How can the method return the longest substring as a String?

Change the return type to String and add int start = 0; beside best in the last-seen version. In place of the Math.max call write if (right - left + 1 > best) { start = left; best = right - left + 1; }, and finish with return s.substring(start, start + best);, remembering that the second argument of substring is exclusive. You get etcar for "streetcar" and world, leading space and all, for "hello world". Because the test is a strict >, a later window of equal length never replaces an earlier one, so "noon" returns no.

Does charAt handle accented letters and emoji?

Accented letters, yes; emoji, not always. Each charAt call returns one UTF-16 code unit. é and ☕ fit in a single unit, so for "café☕é" the char version returns 5 (café☕). An emoji like 😀 lies outside that range and is stored as a surrogate pair, two units: "😀😁".length() is 4, and the char version returns 3, a window that splits an emoji down the middle. To work in code points, take int[] cps = s.codePoints().toArray(), key a HashMap by code point and run the same loop over cps; that gives 5 for "café☕é" and 2 for "😀😁".

Is an int[128] or int[65536] table better than a HashMap?

Faster, since there is no hashing and no boxing, but it fixes the alphabet in advance. Store right + 1 so that the default 0 of a new int[] means not seen; the loop body becomes left = Math.max(left, last[c]); last[c] = right + 1; plus the best update. An int[128] is ASCII only: "café" throws ArrayIndexOutOfBoundsException with the message Index 233 out of bounds for length 128. An int[65536] has a slot for every char and returns 5 for "café☕é", yet allocates 256 KB per call even for a four-letter word. Pick int[128] for known-ASCII input and the HashMap when the alphabet is open or most strings are short.

Why is prev an Integer and not an int?

Because HashMap.get returns null for a missing key, and only a reference type can hold null. Writing int prev = lastSeen.get(c); compiles, but the first unseen character makes Java unbox null and throw a NullPointerException. The same would happen if prev >= left ran before the null check, which is why prev != null comes first. If you prefer a primitive, int prev = lastSeen.getOrDefault(c, -1); works: -1 is never >= left, so one comparison covers both cases.

How does the window change for at most k distinct characters?

The window shape stays; only the shrink rule changes. Count characters in a HashMap and add the incoming one with counts.merge(s.charAt(right), 1, Integer::sum). While counts.size() is above k, lower the count of s.charAt(left) with merge(out, -1, Integer::sum), remove the key once it reaches zero so size() reflects the window, and advance left. "mississippi" with k = 2 gives 7 ("ississi"). To allow each character up to k times instead, shrink while the count of the newly added character exceeds k.

Longest Substring Without Repeating Characters in Java

length_of_longest_substring.java

Java 25.0

Not run yet

+ 8 hidden

Input

s = "streetcar"

Expected

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