LeetCode 521: Longest Uncommon Subsequence I Solution in Python – A Step-by-Step Guide

What Is LeetCode 521: Longest Uncommon Subsequence I?

In LeetCode 521: Longest Uncommon Subsequence I, you’re given two strings a and b, and your task is to find the length of the longest uncommon subsequence—meaning a subsequence present in one string but not the other. A subsequence is a sequence derived by deleting some (or no) characters without changing the order. For example, with a = "aba" and b = "cdc", “aba” itself (length 3) is uncommon because it’s not in “cdc”. This problem tests string comparison, related to LeetCode 522: Longest Uncommon Subsequence II.

Problem Statement

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

Constraints

• 1 <= a.length, b.length <= 100
• a and b contain only lowercase English letters.

Examples

• Input: a = "aba", b = "cdc"
• Output: 3
• “aba” (length 3) is in a but not b.
• Input: a = "aaa", b = "aaa"
• Output: -1
• Identical strings, no uncommon subsequence.
• Input: a = "aaa", b = "bbb"
• Output: 3
• “aaa” (length 3) is in a but not b.

Understanding the Problem: Finding the Uncommon Length

To solve LeetCode 521: Longest Uncommon Subsequence I in Python, we need a method to identify the longest subsequence that exists in one string but not the other. At first glance, you might think of generating all subsequences, but there’s a twist: if the strings differ, one string itself is an uncommon subsequence! We’ll explore:

• Best Solution (Length Comparison): O(1) time, O(1) space—shockingly simple.
• Alternative Solution (Subsequence Generation): O(2ⁿ) time, O(2ⁿ) space—thorough but unnecessary.

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

Best Solution: Simple Length Comparison

Why Length Comparison Wins

The length comparison solution is the best for LeetCode 521 because it leverages a key insight: if two strings are different, the longer one (or either if equal length) is itself an uncommon subsequence, and if they’re the same, no uncommon subsequence exists. It runs in O(1) time (just a string comparison) and O(1) space, making it incredibly efficient. It’s like realizing the puzzle’s answer is hiding in plain sight!

How It Works

Think of this as a string showdown:

• Step 1: Compare Strings:
• Check if a == b.
• Step 2: Apply Insight:
• If equal: No uncommon subsequence, return -1.
• If different: Return the length of the longer string (or a if lengths equal).
• Step 3: Return:
• Simple integer based on the check.
• Why It Works:
• If strings differ, the whole string isn’t a subsequence of the other.
• If identical, all subsequences are common.

It’s like a string uniqueness trick!

Step-by-Step Example

Example: a = "aba", b = "cdc"

• Step 1: a != b (True).
• Step 2:
• len(a) = 3, len(b) = 3.
• Different strings, return max(3, 3) = 3.
• Result: 3.

Example: a = "aaa", b = "aaa"

• Step 1: a == b (True).
• Step 2: Identical, return -1.
• Result: -1.

Example: a = "aaa", b = "aa"

• Step 1: a != b (True).
• Step 2:
• len(a) = 3, len(b) = 2.
• Return max(3, 2) = 3.
• Result: 3.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def findLUSlength(self, a: str, b: str) -> int:
        # Step 1: Compare strings
        if a == b:
            # Step 2: If equal, no uncommon subsequence
            return -1

        # Step 3: If different, return longer length
        return max(len(a), len(b))

• Line 4: Check if strings are identical.
• Line 6: If same, return -1 (no uncommon subsequence).
• Line 9: If different, return max length.
• Time Complexity: O(1)—string comparison is O(min(len(a), len(b))), but treated as constant for small inputs.
• Space Complexity: O(1)—no extra space.

It’s like a string length shortcut!

Alternative Solution: Subsequence Generation Check

Why an Alternative Approach?

The subsequence generation solution explicitly generates all subsequences of both strings, finds the longest one unique to one string, and returns its length. It’s O(2ⁿ) time and O(2ⁿ) space—thorough but wildly inefficient for this problem, though it illustrates the brute-force mindset. Great for understanding subsequences!

How It Works

Picture this as a subsequence scavenger hunt:

• Step 1: Generate all subsequences of a and b.
• Step 2: Find longest subsequence in a not in b (or vice versa).
• Step 3: Return length, or -1 if none.

It’s like a string treasure dig!

Step-by-Step Example

Example: a = "ab", b = "cd"

• Step 1: Subsequences:
• a: “”, “a”, “b”, “ab”.
• b: “”, “c”, “d”, “cd”.
• Step 2: Check a’s subsequences:
• “” in b, skip.
• “a” not in b, length 1.
• “b” not in b, length 1.
• “ab” not in b, length 2.
• Max uncommon = 2.
• Result: 2.

Code for Subsequence Approach

Copy codeclass Solution:
    def findLUSlength(self, a: str, b: 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: Get subsequences of a and b
        subs_a = get_subsequences(a)
        subs_b = get_subsequences(b)

        # Step 3: Find longest uncommon subsequence
        max_len = -1
        for sub in subs_a:
            if sub not in subs_b:
                max_len = max(max_len, len(sub))

        return max_len

• Lines 4-12: Generate subsequences using bitmask.
• Lines 15-16: Get all subsequences for both strings.
• Lines 19-22: Check a’s subsequences, find longest not in b.
• Time Complexity: O(2ⁿ)—exponential subsequence generation.
• Space Complexity: O(2ⁿ)—store all subsequences.

It’s a subsequence overachiever!

Comparing the Two Solutions

• Length Comparison (Best):
• Pros: O(1), O(1), brilliantly simple.
• Cons: Relies on problem insight.
• Subsequence Generation (Alternative):
• Pros: O(2ⁿ), explicit solution.
• Cons: Impractical for large strings.

Length comparison wins by a mile!

Additional Examples and Edge Cases

• ** "a", "b": 1.
• ** "abcd", "ab": 4.
• ** "x", "x": -1.

Length comparison handles them all!

Complexity Recap

• Length Comparison: Time O(1), Space O(1).
• Subsequence: Time O(2ⁿ), Space O(2ⁿ).

Length comparison’s the efficiency champ!

Key Takeaways

• Insight: Simple can be best—learn at Python Basics!
• Brute Force: Overkill here.
• Strings: Subsequences are fun.
• Python: Len() for the win!

Final Thoughts: Find That Uncommon Length!

LeetCode 521: Longest Uncommon Subsequence I in Python is a deceptively simple string challenge. The length comparison is your fast track, while subsequence generation shows the long way. Want more? Try LeetCode 522: Longest Uncommon Subsequence II or LeetCode 392: Is Subsequence. Ready to compare? Head to Solve LeetCode 521 on LeetCode and uncover that length today!