Longest Substring without repeating characters issue with edge case - string

Question

I was trying to solve this problem: Longest substring without repeating characters. The issue is, it's failing in couple test cases, I don't know how to fix it. I would need your help to see where I'm going wrong.
Question:
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.
Given "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.
This is my code:
function longestSubString(arr){
let localSum=0,globalSum=0;
let set = new Set();
for(let i=0; i<arr.length; i++){
let current = arr[i];
//if the key is present in the store.
if(set.has(current)){
set.clear();
localSum = 1;
set.add(current);
} else {
localSum +=1;
set.add(current);
}
if(globalSum < localSum){
globalSum = localSum;
}
}
return globalSum;
}
Tests:
let test = "abcabc"; //returns 3 - correct
let test2 = "bbb"; //returns 1 - correct
let test5 = "dvdf"; //returns 2 - INCORRECT! it should return 3 (i.e for vdf) since I'm doing set.clear() I'm not able to store previous elements.
longestSubString(test5); //incorrect
Live:
https://repl.it/Jo5Z/10

Answer 1

Not fully tested!
function longestSubString(arr){
let localSum=0,globalSum=0;
let set = new Set();
for(let i=0; i<arr.length; i++){
let current = arr[i];
//if the key is present in the store.
if(set.has(current)){
let a = Array.from(set);
a.splice(0,a.indexOf(current)+1);
set = new Set(a);
set.add(current);
localSum = set.size;
} else {
localSum +=1;
set.add(current);
}
if(globalSum < localSum){
globalSum = localSum;
}
}
return globalSum;
}
The idea is that when you get duplicate, you should start from the charachter after the first duplicated character, in your case dvdf, when you reach the second d you should continue from vd not from d!

Answer 2

You have to consider that the substring might start from any character in the string. Erasing the set only when you're finding a duplicate makes you only consider a substring starting from characters that are equal to to the first character.
An O(logn*n^2) solution modifying yours just a bit:
function longestSubString(arr){
let globalSum=0;
for(let i=0; i<arr.length; i++){
let set = new Set();
let localSum=0;
for(let j=i; j<arr.lenght; j++){
let current = arr[j];
//if the key is present in the store.
if(set.has(current)){
break;
} else {
localSum +=1;
set.add(current);
}
}
if(globalSum < localSum){
globalSum = localSum;
}
}
return globalSum;
}
There's also a O(n + d) (almost linear) solution, d being the number of characters in the alphabet. See http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/.

Answer 3

There seem to be a lot of long answers here. The implementation I've thought of is simplified due to two observations:
Whenever you encounter a duplicate character, you need to start the next substring just after the previous occurrence of the current character.
Set() creates an array in insertion order when iterated.
function longestSubstring(str) {
let maxLength = 0
let current = new Set()
for (const character of str) {
if (current.has(character)) {
const substr = Array.from(current)
maxLength = Math.max(maxLength, substr.length)
current = new Set(substr.slice(substr.indexOf(character) + 1))
}
current.add(character)
}
return Math.max(maxLength, current.size)
}
const tests = [
"abcabc",
"bbb",
"pwwkew",
"geeksforgeeks",
"dvdf"
]
tests.map(longestSubstring).forEach(result => console.log(result))
A simple edit allows us to keep the first occurrence of the largest substring instead of the maximum length.
function longestSubstring(str) {
let maxSubstr = []
let current = new Set()
for (const character of str) {
if (current.has(character)) {
const substr = Array.from(current)
maxSubstr = maxSubstr.length < substr.length ? substr: maxSubstr
current = new Set(substr.slice(substr.indexOf(character) + 1))
}
current.add(character)
}
const substr = maxSubstr.length < current.size ? Array.from(current) : maxSubstr
return substr.join('')
}
const tests = [
"abcabc",
"bbb",
"pwwkew",
"geeksforgeeks",
"dvdf"
]
tests.map(longestSubstring).forEach(result => console.log(result))
As we can see, the last test yields vdf, as expected.

Answer 4

Below solution gets the length in O(n+d) time and also prints the longest non repeating character substring as well:
public void longestNonRepeatingLength(String a){
a="dvdf";
int visitedIndex[] = new int[256];
int curr_len = 0, max_len = 0, prev_ind = 0, start = 0, end = 1;
for(int i =0;i<256;i++)
visitedIndex[i] = -1;
visitedIndex[a.charAt(0)] = 0;
curr_len++;
int i = 0;
for( i=1;i<a.length();i++){
prev_ind = visitedIndex[a.charAt(i)];
if(prev_ind == -1 || i > prev_ind + curr_len)
curr_len++;
else{
if(curr_len>max_len){
start = prev_ind + 1;
end = i;
max_len = curr_len;
}
curr_len = i - prev_ind;
}
visitedIndex[a.charAt(i)] = i;
}
if(curr_len>max_len){
end = i-1;
max_len = curr_len;
}
for( i = start;i<=end;i++)
System.out.print(a.charAt(i));
System.out.println("");
System.out.println("Length = "+max_len);
}

Answer 5

As set contains the largest set of non-repeating characters of a String ending on index i, this means that when you encounter a previously seen character, rather than starting over with an empty set as your codes does now, you should just remove all characters from your set until that duplicate one.
Say for example your input is "abXcXdef". When the second "X" is encountered, you'll want to drop "a" and "b" from your set, leaving a set of ("c","X") as the longest set up to that point. Adding all other characters (as none are duplicates) you then end up with a max length of 5.
Something like this should work:
function longestSubString(arr) {
let globalSum = 0;
let set = new Set();
for (let i=0; i<arr.length; i++) {
let current = arr[i];
if (set.has(current)) {
while (true) {
let removeChar = arr[i - set.count];
if (removeChar != current)
set.remove(removeChar);
else
break;
}
} else {
set.add(current);
if (set.count > globalSum)
globalSum = set.count;
}
}
return globalSum;
}
As every character is added at most once and deleted at most once, this is an O(N) algorithm.