Solution Explanation: Longest Uncommon Subsequence II

The problem asks to find the length of the longest uncommon subsequence among a given array of strings. An uncommon subsequence is a string that's a subsequence of one string in the array but not a subsequence of any other string.

Approach:

The solution employs a brute-force approach with a helper function to check if one string is a subsequence of another. It iterates through each string in the input array:

1. Subsequence Check (check(s, t)): This function uses a two-pointer approach to efficiently determine if string s is a subsequence of string t. It iterates through t, and if a character in s matches the current character in t, the pointer for s advances. If the pointer for s reaches the end of s, it means s is a subsequence of t.

2. Uniqueness Check: For each string s, the code iterates through the rest of the strings in the array. If s is found to be a subsequence of any other string, it's not an uncommon subsequence, and its length is discarded.

3. Longest Uncommon Subsequence: If a string s passes the uniqueness check (meaning it's not a subsequence of any other string in the array), its length is compared to the current maximum length (ans). The maximum length is updated if a longer uncommon subsequence is found.

4. No Uncommon Subsequence: If no uncommon subsequence is found after iterating through all strings, the function returns -1.

Time Complexity Analysis:

• The outer loop iterates through each string in the input array (n strings).
• The inner loop iterates through the array again for each string in the outer loop.
• The check function has a time complexity of O(m), where m is the maximum length of a string.

Therefore, the overall time complexity is O(n² * m), where n is the number of strings and m is the maximum length of a string. This is a brute-force solution, and its efficiency degrades significantly as the number of strings or their lengths increase.

Space Complexity Analysis:

The solution uses only a constant amount of extra space for variables like ans, i, and j. Therefore, the space complexity is O(1).

Code Examples (with slight improvements for readability):

The provided code in various languages is already quite efficient for this brute-force approach. However, here's a slightly refined Python version:

def findLUSlength(strs):
    def is_subsequence(s, t):
        i = 0
        for char in t:
            if i < len(s) and char == s[i]:
                i += 1
        return i == len(s)
 
    max_len = -1
    for i, s in enumerate(strs):
        is_unique = True
        for j, t in enumerate(strs):
            if i != j and is_subsequence(s, t):
                is_unique = False
                break
        if is_unique:
            max_len = max(max_len, len(s))
    return max_len
 

This version uses a more concise is_subsequence function and slightly clearer variable names. The other language examples provided are equally valid and efficient for this approach.