence.

  2. Check Each String: For each string in this sorted list, check whether it is an uncommon subsequence.

  3. Determine Subsequence: Define a helper function to check if a string a is a subsequence of another string b.

  4. Mark Uncommon Sequences: Iterate through each string and check against all other strings whether it is an uncommon subsequence. If so, return its length.

  5. Edge Cases: Handle the case where no uncommon subsequence is found.

Code

def findLUSlength(strs):
    def is_subsequence(a, b):
        # Function to check if 'a' is a subsequence of 'b'
        it = iter(b)
        return all(char in it for char in a)

    # Sort strings by length in descending order
    strs.sort(key=lambda x: len(x), reverse=True)

    n = len(strs)
    
    # Check each string
    for i in range(n):
        uncommon = True
        for j in range(n):
            if i != j and is_subsequence(strs[i], strs[j]):
                uncommon = False
                break
        if uncommon:
            return len(strs[i])

    return -1

Time Complexity

The time complexity of the solution can be broken down as follows:

  1. Sorting: Sorting the list of strings takes (O(n \log n)), where (n) is the number of strings.

  2. Subsequence Check: Checking if one string is a subsequence of another takes (O(m + k)), where (m) and (k) are the lengths of the strings involved. Since we perform this check for each pair of strings possible, this contributes to (O(n^2 \cdot L)), where (L) is the average length of the strings.

Thus, the overall time complexity can be approximated to (O(n^2 \cdot L)), which checks each string against every other string for subsequences after sorting. The sorting step is dominated by the subsequence checking step due to their polynomial versus logarithmic nature.

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