Longest Uncommon Subsequence I - Video Solutions

Problem Overview: Given two strings a and b, return the length of the longest string that is a subsequence of one string but not the other. If both strings are identical, no such subsequence exists and the answer is -1.

Approach 1: Subsequence Verification (O(n + m) time, O(1) space)

The direct interpretation is to verify whether one string is a subsequence of the other. Use a two‑pointer scan: iterate through the larger string while advancing a pointer on the candidate subsequence when characters match. Perform this check twice—once to see if a is a subsequence of b and once for the reverse. If a is not a subsequence of b, then the entire string a itself becomes an uncommon subsequence. The same logic applies to b. This approach demonstrates understanding of subsequence matching using two pointers and basic string traversal.

Approach 2: Direct String Comparison (O(n) time, O(1) space)

The key observation simplifies the problem drastically. If the two strings are exactly equal, every subsequence of one will also appear in the other, so the answer must be -1. If they differ, the longer string itself cannot be a subsequence of the shorter one, which means the entire longer string is the longest uncommon subsequence. You only need a single equality check and a length comparison. This turns what initially looks like a subsequence problem into a simple comparison on string lengths.

Recommended for interviews: Interviewers expect the direct comparison insight. Many candidates initially attempt subsequence checks with two pointers, which works but adds unnecessary logic. Recognizing that identical strings are the only failing case leads to the optimal O(n) solution with constant space. Showing the subsequence verification approach first can demonstrate reasoning, but the direct comparison method shows strong problem‑solving intuition.