Java Program To Find Length Of The Longest Substring Without Repeating Characters
Given a string str, find the length of the longest substring without repeating characters.
- For “ABDEFGABEF”, the longest substring are “BDEFGA” and “DEFGAB”, with length 6.
- For “BBBB” the longest substring is “B”, with length 1.
- For “GEEKSFORGEEKS”, there are two longest substrings shown in the below diagrams, with length 7
The desired time complexity is O(n) where n is the length of the string.
Method 1 (Simple : O(n3)): We can consider all substrings one by one and check for each substring whether it contains all unique characters or not. There will be n*(n+1)/2 substrings. Whether a substring contains all unique characters or not can be checked in linear time by scanning it from left to right and keeping a map of visited characters. Time complexity of this solution would be O(n^3).
Java
import java.io.*;
import java.util.*;
class GFG{
public static Boolean areDistinct(String str,
int i, int j)
{
boolean [] visited = new boolean [ 26 ];
for ( int k = i; k <= j; k++)
{
if (visited[str.charAt(k) - 'a' ] == true )
return false ;
visited[str.charAt(k) - 'a' ] = true ;
}
return true ;
}
public static int longestUniqueSubsttr(String str)
{
int n = str.length();
int res = 0 ;
for ( int i = 0 ; i < n; i++)
for ( int j = i; j < n; j++)
if (areDistinct(str, i, j))
res = Math.max(res, j - i + 1 );
return res;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
System.out.println( "The input string is " + str);
int len = longestUniqueSubsttr(str);
System.out.println( "The length of the longest " +
"non-repeating character " +
"substring is " + len);
}
}
|
Output
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Method 2 (Better : O(n2)) The idea is to use window sliding. Whenever we see repetition, we remove the previous occurrence and slide the window.
Java
import java.io.*;
import java.util.*;
class GFG{
public static int longestUniqueSubsttr(String str)
{
int n = str.length();
int res = 0 ;
for ( int i = 0 ; i < n; i++)
{
boolean [] visited = new boolean [ 256 ];
for ( int j = i; j < n; j++)
{
if (visited[str.charAt(j)] == true )
break ;
else
{
res = Math.max(res, j - i + 1 );
visited[str.charAt(j)] = true ;
}
}
visited[str.charAt(i)] = false ;
}
return res;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
System.out.println( "The input string is " + str);
int len = longestUniqueSubsttr(str);
System.out.println( "The length of the longest " +
"non-repeating character " +
"substring is " + len);
}
}
|
Output
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Method 3 ( Linear Time ): Using this solution the problem can be solved in linear time using the window sliding technique. Whenever we see repetition, we remove the window till the repeated string.
Java
import java.io.*;
class GFG {
public static int longestUniqueSubsttr(String str)
{
String test = "" ;
int maxLength = - 1 ;
if (str.isEmpty()) {
return 0 ;
}
else if (str.length() == 1 ) {
return 1 ;
}
for ( char c : str.toCharArray()) {
String current = String.valueOf(c);
if (test.contains(current)) {
test = test.substring(test.indexOf(current)
+ 1 );
}
test = test + String.valueOf(c);
maxLength = Math.max(test.length(), maxLength);
}
return maxLength;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
System.out.println( "The input string is " + str);
int len = longestUniqueSubsttr(str);
System.out.println( "The length of the longest "
+ "non-repeating character "
+ "substring is " + len);
}
}
|
Output
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Method 4 (Linear Time): Let us talk about the linear time solution now. This solution uses extra space to store the last indexes of already visited characters. The idea is to scan the string from left to right, keep track of the maximum length Non-Repeating Character Substring seen so far in res. When we traverse the string, to know the length of current window we need two indexes.
1) Ending index ( j ) : We consider current index as ending index.
2) Starting index ( i ) : It is same as previous window if current character was not present in the previous window. To check if the current character was present in the previous window or not, we store last index of every character in an array lasIndex[]. If lastIndex[str[j]] + 1 is more than previous start, then we updated the start index i. Else we keep same i.
Below is the implementation of the above approach :
Java
import java.util.*;
public class GFG {
static final int NO_OF_CHARS = 256 ;
static int longestUniqueSubsttr(String str)
{
int n = str.length();
int res = 0 ;
int [] lastIndex = new int [NO_OF_CHARS];
Arrays.fill(lastIndex, - 1 );
int i = 0 ;
for ( int j = 0 ; j < n; j++) {
i = Math.max(i, lastIndex[str.charAt(j)] + 1 );
res = Math.max(res, j - i + 1 );
lastIndex[str.charAt(j)] = j;
}
return res;
}
public static void main(String[] args)
{
String str = "geeksforgeeks" ;
System.out.println( "The input string is " + str);
int len = longestUniqueSubsttr(str);
System.out.println( "The length of "
+ "the longest non repeating character is " + len);
}
}
|
Output
The input string is geeksforgeeks
The length of the longest non-repeating character substring is 7
Time Complexity: O(n + d) where n is length of the input string and d is number of characters in input string alphabet. For example, if string consists of lowercase English characters then value of d is 26.
Auxiliary Space: O(d)
Alternate Implementation :
Java
import java.util.*;
class GFG {
static int longestUniqueSubsttr(String s)
{
HashMap<Character, Integer> seen = new HashMap<>();
int maximum_length = 0 ;
int start = 0 ;
for ( int end = 0 ; end < s.length(); end++)
{
if (seen.containsKey(s.charAt(end)))
{
start = Math.max(start, seen.get(s.charAt(end)) + 1 );
}
seen.put(s.charAt(end), end);
maximum_length = Math.max(maximum_length, end-start + 1 );
}
return maximum_length;
}
public static void main(String []args)
{
String s = "geeksforgeeks" ;
System.out.println( "The input String is " + s);
int length = longestUniqueSubsttr(s);
System.out.println( "The length of the longest non-repeating character substring is " + length);
}
}
|
Output
The input String is geeksforgeeks
The length of the longest non-repeating character substring is 7
As an exercise, try the modified version of the above problem where you need to print the maximum length NRCS also (the above program only prints the length of it).
Please refer complete article on Length of the longest substring without repeating characters for more details!
Last Updated :
20 Dec, 2021
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