Why can I not write s[i] on a Swift String?

A String is indexed by String.Index, not Int, because characters have different byte lengths and finding the i-th one means walking from the start. s[s.index(s.startIndex, offsetBy: i)] works but is O(i) per access, which turns the window into O(n²). Copying once with Array(s) gives a [Character] with O(1) subscripts.

What counts as one Character in Swift?

A Character is a whole grapheme cluster, what a reader sees as one symbol. "😀😁".count is 2, an emoji with a skin-tone modifier is one Character, and "e\u{301}" (e plus a combining accent) is one Character that compares equal to the precomposed "é". Other languages count code units or code points instead; for the ASCII test inputs they all agree.

What does insert(_:) return?

A tuple (inserted: Bool, memberAfterInsert: Element). inserted is false when an equal element was already in the set, so !seen.insert(x).inserted tests and inserts in one call. The result is marked @discardableResult, which is why a plain seen.insert(x) compiles without a warning.

Is Array(s.utf8) faster than Array(s)?

Yes. Array(s.utf8) gives a [UInt8], and bytes are much cheaper to hash and compare than grapheme clusters; with a fixed table var last = [Int](repeating: 0, count: 256) storing right + 1 there is no hashing at all. The catch is that it counts UTF-8 bytes, so it is only right when the input is ASCII, as it is here.

Do I need import Foundation for this problem?

No. Array, Set, Dictionary, max and enumerated() all come from the standard library. Importing Foundation would also work but adds startup time to every Run on this site.

Longest Substring Without Repeating Characters in Swift: Brute Force and Sliding Window Solutions

length_of_longest_substring.swift

Swift 6.2

Not run yet

+ 8 hidden

Input

s = "streetcar"

Expected

The double e breaks the word up. After it, "etcar" has five different characters.