Find longest unique substring

Given a string s, find the length and one example of the longest substring of s that contains no repeated characters. If multiple answers exist, return any valid one. Explain your algorithm, prove its correctness at a high level, analyze time and space complexity, and discuss edge cases (e.g., empty string, all identical characters, presence of Unicode code points).

What a Strong Answer Covers

• A clear algorithm with the right data structures and enough pseudocode or code-level detail to implement it.
• A correctness argument that explains why the algorithm covers all required cases.
• Time and space complexity, plus at least one alternative approach when relevant.
• Focused tests for normal cases, edge cases, and failure modes.

Follow-up Questions

• How would the approach change if the input were streaming or too large for memory?
• What invariants would you assert in production code?
• Which tests would catch off-by-one, duplicate, or tie-breaking bugs?

Find Longest Unique Substring

Given a string s, find the length and one example of the longest substring of s that contains no repeated characters. Return the result as a two-element list [length, example_substring]. If multiple substrings achieve the maximum length, return any one of them.

A substring is a contiguous run of characters within the string. Treat the empty string as having a longest-unique-substring length of 0 with an empty example.

Examples

• s = "abcabcbb" -> [3, "abc"] (the answer "bca" or "cab" would also be acceptable in length).
• s = "bbbbb" -> [1, "b"].
• s = "pwwkew" -> [3, "wke"] (note "pwke" is a subsequence, not a substring, so it does not count).
• s = "" -> [0, ""].

Constraints

• 0 <= len(s) <= 10^5
• s may contain any Unicode code points (ASCII letters, digits, symbols, spaces).
• Characters are compared by exact code-point equality (case-sensitive: 'A' != 'a').
• If s is empty, return [0, ""].

Hints

1. Use a sliding window with two pointers: a left boundary start and a right boundary i scanning the string.
2. Keep a hash map from each character to the most recent index where you saw it. When the current character was last seen at index >= start, jump start to one past that index to drop the duplicate from the window.
3. Track the best (longest) window length and its start index as you go, so you can slice out one example substring at the end. The left pointer never moves backward, which is why the whole scan is O(n).

Solution

Solution Alignment

The prompt asks for an implementation-level answer. The safest way to present it is to define the state, maintain clear invariants, then walk through complexity and tests.

Problem Restatement

Given a string s, find the length and one example of the longest substring of s that contains no repeated characters. If multiple answers exist, return any valid one. Explain your algorithm, prove its correctness at a high level, analyze time and space complexity, and discuss edge cases (e.g., empty string, all identical characters, presence of Unicode code points).

Recommended Approach

Choose traversal based on the required output. DFS is natural for subtree computations, reconstruction, and range pruning; BFS is natural for level order or side views. Keep per-depth or per-position state when the output depends on columns, rows, or depths.

Correctness

The implementation should maintain an invariant after each loop or operation that directly matches the problem statement. At termination, that invariant implies the returned value has considered every valid candidate exactly once, or has preserved the required data-structure state after every API call.

Complexity

Most tree traversals are O(n) time and O(h) recursion stack for DFS or O(w) queue space for BFS, where h is height and w is maximum width.

Edge Cases and Tests

Empty tree, one node, skewed tree, duplicate values when reconstruction assumes uniqueness, deep recursion, and tie-breaking for same row/column nodes.