Longest Uncommon Subsequence II

Given an array of strings strs, return the length of the longest uncommon subsequence between them. If the longest uncommon subsequence does not exist, return -1.

An uncommon subsequence between an array of strings is a string that is a subsequence of one string but not the others.

A subsequence of a string s is a string that can be obtained after deleting any number of characters from s.

• For example, "abc" is a subsequence of "aebdc" because you can delete the underlined characters in "a<u>e</u>b<u>d</u>c" to get "abc". Other subsequences of "aebdc" include "aebdc", "aeb", and "" (empty string).

Example 1:

Input: strs = ["aba","cdc","eae"]
Output: 3

Example 2:

Input: strs = ["aaa","aaa","aa"]
Output: -1

Example 3:

Input: strs = ["abcd","abc","ab"]
Output: 4

Constraints:

• 2 <= strs.length <= 50
• 1 <= strs[i].length <= 10
• strs[i] consists of lowercase English letters.

def is_subsequence(s, t):
    """Checks if string s is a subsequence of string t."""
    i = 0  # Pointer for s
    j = 0  # Pointer for t
    while i < len(s) and j < len(t):
        if s[i] == t[j]:
            i += 1
        j += 1
    return i == len(s)


def longest_uncommon_subsequence(strs):
    """Finds the length of the longest uncommon subsequence among strings."""
    n = len(strs)
    max_len = -1

    for i in range(n):
        is_uncommon = True
        for j in range(n):
            if i != j and is_subsequence(strs[i], strs[j]):
                is_uncommon = False
                break

        if is_uncommon:
            max_len = max(max_len, len(strs[i]))

    return max_len


# Example Usage
strs1 = ["aba", "cdc", "eae"]
print(f"Longest uncommon subsequence length for {strs1}: {longest_uncommon_subsequence(strs1)}")

strs2 = ["aaa", "aaa", "aa"]
print(f"Longest uncommon subsequence length for {strs2}: {longest_uncommon_subsequence(strs2)}")

strs3 = ["abcd", "abcd", "abcde"]
print(f"Longest uncommon subsequence length for {strs3}: {longest_uncommon_subsequence(strs3)}")

strs4 = ["a", "b", "c"]
print(f"Longest uncommon subsequence length for {strs4}: {longest_uncommon_subsequence(strs4)}")

Naive Approach:

The naive approach involves checking every string against all other strings to see if it's a subsequence of any of them. If a string is not a subsequence of any other string, it's a candidate for the longest uncommon subsequence. We keep track of the maximum length of such strings.

Optimal Approach:

The provided code directly implements the optimal approach within the given constraints. Since the constraints limit the number of strings to 50 and the maximum length of each string to 10, the time complexity is acceptable. The algorithm iterates through each string and checks if it's a subsequence of any other string. If it's not, the length of the string is compared with the current maximum length to update it.

Big(O) Runtime Analysis:

• Outer Loop: The code iterates through each string in the strs array once, which contributes a factor of n, where n is the number of strings.
• Inner Loop: For each string, the code iterates through all other strings to check for subsequences, contributing another factor of n.
• is_subsequence Function: The is_subsequence function compares two strings. In the worst case, it might iterate through both strings completely, contributing a factor of m, where m is the maximum length of the strings.
• Overall: The overall time complexity is O(n^2 * m), where n is the number of strings and m is the maximum length of a string. Given that the constraints specify that n <= 50 and m <= 10, this complexity is manageable.

Big(O) Space Usage Analysis:

• The primary space usage comes from the variables used within the loops and the is_subsequence function. These variables use a constant amount of space.
• The space used by the input strs is not considered as part of the algorithm's space complexity.
• The is_subsequence function uses a constant amount of space.
• Overall: The space complexity is O(1), which means the algorithm uses a constant amount of extra space regardless of the input size.

Edge Cases and Handling:

1. Empty Input: If the input array strs is empty, the code will not execute the loops, and the initial value of max_len (-1) will be returned. Although the problem statement says 2 <= strs.length <= 50, it's good to be aware of this case.
2. All Strings Are Identical: If all strings in the input array are identical, no string can be an uncommon subsequence because each string is a subsequence of every other string. The function will return -1.
3. Strings Are Subsequences of Each Other: If some strings are subsequences of others, the algorithm correctly identifies and excludes them from consideration. For example, if `strs = [