LeetCode – Longest Substring Without Repeating Characters (Java)
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.
Java Solution 1
The first solution is like the problem of "determine if a string has all unique characters" in CC 150. We can use a flag array to track the existing characters for the longest substring without repeating characters.
public int lengthOfLongestSubstring(String s) { if(s==null) return 0; boolean[] flag = new boolean[256]; int result = 0; int start = 0; char[] arr = s.toCharArray(); for (int i = 0; i < arr.length; i++) { char current = arr[i]; if (flag[current]) { result = Math.max(result, i - start); // the loop update the new start point // and reset flag array // for example, abccab, when it comes to 2nd c, // it update start from 0 to 3, reset flag for a,b for (int k = start; k < i; k++) { if (arr[k] == current) { start = k + 1; break; } flag[arr[k]] = false; } } else { flag[current] = true; } } result = Math.max(arr.length - start, result); return result; } |
Java Solution 2 - HashSet
public int lengthOfLongestSubstring(String s) { if(s==null || s.length()==0) return 0; HashSet<Character> set = new HashSet<Character>(); int max=0; int i=0; int start=0; while(i<s.length()){ char c = s.charAt(i); if(!set.contains(c)){ set.add(c); }else{ max = Math.max(max, set.size()); while(start<i&&s.charAt(start)!=c){ set.remove(s.charAt(start)); start++; } start++; } i++; } max = Math.max(max, set.size()); return max; } |
or
public int lengthOfLongestSubstring(String s) { if(s==null || s.length()==0){ return 0; } int start=0; int max = 0; HashSet<Character> set = new HashSet<Character>(); for(int i=0; i<s.length(); i++){ char c = s.charAt(i); if(!set.contains(c)){ set.add(c); max = Math.max(max, i-start+1); }else{ for(int j=start; j<i; j++){ set.remove(s.charAt(j)); if(s.charAt(j)==c){ start=j+1; break; } } set.add(c); } } return max; } |
<pre><code> String foo = "bar"; </code></pre>
Leave a comment
Here is my solution with hash:
public int solution(String s) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
char[] target = s.toCharArray();
int pre = 0;
Hashtable h = new Hashtable();
for(int i =0; i h.size() ? pre : h.size();
i = h.get(target[i]);
h.clear();
}
}
return Math.max(pre, h.size());
}
public static int LongestSubStrWithUniqueChars(String s){
if(s==null || s.isEmpty())
return 0;
char[] chars = s.toCharArray();
int current = 0;//current pos in string
int length = 0; //current longest substr length
int pos = 0; //current longest substr starting pos
for(current =0 ; current<s.length(); current++){
pos = current;
boolean[] found = new boolean[256];
while(current<s.length() && !found[chars[current]]){
found[chars[current]] = true;
current++;
}
length = Math.max(length, current – pos);
current = pos;
}
return length;
}
string longestStr = string.Empty;
int length = str.Length;
int i = 0;
Loop:
string tempLongestStr = string.Empty;
for (int j = i; j longestStr.Length)
{
longestStr = tempLongestStr;
}
i = i + tempLongestStr.IndexOf(temp) + 1;
goto Loop;
}
else
{
tempLongestStr += temp;
if (tempLongestStr.Length > longestStr.Length)
{
longestStr = tempLongestStr;
}
}
}
Console.WriteLine(longestStr);
Console.ReadLine();
It’s the same.
good solution, thanks.
this solution has time limited issue in leetcode.
Tia’s solution is good.
I have the same solution but something different.
The time : O(n)
The space may be : O(n)
and it can output the maxlength substring. (you need to delete the comment tag, and run it)
like this:
public static int lengthOfLongestSubstring(String s) {
char[] arr = s.toCharArray();
int maxlength = 0;
int oldstart = 0;
int newstart = 0;
// int compare_times = 0; // compare times
HashMap num = new HashMap(); // recode char and the char index
HashMap map = new HashMap(); // recode index and the index of char
for (int i = 0; i < arr.length; i++) {
// compare_times++; //
if (num.get(arr[i]) != null) {
int end = num.get(arr[i]) + 1;
for (int j = newstart; j < end; j++)
{ // compare times will less than s.length()
num.remove(map.get(j));
map.remove(j);
}
newstart = end;
if (maxlength < num.size()) {
maxlength = num.size();
oldstart = newstart;
}
}
num.put(arr[i], i);
map.put(i, arr[i]);
}
if (maxlength < num.size()) {
maxlength = num.size();
oldstart = newstart;
}
// System.out.println("compare_times:" + compare_times); // output the compare times
// System.out.println(s.substring(oldstart, oldstart + maxlength)); // output the longest substring
return maxlength;
}
Updated Tia solution to
1. Find all patterns with no repeating characters.
2. Find the longest pattern and its size.
public static void printPatternWithNoReaptingCharsInString(String s) {
char[] arr = s.toCharArray();
Map charMap = new HashMap();
int size = 0;
List uniquePatterns = new ArrayList();
StringBuilder str = new StringBuilder();
for (int i = 0; i < arr.length; i++) {
if (!charMap.containsKey(arr[i])) {
charMap.put(arr[i], i);
} else {
if (size <= charMap.size()) {
size = charMap.size();
if (str.length() <= size) {
str.delete(0, str.length()).append(
s.substring(i – size, i));
uniquePatterns.add(str.toString());
}
}
i = charMap.get(arr[i]);
charMap.clear();
}
}
System.out.println("Longest Pattern with no repeating characters:"
+ str.toString() + " with size " + size);
System.out.println("There are in total " + uniquePatterns.size()
+ " patterns with no repeating characters.");
}
Below method call:
printUniquePatterninString("abcdaaabcdaabcdebbpqrstuvaaa");
will print:
Longest Pattern with no repeating characters:bpqrstuva with size 9
There are in total 6 patterns with no repeating characters.
Solution 2 cannot pass the OJ because of the Time limit.
My solution but no idea why it is not correct:
public class Solution {
public int lengthOfLongestSubstring(String s) {
// if the string is empty or has only one character
if (s.length() < 2) return s.length();
HashMap hashMap = new HashMap();
int j = 0;
int maxLen = 1;
for (int i = 0; i < s.length(); i++) {
if (!hashMap.containsKey(s.charAt(i))) {
hashMap.put(s.charAt(i), i);
} else {
maxLen = Math.max(maxLen, i – j);
j = hashMap.get(s.charAt(i)) + 1;
hashMap.put(s.charAt(i), i);
}
}
return Math.max(maxLen, s.length() – j);
}
}
Hi guys My solution simple and works fine.
public String getLongestSubStringWithoutRepeatedChar(String str){
ArrayList list=new ArrayList();
String longestSoFar=””;
int count=0;
for(int i=0;i<str.length();i++){
if(!list.contains(str.charAt(i))){
list.add(str.charAt(i));
}else{
if(count<list.size()){
count=list.size();
longestSoFar=getLongestString(list);
list.clear();
}
}
}
return longestSoFar;
}
private String getLongestString(ArrayList list) {
StringBuilder sb=new StringBuilder();
for(Character c:list){
sb.append(c);
}
return sb.toString();
}
whats the need for ” i = h.get(target[i]); ” ? It should work without it
I have written a much simpler version than version 2 (python).. easier to understand. its here, http://ideone.com/BHJlvS
This doesn’t work if you pass “abcadeftgh” to your method. It should output 9 instead of 6
I found the mistake, see this http://ideone.com/WMXFOw
the LIS for “abcadeftgh” is 7
abcadeftgh ==> bcadeftgh (this is the longest substring without repeating element, and its length is 9)
I am sorry.. i misunderstood LIS.. this http://ideone.com/hcLYgy should fix it
How about this O(n) solution? It’s easy to understand and allows for the retrieval of the longest substring as well:
public static int lengthOfLongestSubstring(String word) {
int start = 0;
int end = 0;
int maxStart = 0;
int maxEnd = 0;
int[] seen = new int[256];
for (int i = 0; i < 256; i++) {
seen[i] = -1;
}
for (int i = 0; i = start) {
start = seen[c] + 1;
}
end++;
seen[c] = i;
if (end – start > maxEnd – maxStart) {
maxEnd = end;
maxStart = start;
}
}
// Return word.substring(maxStart, maxEnd) for actual substring
return maxEnd – maxStart;
}
Does not work for string “eadsabcadcfef”
corrected one:
public String getLongestSubStringWithoutRepeatedChar (String str) {
{
List list = new ArrayList();
String longestSoFar =””;
int count = 0;
int longestSize=0;
for (int i = 0; i longestSize) {
longestSize=count;
longestSoFar = getLongestString(list);
System.out.println(longestSoFar);
}
count=0;
list.clear();
}
}
return longestSoFar;
}
}
private String getLongestString (List list) {
StringBuilder sb = new StringBuilder();
for (Character c : list) {
sb.append(c);
}
return sb.toString();
}
this code handles:
public String getLongestSubStringWithoutRepeatedChar (String str) {
{
List list = new ArrayList();
String longestSoFar =””;
int count = 0;
int longestSize=0;
//str.length()+1 for capturing longest string at the end.
for (int i = 0; i < str.length()+1; i++) {
if (i longestSize) {
longestSize=count;
longestSoFar = getLongestString(list);
System.out.println(longestSoFar);
}
count=0;
list.clear();
}
}
return longestSoFar;
}
}
private String getLongestString (List list) {
StringBuilder sb = new StringBuilder();
for (Character c : list) {
sb.append(c);
}
return sb.toString();
}
Proposed solution could be improved. You don’t need to clear the whole map and change i.
private static String calculateFast(String text) {
Map map = new HashMap();
int maxStart = 0;
int maxEnd = 1;
int currentStart = 0;
for (int i = 0; i maxEnd – maxStart) {
maxStart = currentStart;
maxEnd = i;
}
currentStart = firstOccurrenceIndex + 1;
}
map.put(c, i);
}
if (text.length() – currentStart > maxEnd – maxStart) {
maxStart = currentStart;
maxEnd = text.length();
}
return text.substring(maxStart, maxEnd);
}
this is simpler.
public static lls(String s) {
HashMap map = new HashMap();
int maxStart = 0;
int pre = 0;
for(int i = 0; i pre) {
pre = map.size();
maxStart = map.get(c);
}
}
map.put(c, i);
}
return s.substring(maxStart, maxStart+pre);
}
Hey guys,
Isn’t the best solution just a simple for loop with an if and some counters? Perhaps my solution is too simple and crashes or returns the results incorrectly somehow??? I tested it for the longest string and the beginning/end/middle and I believe it works.
public static int getNoRepeatLen (String str) {
int max=0,tLen=0;
for(int i=1;i<str.length();i++) {
if(str.charAt(i-1) == str.charAt(i)){
max=Math.max(tLen+1,max);
tLen=0;
} else {
tLen++;
}
}
return Math.max(max,tLen);
}
public static int getNoRepeatLen (String str) {
int max=0,tLen=0;
for(int i=1;i<str.length();i++) {
if(str.charAt(i-1) == str.charAt(i)){
max=Math.max(tLen+1,max);
tLen=0;
} else {
tLen++;
}
}
return Math.max(max,tLen);
}
Hi Ryan,
I like your site a lot, but I noticed that the recently-added social area on every post is blocking the sight.
Hopefully this can be fixed and I can enjoy reading your site again.
So I came up with an O(n) solution in python, please help me detect in what cases it can possibly fail
#put test string here
string = “eadsabcadcfef”
indexOfSubsequence = 0
tempIndexOfSubsequence = 0
lenghtOfSubsequence = 0
tempLenghtOfSubsequence = 0
tableOfCharsIndex = [-1] * 26
for i in xrange(0, len(string)):
indexOfChar = ord(string[i]) – 97
if tableOfCharsIndex[ indexOfChar ] == -1 or tableOfCharsIndex[ indexOfChar ] lenghtOfSubsequence :
lenghtOfSubsequence = tempLenghtOfSubsequence
indexOfSubsequence = tempIndexOfSubsequence
else:
if tempLenghtOfSubsequence > lenghtOfSubsequence :
lenghtOfSubsequence = tempLenghtOfSubsequence
indexOfSubsequence = tempIndexOfSubsequence
tempIndexOfSubsequence = i
tempLenghtOfSubsequence = 1
tableOfCharsIndex[ indexOfChar ] = i
print string[ indexOfSubsequence : indexOfSubsequence + lenghtOfSubsequence ]
print lenghtOfSubsequence
how is solution two O(n^3)?
I think Following is O(n) Solution 😀
public static int lengthOfLongestSubstring(String s) {
char[] arr = s.toCharArray();
int pre = 0;
int k=0;
boolean flag = true;
HashMap map = new HashMap();
for (int i = 0; i pre)
pre=l;
k = map.get(arr[i]);
map.put(arr[i],i);
k++;
}
}
if(flag) pre = arr.length;
return pre;
}
“bbabceabcdbb” fails with this string returns a wrong string
Your code only compares the two consecutive characters, it can not handle case like “a”.
Thanks. I think the problem is fixed now.
Your code is not right, for example, “aab”.
You solution is wrong. maxStart is not updated correctly.
I think the following works too:
static public String nonRepeated(String input){
char[] charAInput = input.toCharArray();
String aux = “”;
String resultSubStr = “”;
if(input.length() == 1)
return input;
for(int i =0; i < charAInput.length; i++ ){
if(!aux.contains("" + charAInput[i])){
aux = aux.concat("" + charAInput[i]);
}
else
aux = "" + charAInput[i];
if(resultSubStr.length() < aux.length())
resultSubStr = aux;
}
System.out.println(resultSubStr);
return resultSubStr;
}
No, it doesn’t. For example, given “abac”, when it meets the second “a”, aux should not be set to “a”, instead it should be set to “ba”.
What is the time and space complexity of first solution?
Better solution: http://www.geeksforgeeks.org/length-of-the-longest-substring-without-repeating-characters/
For the solution using the HashMap, I think the following line :
i = map.get(arr[i]);
should be changed to :
i = map.get(arr[i])+1;
Otherwise you will get an infinite loop ?
can you explain why second one in On3 .. i cant get it … thanks
I believe this is O(n):
public static int lengthOfLongestSubstring(String s) {
if(s==null)
return 0;
char[] arr = s.toCharArray();
int pre = 0;
HashMap map = new HashMap();
int j = 0;
int curr = 0;
for (int i = 0; i < arr.length; i++) {
if (!map.containsKey(arr[i]) || map.get(arr[i]) < j) {
map.put(arr[i], i);
curr++;
} else {
pre = Math.max(pre, curr);
j = map.get(arr[i]) + 1;
curr = i + 1 – j;
map.put(arr[i], i);
}
}
return Math.max(pre, curr);
}
I don’t think you need add the statement: ‘map.get(arr[i]) < j'
A simpler O(n) solution.
public int lengthOfLongestSubstring(String s) {
HashMap map = new HashMap();
if (s == null || s.length() == 0) return 0;
int first = 0, second = -1, answer = 0;
while (first != s.length()) {
Integer previousOccurrence = map.put(s.charAt(first), first);
if (previousOccurrence != null) second = Math.max(second, previousOccurrence);
answer = Math.max(answer, first – second);
first++;
}
return answer;
}
second = -1 is pretty weird.
Thanks. Agree. Updated the code with some refactoring 🙂 Intention was to initialize it one prior to the rightPointer. Comments are welcome!
public static String uniqueCharSubstring(String str) {
String result = “”;
int len = str.length();
HashMap map = new HashMap();
char[] c = str.toCharArray();
int right = 0, max = 0;
for (int left = 0; left < len; left++) {
while (right max) {
max = right – left;
result = str.substring(left, right + 1);
}
right++;
}
}
return result;
}
List list = new ArrayList();
for(int i=0 ; i<a.length();i++){
if(!list.contains(a.charAt(i))){
list.add(a.charAt(i));
}
else
break;
}
System.out.println("The String Lengtht" +a.length());
It is required. Consider the case “dabcabcde”. If ‘map.get(arr[i]) < j' is not done then, for the iteration with "i=7" it would be computed that "d" has already contained in the HashMap and also the logic inside the "else" will fail and compute to give 'curr=7'.
void LargestNonRepeatedSubStr() {
string str = "abcabcbb";
int max_length = 0;
int i = 0;
while (i < str.length()) {
set check_set;
int j = i;
int length = 0;
while (j < str.length()) {
if (check_set.find(str[j]) != check_set.end()) {
break;
} else {
check_set.insert(str[j]);
length++;
}
j++;
}
check_set.clear();
max_length = std::max(max_length, length);
i++;
}
cout << "The max length is: " << max_length << "n";
}
Why time complexity of second approach is O(n^3)?
O(n) solution with LinkedHashSet JAVA !
public class LongestPalindromeWithoutRepChar {
public static int lengthOfLongestSubstring(String s) {
int maxCount=0;
if(s.length() < 2) return s.length();
Set hs = new LinkedHashSet();
for(int i=0;i maxCount) maxCount = hs.size();
Iterator itr = hs.iterator();
Character curr = null;
while(c!=curr)
{
curr = itr.next();
itr.remove();
}
}
}
return maxCount;
}
}
public class LCSWIthoutRepeating{
public static void main(String args[]){
String str = “aaaaapritikamehta”;
LinkedHashMap hash = new LinkedHashMap();
int maxLen =0;
int max =0;
int num =0;
int arr[] = new int[26];
for(int i =0;i<str.length();i++){
char c = str.charAt(i);
if(hash.get(c) == null){
hash.put(c, i);
max++;
}else{
maxLen = i – hash.get(c)max?maxLen:max);
}
}
Why is the time complexity mentioned as n^3 for Java Solution 2 above? It should be O(n). The first loop is O(n) and in that main loop i is getting pushed back to an already visited location. So effectively we may end up looping through the whole loop 2n times. That still means O(n), then why n^3 is mentioned.
Please help!
Following works and it also prints the substring
Here is my O(N) time solution in Java http://www.capacode.com/string/longest-substring-without-repeating-characters/
Following code is O(N) time complexity and constant space complexity O(1).
public int LongestSubstringWithoutRepeatingCharacters(String src)
{
if (src == null || src.Length == 0)
return 0;
// v[j] stores i position of the src[i] character, where j = src[i]
int[] v = new int[256];
// Initialize visited state
for (int i = 0; i < 256; ++i)
v[i] = -1;
v[src[0]] = 0;
// start position of the longest string of non repeating characters
int start = 0;
// max length of string with non repeating characters so far
int max_len = 1;
for (int i = 1; i < src.Length; ++i)
{
if (v[src[i]] == -1)
{
v[src[i]] = i;
++max_len;
}
else
{
int cur_len = i - start;
if (max_len < cur_len)
max_len = cur_len;
// move start of the substring with non repeating characters
// after already visited(repeated) character
start = v[src[i]] + 1;
}
}
return max_len;
}
Simplest solution:
//Longest Substring Without Repeating Characters (Java)
public int solution(String s) {
Set set = new LinkedHashSet();
for (char c : s.toCharArray()) {
set.add(c);
}
return set.size();
}
Lol! Are you kidding?
simple solution
import java.util.*;
public class longest_substring{
/*
Given a string, find the length of the longest substring without repeating characters.
For example, the longest substring without repeating letters for "abcabcbb" is "abc",
which the length is 3. For "bbbbb" the longest substring is "b",
with the length of 1.
*/
public static void main(String[] args){
System.out.println("longest " +get_longest(args[0]));
}
public static int get_longest(String str){
HashSet wordSet = new HashSet();
int longest = 0;
for(int i = 0; i < str.length(); i++){
for(int j = i; j longest){
longest = wordSet.size();
}
wordSet.clear();
break;
}
}
}
return longest;
}
}
Yep. That was my misunderstanding… Hope now it is better…
public class longestsubstring {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in=new Scanner(System.in);
ArrayList a=new ArrayList();
String s=in.next();
String temp=””;
String target=””;
for(int i=0;i
target.length())target=temp;
a.clear();
temp=””;
}
}
System.out.print(target);
}
//time : O(n) space : O(n)
public class Solution {
public int lengthOfLongestSubstring(String s) {
int start = 0,max = 0,i =0;
Map lookup = new HashMap();
for (;i
= start) {
max=Math.max(max,i-start);
start=lookup.get(ch)+1;
}
lookup.put(ch,i);
}
max=Math.max(max,i-start);
return max;
}
}
simple Solution is to add every character to Hash set and get the size of it. As Set doesn’t store duplicates this is the easiest and simple way to get the count.
below is the code
static int longest(String s){
int result = 0;
int a = s.length();
Set set = new HashSet();
for(int i=0; i<a-1 ; i++){
set.add(s.charAt(i));
}
System.out.println("Length of the string is" + set.size());
return result;
}
static int lengthOfLongestSubstring(String s){
if(s==null || s.length()==0)
return 0;
boolean []flag=new boolean[256];
int count=0;
int max=Integer.MIN_VALUE;
for(int i=0;i
max){max=count;
}
flag[c]=false;
count=0;
}
}
return max;
}
This is My Solution
public class Solution {
public int lengthOfLongestSubstring(String s) {
if(s.trim().length() == 0) {
return 0;
}
char [] strArr = s.toCharArray();
HashMap map = new HashMap();
int maxLen = 0;
int firstIndex = 0;
int lastIndex = 0;
for(int i=0; i maxLen){
maxLen = i – firstIndex;
}
if(map.get(strArr[i]) >= firstIndex){
firstIndex = map.get(strArr[i]) + 1;
}
map.remove(map.get(strArr[i]));
}
map.put(strArr[i],i);
lastIndex = i;
}
if((lastIndex – firstIndex + 1) > maxLen){
maxLen = lastIndex – firstIndex + 1;
}
return maxLen;
}
}
Here is my javascript solution with great details:
var lengthOfLongestSubstring = function(s) {
const valueIdxHash = {};
var longest = 0;
var startIdx = 0;
for(var i = 0; i < s.length; i++) {
if(s[i] in valueIdxHash) {
startIdx = Math.max(startIdx, valueIdxHash[s[i]] + 1);
}
valueIdxHash[s[i]] = i;
longest = Math.max(longest, i – startIdx + 1);
}
return longest;
};
Checkout my blog for detailed explanation: https://algorithm.pingzhang.io/String/longest_substring_without_repeating_characters.html
Below is a solution that is in O(n) time complexity and O(1) space.
This solution uses hash table of size 26 (number of alphabets).
I have tested it for as many cases as I could think of. Have a look:
#include
#include
main(){
char st[] = “geeksforgeeks”;
int hash[26] = {0};
int i, j = -1, max = 0, res = 1;
for (i = 0; i < strlen(st); i++) {
if ( (j != -1 && hash[st[i] – 'a'] < j) || (j == -1 && hash[st[i] – 'a'] == 0)) {
if (res < i – j) res = i – j;
} else {
j = hash[st[i] – 'a'];
}
hash[st[i] – 'a'] = i;
}
printf("%dn", res);
}