Given a string s, find the length of the longest substring without repeating characters.
Example 1:
Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
Example 4:
Input: s = ""
Output: 0
Constraints:
0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.
Given a string s, find the length of the longest substring without repeating characters.
Example 1:
Input: s = "abcabcbb"
Output: 3
Explanation: The answer is "abc", with the length of 3.
Example 2:
Input: s = "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:
Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
Example 4:
Input: s = ""
Output: 0
Constraints:
0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces.
public int lengthOfLongestSubstring(String s) {
int i=0;
int j=0;
int max = 0;
HashSet<Character> set = new HashSet<Character>();
while(j<s.length() && i<s.length()){
char sc = s.charAt(i);
char ec = s.charAt(j);
if(set.contains(ec)){
set.remove(sc);
i++;
}
else{
set.add(ec);
max = Math.max(max, set.size());
j++;
}
}
return max;
}
Entire Code
/* package whatever; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class AskGif
{
public static void main (String[] args) throws java.lang.Exception
{
AskGif askgif = new AskGif();
Scanner scn = new Scanner(System.in);
String str1 = scn.nextLine();
str1 = stringToString(str1);
System.out.println(askgif.lengthOfLongestSubstring(str1));
}
private static String stringToString(String str){
str = str.substring(1, str.length() - 1);
return str;
}
{{codeFunction}}
}
The basic idea of solving this problem is to keep track of all the characters we have already visited and maintain a window size in which each character appeared only once. So throughout the traversal whenever the window size is maximum, that will give us the result.
Time Complexity: O(n) // For traversing all the items in the array.
Space Complexity: O(n) // for keeping track of already visited items in the window, in the worst case all items will be present in the HashSet.
