LeetCode 521 - Longest Uncommon Subsequence I
The problem asks us to find the length of the longest uncommon subsequence between two strings a and b. To understand the problem clearly, we first need to understand what an uncommon subsequence means.
Difficulty: 🟢 Easy
Topics: String
Solution
Problem Understanding
The problem asks us to find the length of the longest uncommon subsequence between two strings a and b.
To understand the problem clearly, we first need to understand what an uncommon subsequence means. A subsequence is a sequence of characters that can be obtained from a string by deleting zero or more characters without changing the relative order of the remaining characters. For example, "ace" is a subsequence of "abcde".
An uncommon subsequence is a subsequence that appears in exactly one of the two strings. In other words, we want a string that can be formed as a subsequence of one input string but cannot be formed as a subsequence of the other.
The goal is to return the maximum possible length of such an uncommon subsequence. If no uncommon subsequence exists, we return -1.
The input consists of two lowercase English strings, a and b, each with length between 1 and 100. Since the strings are very short, even somewhat inefficient solutions could work, but there is an elegant observation that leads to an extremely simple optimal solution.
The key challenge is recognizing what the "longest" uncommon subsequence could be. A naive interpretation may suggest generating all subsequences and comparing them, but that quickly becomes expensive because each string of length n has 2^n subsequences.
Several edge cases are important to identify upfront:
- If both strings are exactly equal, then every subsequence of one is also a subsequence of the other. In this case, there is no uncommon subsequence, so the answer is
-1. - If the strings are different but have the same length, then the entire longer string itself is automatically an uncommon subsequence, because identical-length strings cannot be subsequences of each other unless they are equal.
- If the strings have different lengths, then the longer string itself must be an uncommon subsequence, because a shorter string cannot contain a longer string as a subsequence.
Approaches
Brute Force Approach
A brute force solution would generate all possible subsequences of both strings and compare them.
We could enumerate every subsequence of a, check whether it is absent from b, and track the longest valid candidate. Then we would repeat the same process for subsequences of b.
To verify whether a candidate subsequence belongs to the other string, we could use a two-pointer subsequence check. Since each string of length n has 2^n subsequences, this quickly becomes expensive.
For example, if a string has length 100, there are 2^100 possible subsequences, which is astronomically large. Even though the constraints are small, brute force is completely impractical.
The brute force method is correct because it explicitly checks every possible uncommon subsequence and chooses the longest valid one. However, its exponential complexity makes it infeasible.
Key Insight for the Optimal Approach
The crucial observation is surprisingly simple:
If the two strings are not equal, then the answer is always the length of the longer string.
Why does this work?
Suppose the strings are different.
- If one string is longer, the entire longer string itself cannot possibly be a subsequence of the shorter string. Therefore, it is automatically an uncommon subsequence, and since it is the entire string, it is also the longest possible candidate.
- If the strings have the same length but are different, then neither string can be a subsequence of the other unless they are identical. Therefore, either whole string itself becomes an uncommon subsequence.
The only time no uncommon subsequence exists is when the strings are exactly equal. In that case, every subsequence of one string also appears in the other.
| Approach | Time Complexity | Space Complexity | Notes |
|---|---|---|---|
| Brute Force | O(2^(n+m) × max(n,m)) | O(2^(n+m)) | Generates all subsequences and compares them |
| Optimal | O(1) | O(1) | Simple comparison based on equality and length |
Algorithm Walkthrough
Optimal Algorithm
- Compare the two input strings
aandb.
We first determine whether the strings are exactly identical. This check is essential because identical strings share all subsequences.
2. If a == b, return -1.
Since the strings are identical, every subsequence of a is also a subsequence of b, and vice versa. Therefore, there is no uncommon subsequence.
3. Otherwise, return max(len(a), len(b)).
If the strings differ, the longer string itself must be an uncommon subsequence. If they have the same length but differ, either full string works as an uncommon subsequence.
Why it works
The correctness relies on a simple property of subsequences:
A string can only be a subsequence of another string with the same length if the two strings are identical. Additionally, a longer string can never be a subsequence of a shorter one.
Therefore, when a != b, at least one entire string is guaranteed to be an uncommon subsequence, and choosing the longer one gives the maximum possible length. If a == b, no uncommon subsequence exists.
Python Solution
class Solution:
def findLUSlength(self, a: str, b: str) -> int:
if a == b:
return -1
return max(len(a), len(b))
The implementation follows the exact logic of the optimal algorithm.
The first condition checks whether the two strings are identical. If they are, the function immediately returns -1 because no uncommon subsequence can exist.
If the strings are different, the function returns the larger of the two lengths. This works because the longer string itself is guaranteed to be an uncommon subsequence. When both strings have equal length but differ, max(len(a), len(b)) simply returns that shared length.
The implementation is concise because the key insight eliminates the need for subsequence generation or comparison logic.
Go Solution
func findLUSlength(a string, b string) int {
if a == b {
return -1
}
if len(a) > len(b) {
return len(a)
}
return len(b)
}
The Go implementation mirrors the Python logic closely. Since Go does not provide a built-in max function for integers in older standard library versions, we manually compare the lengths using an if statement.
There are no concerns about integer overflow because the maximum string length is only 100. Empty strings are not possible due to the problem constraints, but the implementation would still behave correctly if they appeared.
Worked Examples
Example 1
Input: a = "aba", b = "cdc"
Since the strings are not equal, we return the maximum length.
| Step | a | b | Equal? | Result |
|---|---|---|---|---|
| 1 | "aba" |
"cdc" |
No | Continue |
| 2 | len(a)=3 |
len(b)=3 |
- | Return 3 |
The answer is 3.
One valid uncommon subsequence is "aba" because it appears in "aba" but not in "cdc".
Example 2
Input: a = "aaa", b = "bbb"
The strings differ, so we return the larger length.
| Step | a | b | Equal? | Result |
|---|---|---|---|---|
| 1 | "aaa" |
"bbb" |
No | Continue |
| 2 | len(a)=3 |
len(b)=3 |
- | Return 3 |
The answer is 3.
Both "aaa" and "bbb" are valid longest uncommon subsequences.
Example 3
Input: a = "aaa", b = "aaa"
The strings are identical.
| Step | a | b | Equal? | Result |
|---|---|---|---|---|
| 1 | "aaa" |
"aaa" |
Yes | Return -1 |
Since both strings are the same, every subsequence exists in both strings.
Complexity Analysis
| Measure | Complexity | Explanation |
|---|---|---|
| Time | O(1) | String comparison and length retrieval are constant time under fixed constraints |
| Space | O(1) | No extra data structures are used |
Although string comparison technically takes O(n) time in the general case, the input size is capped at 100, making it effectively constant for this problem. More formally, the runtime can be written as O(n) where n = max(len(a), len(b)).
Test Cases
solution = Solution()
assert solution.findLUSlength("aba", "cdc") == 3 # provided example, different strings
assert solution.findLUSlength("aaa", "bbb") == 3 # provided example, same length but different
assert solution.findLUSlength("aaa", "aaa") == -1 # provided example, identical strings
assert solution.findLUSlength("a", "b") == 1 # minimum length, different strings
assert solution.findLUSlength("a", "a") == -1 # minimum length, identical
assert solution.findLUSlength("abcd", "abc") == 4 # different lengths
assert solution.findLUSlength("abc", "abcd") == 4 # reversed length ordering
assert solution.findLUSlength("abc", "def") == 3 # same length, completely different
assert solution.findLUSlength("abc", "abd") == 3 # same length, one character difference
assert solution.findLUSlength("aaaa", "aaa") == 4 # repeated characters, different lengths
assert solution.findLUSlength("xyzxyz", "xyzxyz") == -1 # longer identical strings
| Test | Why |
|---|---|
("aba", "cdc") |
Validates provided example with different strings |
("aaa", "bbb") |
Validates same-length unequal strings |
("aaa", "aaa") |
Validates identical strings return -1 |
("a", "b") |
Tests minimum input size |
("a", "a") |
Tests smallest identical case |
("abcd", "abc") |
Tests different lengths |
("abc", "abcd") |
Ensures ordering does not matter |
("abc", "def") |
Tests unrelated same-length strings |
("abc", "abd") |
Tests minor difference case |
("aaaa", "aaa") |
Tests repeated characters |
("xyzxyz", "xyzxyz") |
Tests larger identical strings |
Edge Cases
Identical Strings
When a and b are exactly the same, every subsequence in one string automatically exists in the other. A naive implementation might mistakenly return the string length because it sees a valid subsequence candidate, but this would be incorrect. The implementation explicitly checks a == b first and returns -1.
Different Lengths
If one string is longer than the other, the entire longer string is guaranteed not to fit inside the shorter string as a subsequence. A buggy solution might overcomplicate the logic by trying to compare subsequences unnecessarily. The implementation correctly handles this by directly returning the longer length.
Same Length but Different Strings
This case can be unintuitive. Two strings of equal length may look similar, but if they differ even by one character, neither full string can be a subsequence of the other. A naive implementation might incorrectly attempt character-by-character matching and miss this insight. The solution handles this naturally by returning the shared length whenever the strings are not equal.