LeetCode 522: Longest Uncommon Subsequence II Solution in Python – A Step-by-Step Guide

What Is LeetCode 522: Longest Uncommon Subsequence II?

In LeetCode 522: Longest Uncommon Subsequence II, you’re given an array of strings, and your task is to find the length of the longest uncommon subsequence—a subsequence that appears in exactly one string and not in any others. A subsequence is formed by deleting some (or no) characters without changing the order. For example, with ["aba", "cdc", "eae"], “aba” (length 3) is uncommon because it’s only in the first string. This problem builds on LeetCode 521: Longest Uncommon Subsequence I, extending to multiple strings.

Problem Statement

• Input: List of strings strs.
• Output: Integer—length of the longest uncommon subsequence, or -1 if none exists.
• Rules: Subsequence must be unique to one string; not contiguous.

Constraints

• 1 <= strs.length <= 50
• 1 <= strs[i].length <= 50
• strs[i] contains only lowercase English letters.

Examples

• Input: ["aba","cdc","eae"]
• Output: 3
• “aba” (length 3) is in first string only.
• Input: ["aaa","aaa","aa"]
• Output: -1
• No uncommon subsequence (all overlap).
• Input: ["aabb","aab"]
• Output: 4
• “aabb” (length 4) is uncommon.

Understanding the Problem: Seeking the Unique Subsequence

To solve LeetCode 522: Longest Uncommon Subsequence II in Python, we need to find the longest subsequence that appears in one string but not in any others in the array. Unlike LeetCode 521, where two strings made it simple, here we must check multiple strings, and a string itself can be the answer if it’s unique. A naive approach might generate all subsequences, but with up to 50 strings of length 50, we need efficiency. We’ll explore:

• Best Solution (String Comparison with Subsequence Check): O(n² * m) time, O(1) space (n = number of strings, m = avg string length)—smart and practical.
• Alternative Solution (Brute-Force Subsequence Generation): O(n * 2ᵐ) time, O(2ᵐ) space—exhaustive but slow.

Let’s dive into the best solution with a friendly breakdown!

Best Solution: String Comparison with Subsequence Check

Why String Comparison Wins

The string comparison with subsequence check is the best for LeetCode 522 because it leverages a key insight: we can test each string as a candidate for the longest uncommon subsequence, checking if it’s a subsequence of any other string, and it runs efficiently at O(n² * m). It uses O(1) space beyond input, making it scalable. It’s like scanning a lineup of strings to find the standout!

How It Works

Think of this as a string uniqueness contest:

• Step 1: Sort by Length:
• Sort strings by decreasing length (longest first).
• Step 2: Check Each String:
• For each string s, test if it’s a subsequence of any other string.
• If not, it’s uncommon; track its length.
• Step 3: Subsequence Check:
• Use a two-pointer method to see if s is in another string.
• Step 4: Return:
• Longest uncommon length, or -1 if none.
• Why It Works:
• Longest uncommon must be a full string (subsequences can’t exceed it).
• Early sorting optimizes search.

It’s like a string standout detector!

Step-by-Step Example

Example: ["aba", "cdc", "eae"]

• Step 1: Sort by length: ["aba", "cdc", "eae"] (all 3).
• Step 2: Check each:
• s = "aba":
• vs “cdc”: Not a subsequence (a→c fails).
• vs “eae”: Not a subsequence (a→e fails).
• Uncommon, length 3, stop (longest first).
• Result: 3.

Example: ["aaa", "aaa", "aa"]

• Step 1: Sort: ["aaa", "aaa", "aa"].
• Step 2:
• “aaa” vs “aaa”: Subsequence, skip.
• “aaa” vs “aa”: Subsequence, skip.
• “aa” vs both “aaa”: Subsequence, skip.
• Result: -1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def findLUSlength(self, strs: List[str]) -> int:
        # Step 1: Sort by length descending
        strs.sort(key=len, reverse=True)

        # Step 2: Subsequence check helper
        def is_subsequence(s, t):
            i = 0
            for char in t:
                if i < len(s) and s[i] == char:
                    i += 1
            return i == len(s)

        # Step 3: Check each string
        for i, s in enumerate(strs):
            is_uncommon = True
            for j, t in enumerate(strs):
                if i != j and is_subsequence(s, t):
                    is_uncommon = False
                    break
            if is_uncommon:
                return len(s)

        # Step 4: No uncommon found
        return -1

• Line 4: Sort strings by length, longest first.
• Lines 7-12: is_subsequence:
• Two-pointer: Match s in t, return True if all match.
• Lines 15-21: For each string s:
• Check against all others; if subsequence elsewhere, not uncommon.
• If unique, return its length.
• Line 24: Default to -1 if no uncommon subsequence.
• Time Complexity: O(n² * m)—n strings, m avg length for subsequence check.
• Space Complexity: O(1)—no extra space beyond input.

It’s like a string uniqueness sleuth!

Alternative Solution: Brute-Force Subsequence Generation

Why an Alternative Approach?

The brute-force subsequence generation solution generates all subsequences of each string, checks for uniqueness, and finds the longest uncommon one. It’s O(n * 2ᵐ) time and O(2ᵐ) space—thorough but impractical for larger strings, though it shows the full process. Great for learning subsequences!

How It Works

Picture this as a subsequence treasure hunt:

• Step 1: Generate all subsequences for each string.
• Step 2: Count occurrences across all strings.
• Step 3: Find longest with count 1.

It’s like a string excavation!

Step-by-Step Example

Example: ["ab", "cd"]

• Step 1: Subsequences:
• “ab”: “”, “a”, “b”, “ab”.
• “cd”: “”, “c”, “d”, “cd”.
• Step 2: Count:
• “”: 2, “a”: 1, “b”: 1, “ab”: 1, “c”: 1, “d”: 1, “cd”: 1.
• Step 3: Longest with count 1: “ab”, “cd” (length 2).
• Result: 2.

Code for Brute-Force Approach

Copy codeclass Solution:
    def findLUSlength(self, strs: List[str]) -> int:
        # Step 1: Generate all subsequences
        def get_subsequences(s):
            subs = set()
            n = len(s)
            for mask in range(1 << n):
                sub = ""
                for i in range(n):
                    if mask & (1 << i):
                        sub += s[i]
                subs.add(sub)
            return subs

        # Step 2: Count all subsequences
        sub_count = {}
        for s in strs:
            for sub in get_subsequences(s):
                sub_count[sub] = sub_count.get(sub, 0) + 1

        # Step 3: Find longest uncommon
        max_len = -1
        for sub, count in sub_count.items():
            if count == 1:
                max_len = max(max_len, len(sub))

        return max_len

• Lines 4-12: Generate subsequences with bitmask.
• Lines 15-18: Count occurrences across all strings.
• Lines 21-24: Find longest with count 1.
• Time Complexity: O(n * 2ᵐ)—exponential per string.
• Space Complexity: O(2ᵐ)—store subsequences.

It’s a subsequence overachiever!

Comparing the Two Solutions

• String Comparison (Best):
• Pros: O(n² * m), O(1), efficient.
• Cons: Requires subsequence logic.
• Brute-Force (Alternative):
• Pros: O(n * 2ᵐ), explicit.
• Cons: Too slow for large strings.

String comparison wins for practicality!

Additional Examples and Edge Cases

• ** ["a", "b", "c"]: 1.
• ** ["abcd", "abc"]: 4.
• ** ["x", "x"]: -1.

String comparison handles them all!

Complexity Recap

• String Comparison: Time O(n² * m), Space O(1).
• Brute-Force: Time O(n * 2ᵐ), Space O(2ᵐ).

String comparison’s the efficiency champ!

Key Takeaways

• Comparison: Clever check—learn at Python Basics!
• Brute Force: Full but slow.
• Strings: Uniqueness is key.
• Python: Loops and logic shine!

Final Thoughts: Uncover That Uncommon Length!

LeetCode 522: Longest Uncommon Subsequence II in Python is a clever string challenge. String comparison with subsequence check is your fast track, while brute-force generation shows the long way. Want more? Try LeetCode 521: Longest Uncommon Subsequence I or LeetCode 392: Is Subsequence. Ready to hunt? Head to Solve LeetCode 522 on LeetCode and find that unique length today!