Find longest unique-character substring

Given a string s, return the length of the longest substring without repeating characters. Explain and implement an O(n) sliding-window solution using a hash map or set. Discuss time and space complexity, and note any changes needed to correctly handle Unicode characters.

Find longest unique-character substring

Given a string s, return the length of the longest substring of s that contains no repeating characters.

A substring is a contiguous run of characters within the string. Implement an O(n) sliding-window solution using a hash map (character -> last seen index) or a set. As the right edge of the window expands one character at a time, advance the left edge just past the previous occurrence of any character that would otherwise repeat, and track the maximum window width seen.

Note on Unicode: iterating in Python over str already yields Unicode code points, so the algorithm is correct as written for the basic multilingual characters most inputs use. For full correctness over user-perceived characters (grapheme clusters such as emoji with combining marks or skin-tone modifiers), you would segment on grapheme boundaries rather than code points before applying the same window logic.

Return 0 for an empty string.

Constraints

• 0 <= len(s) <= 5 * 10^4
• s may contain letters, digits, symbols, and spaces (any Unicode code points)
• Return 0 for the empty string

Examples

Example 1

Input: ("abcabcbb",)
Output: 3

Explanation: The answer is "abc", length 3.

Example 2

Input: ("bbbbb",)
Output: 1

Explanation: The answer is "b", length 1 — every character repeats.

Hints

1. Use two pointers forming a window [start, i]; expand i one step at a time and shrink from the left only when you hit a repeat.
2. Store the last index at which each character was seen so you can jump start directly past the previous occurrence instead of stepping one-by-one.
3. Guard the jump with last_seen[ch] >= start — a stale occurrence that is already outside the window (e.g. the second 'a' in "abba") must not move start backward.