# Length of Longest Substring Without Repeating Characters

## Problem (from Leetcode)

Given a string, find the length of the longest substring without repeating characters.

## Examples

Given abcabcbb, the answer is abc, which the length is 3.

bbbbb, the answer is b, with the length of 1.

Given pwwkew, the answer is wke, with the length of 3. Note that the answer must be a substring, pwke is a subsequence and not a substring.

## Discussion

### Approach

1. Keep track of the substring characters in a Set.
2. Keep track of the evaluated ("seen") characters in a Queue.
3. Iterate through the characters in the input String.
• If the substring characters include the current character
1. Iterate through the head of the seen characters Queue until you find a match in the Queue.
2. For each non-matching character from the Queue, remove it from the Queue and remove it from the Set.
3. Once you find a matching character, remove it from the head of the Queue and add it to the tail of the Queue.
• Else
1. Add the current character to the Set
2. Add the current character to the tail of the Queue.
• If the size of the Set is greater than the current longest substring length, replace the current longest substring length with the Set size.

### General Questions

• Naming sucks - open to a better name - there's gotta be something better than LongestSubstringWithoutRepeatingCharactersLengthIdentifier...right?
• Is there a cleaner implementation that uses other data structures?

## Implementation

import java.util.HashSet;
import java.util.Queue;
import java.util.Set;

public class LongestSubstringWithoutRepeatingCharactersLengthIdentifier {
public static int identify(String s) {
int longestSubstringLength = 0;
Set<Character> substringCharacters = new HashSet<>();

for (char currentCharacter : s.toCharArray()) {
if (substringCharacters.contains(currentCharacter)) {
while (seenCharacters.peek() != currentCharacter) {
substringCharacters.remove(seenCharacters.poll());
}
seenCharacters.poll();
} else {
}

if (substringCharacters.size() > longestSubstringLength) {
longestSubstringLength = substringCharacters.size();
}
}
return longestSubstringLength;
}
}

• You asked about naming. Perhaps use max... instead of longest...? – rossum Oct 8 '17 at 20:02

Queue<Character> seenCharacters = new LinkedBlockingQueue<>();


java.util.concurrent.LinkedBlockingQueue is a concurrent data structure; I suggest you use java.util.ArrayDeque. Also, I think the better name would be characterWindow.

Opinion 1

Set<Character> substringCharacters = new HashSet<>();


I would rename the above to substringCharacterFilter.

Opinion 2

if (substringCharacters.size() > longestSubstringLength) {
longestSubstringLength = substringCharacters.size();
}


I think the above would be more readable like this:

if (longestSubstringLength < substringCharacterFilter.size()) {
longestSubstringLength = substringCharacterFilter.size();
}


Alternative implementation

public static int identify(String s) {
int longestSubstringLength = 0;
Set<Character> substringCharacterFilter = new HashSet<>();
Deque<Character> characterWindow = new ArrayDeque<>();

for (char currentCharacter : s.toCharArray()) {
if (substringCharacterFilter.contains(currentCharacter)) {
while (characterWindow.getFirst() != currentCharacter) {
substringCharacterFilter.remove(characterWindow.removeFirst());
}

characterWindow.removeFirst();
} else {
}

if (longestSubstringLength < substringCharacterFilter.size()) {
longestSubstringLength = substringCharacterFilter.size();
}
}

return longestSubstringLength;
}

1. LinkedBlockingQueue is not the best choice. It's based on a linked list, which lacks spacial locality. An continuous array-backed queue would be more efficient (there're a few options in the standard library).

2.  if (substringCharacters.size() > longestSubstringLength) {
longestSubstringLength = substringCharacters.size();
}


I'd use the Math.max method here.

3. You can avoid using the queue altogether. You can keep the leftmost index of the current "window" and the set instead.

        Queue<Character> seenCharacters = new LinkedBlockingQueue<>();


Why the concurrent version? If you really want this, just use the regular version.

        Queue<Character> seenCharacters = new LinkedList<>();


It's not like the method is reentrant. There's no state for threading to confuse.

I'm ignoring the question of whether LinkedList or ArrayDeque is better here. Because ...

The truth is that you don't need it. You're just duplicating the string. Instead, iterate over the string by index.

        for (int i = 0; i < s.length(); i++) {
char currentCharacter = s.charAt(i);


Also change the Set to a Map (or even an array).

        Set<Character> substringCharacters = new HashSet<>();


becomes

        Map<Character, Integer> lastSeen = new HashMap<>();


        int lastDuplicate = 0;


        if (s.isEmpty()) {
return 0;
}

int longestSubstringLength = 1;
int lastDuplicateIndex = -1;
Map<Character, Integer> lastSeen = new HashMap<>();

lastSeen.put(s.charAt(0), 0);

for (int i = 1; i < s.length(); i++) {
char current = s.charAt(i);

Integer lastIndex = lastSeen.get(current);
if (lastIndex != null && lastDuplicateIndex < lastIndex) {
lastDuplicateIndex = lastIndex;
}

lastSeen.put(current, i);

if (longestSubstringLength < i - lastDuplicateIndex) {
longestSubstringLength = i - lastDuplicateIndex;
}
}


I used a less Hungarian scheme for naming. I.e. no types in the names.

This uses the map to track where the current character was seen last. If that's after the last duplicate, point the last duplicate index to the place where the current character was seen last. That's the new last duplicate.

I tested this on pwwkew (3), abcabcab (3), and bbbbb (1).

• hmmm...I'm not sure this would work for aaa (which should return 1 but I think this method would return 3 due to the longestSubstringLength = i - lastDuplicateIndex for the first a). – Jae Bradley Oct 6 '17 at 21:17
• It works for bbbbb (returns 1). Note that it updates lastDuplicateIndex for each a. I did make a copying error that I've fixed now. – mdfst13 Oct 7 '17 at 1:02