Problem Description

You are given two strings a and b. You need to find the length of the longest uncommon subsequence between these two strings. If no uncommon subsequence exists, return -1.

An uncommon subsequence is defined as a string that is a subsequence of exactly one of the two given strings (not both).

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

The key insight here is understanding what makes a subsequence "uncommon":

• It must be a subsequence of either string a OR string b
• It cannot be a subsequence of both strings simultaneously

Let's think through this problem:

• If the two strings a and b are identical (e.g., both are "abc"), then any subsequence of a will also be a subsequence of b, meaning no uncommon subsequence exists. In this case, return -1.
• If the two strings are different, then the entire string a itself is a subsequence of a but not of b (since they're different strings). Similarly, the entire string b is a subsequence of b but not of a. Therefore, both complete strings are uncommon subsequences, and we return the length of the longer one.

For example:

• If a = "aba" and b = "cdc", the longest uncommon subsequence has length 3 (either entire string qualifies)
• If a = "aaa" and b = "aaa", return -1 since no uncommon subsequence exists

Intuition

At first glance, this problem might seem complex because we need to consider all possible subsequences. However, let's think about what happens with the strings themselves.

The crucial observation is that if two strings are different, then each complete string is automatically an uncommon subsequence. Why? Because if a ≠ b, then:

• The string a is a subsequence of itself (by taking all characters)
• The string a cannot be a subsequence of b (since they're different complete strings)
• Similarly, b is a subsequence of itself but not of a

This means both a and b are uncommon subsequences when they're different. Since we want the longest one, we simply take max(len(a), len(b)).

What if a == b? In this case, any subsequence we can form from a can also be formed from b in exactly the same way (since they're identical strings). Therefore, no subsequence can belong to just one string - every subsequence belongs to both or neither. This means no uncommon subsequence exists, so we return -1.

The beauty of this solution is its simplicity: we don't need to generate or check any subsequences at all. The answer depends solely on whether the strings are equal:

• If a == b: return -1 (no uncommon subsequence exists)
• If a ≠ b: return max(len(a), len(b)) (the longer string itself is the longest uncommon subsequence)

This transforms what initially appears to be a complex subsequence problem into a simple string comparison.

Solution Approach

The implementation is remarkably straightforward once we understand the key insight about the problem.

The algorithm follows a simple decision tree:

1. Check if the strings are equal: Compare a and b directly
2. Return the appropriate result:
   • If a == b, return -1 (no uncommon subsequence exists)
   • Otherwise, return max(len(a), len(b)) (the length of the longer string)

Here's how the solution works step by step:

def findLUSlength(self, a: str, b: str) -> int:
    return -1 if a == b else max(len(a), len(b))

The implementation uses a single conditional expression (ternary operator):

• First, we check the condition a == b
• If true, we return -1 immediately
• If false, we calculate and return max(len(a), len(b))

Why this works:

• When strings are different, each complete string is a valid uncommon subsequence of itself but not of the other
• When strings are identical, every possible subsequence appears in both strings, making uncommon subsequences impossible

Time Complexity: O(min(m, n)) where m and n are the lengths of strings a and b. This is because string comparison takes linear time in the length of the shorter string.

Space Complexity: O(1) as we only use a constant amount of extra space.

The elegance of this solution lies in recognizing that we don't need to examine any actual subsequences - the answer depends entirely on whether the two strings are identical or not.

Example Walkthrough

Let's walk through two examples to illustrate the solution approach:

Example 1: Different strings

• Input: a = "aba", b = "cdc"
• Step 1: Check if a == b?
  • "aba" ≠ "cdc", so the strings are different
• Step 2: Since they're different, each complete string is an uncommon subsequence
  • "aba" is a subsequence of a (take all characters) but NOT of b
  • "cdc" is a subsequence of b (take all characters) but NOT of a
• Step 3: Return the length of the longer string
  • max(len("aba"), len("cdc")) = max(3, 3) = 3
• Output: 3

Example 2: Identical strings

• Input: a = "aaa", b = "aaa"
• Step 1: Check if a == b?
  • "aaa" == "aaa", so the strings are identical
• Step 2: Since they're identical, any subsequence we form from a can be formed the exact same way from b
  • For instance: "a" (taking first character) exists in both
  • "aa" (taking first two characters) exists in both
  • "aaa" (taking all characters) exists in both
  • No subsequence can belong to just one string
• Step 3: Return -1 since no uncommon subsequence exists
• Output: -1

The key insight: when strings differ, we don't need to check subsequences at all - the complete strings themselves serve as the longest uncommon subsequences!

Solution Implementation

Time and Space Complexity

The time complexity is O(n), where n is the length of the longer string among a and b. This is because the string comparison a == b requires iterating through both strings character by character, which takes O(min(len(a), len(b))) time in the worst case when the strings are equal. However, since we need to compare up to the length of the shorter string and potentially all characters match, this is effectively O(n) where n = max(len(a), len(b)).

The space complexity is O(1). The algorithm only uses a constant amount of extra space regardless of the input size, as it performs a comparison operation and returns either -1 or the maximum length without creating any additional data structures.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Overthinking the Problem

The most common pitfall is overcomplicating the solution by trying to generate and compare actual subsequences. Many developers initially attempt to:

• Generate all possible subsequences of both strings
• Compare them to find uncommon ones
• Track the longest uncommon subsequence

Why this happens: The term "subsequence" triggers algorithmic thinking about dynamic programming or recursive generation of subsequences.

Solution: Remember that when two strings are different, the strings themselves are valid subsequences that don't appear in each other. No complex subsequence generation is needed.

2. Misunderstanding "Uncommon Subsequence"

Some developers incorrectly interpret an uncommon subsequence as:

• A subsequence that appears a different number of times in each string
• A subsequence that has no common characters between the strings
• The longest subsequence that differs between the strings

Example of misunderstanding:

# INCORRECT: Trying to find subsequences with different frequencies
def findLUSlength(self, a: str, b: str) -> int:
    subsequences_a = generate_all_subsequences(a)
    subsequences_b = generate_all_subsequences(b)
    # Complex comparison logic...

Solution: An uncommon subsequence must exist in exactly ONE string, not both. If a ≠ b, then a is a subsequence of a but not of b.

3. Incorrect Handling of Equal Strings

Some implementations incorrectly return 0 or the length of the string when both strings are equal:

# INCORRECT implementations:
if a == b:
    return 0  # Wrong: should be -1
  
if a == b:
    return len(a)  # Wrong: should be -1

Solution: When strings are identical, return -1 as specified, since every subsequence of one string is also a subsequence of the other.

4. Edge Case Confusion

Developers sometimes add unnecessary edge case handling:

# UNNECESSARY complexity:
def findLUSlength(self, a: str, b: str) -> int:
    if not a and not b:  # Unnecessary check
        return -1
    if not a or not b:   # Unnecessary check
        return max(len(a), len(b))
    # ... rest of logic

Solution: The simple a == b check handles all edge cases correctly, including empty strings. If both are empty ("" == ""), it returns -1. If one is empty and the other isn't, it returns the length of the non-empty string.