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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

 `// Java program to find the length of the``// longest substring without repeating``// characters``import` `java.io.*;``import` `java.util.*;`` ` `class` `GFG{`` ` `// This function returns true if all characters in``// str[i..j] are distinct, otherwise returns false``public` `static` `Boolean areDistinct(String str, ``                                  ``int` `i, ``int` `j)``{``     ` `    ``// Note : Default values in visited are false``    ``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``;``}`` ` `// Returns length of the longest substring``// with all distinct characters.``public` `static` `int` `longestUniqueSubsttr(String str)``{``    ``int` `n = str.length();``     ` `    ``// Result``    ``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;``}`` ` `// Driver code``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);``}``}`` ` `// This code is contributed by akhilsaini`

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

 `// Java program to find the length of the ``// longest substring without repeating``// characters``import` `java.io.*;``import` `java.util.*;`` ` `class` `GFG{`` ` `public` `static` `int` `longestUniqueSubsttr(String str)``{``    ``int` `n = str.length();``     ` `    ``// Result``    ``int` `res = ``0``;``     ` `    ``for``(``int` `i = ``0``; i < n; i++)``    ``{``         ` `        ``// Note : Default values in visited are false``        ``boolean``[] visited = ``new` `boolean``[``256``];``         ` `        ``for``(``int` `j = i; j < n; j++)``        ``{``             ` `            ``// If current character is visited``            ``// Break the loop``            ``if` `(visited[str.charAt(j)] == ``true``)``                ``break``;`` ` `            ``// Else update the result if``            ``// this window is larger, and mark``            ``// current character as visited.``            ``else``            ``{``                ``res = Math.max(res, j - i + ``1``);``                ``visited[str.charAt(j)] = ``true``;``            ``}``        ``}`` ` `        ``// Remove the first character of previous``        ``// window``        ``visited[str.charAt(i)] = ``false``;``    ``}``    ``return` `res;``}`` ` `// Driver code``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);``}``}`` ` `// This code is contributed by akhilsaini`

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 = ``""``;`` ` `        ``// Result``        ``int` `maxLength = -``1``;`` ` `        ``// Return zero if string is empty``        ``if` `(str.isEmpty()) {``            ``return` `0``;``        ``}``        ``// Return one if string length is one``        ``else` `if` `(str.length() == ``1``) {``            ``return` `1``;``        ``}``        ``for` `(``char` `c : str.toCharArray()) {``            ``String current = String.valueOf(c);`` ` `            ``// If string already contains the character``            ``// Then substring after repeating character``            ``if` `(test.contains(current)) {``                ``test = test.substring(test.indexOf(current)``                                      ``+ ``1``);``            ``}``            ``test = test + String.valueOf(c);``            ``maxLength = Math.max(test.length(), maxLength);``        ``}`` ` `        ``return` `maxLength;``    ``}`` ` `    ``// Driver code``    ``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);``    ``}``}`` ` `// This code is contributed by Alex Bennet`

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

 `// Java program to find the length of the longest substring``// without repeating characters``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``; ``// result`` ` `        ``// last index of all characters is initialized``        ``// as -1``        ``int` `[] lastIndex = ``new` `int``[NO_OF_CHARS];``        ``Arrays.fill(lastIndex, -``1``);`` ` `        ``// Initialize start of current window``        ``int` `i = ``0``;`` ` `        ``// Move end of current window``        ``for` `(``int` `j = ``0``; j < n; j++) {`` ` `            ``// Find the last index of str[j]``            ``// Update i (starting index of current window)``            ``// as maximum of current value of i and last``            ``// index plus 1``            ``i = Math.max(i, lastIndex[str.charAt(j)] + ``1``);`` ` `            ``// Update result if we get a larger window``            ``res = Math.max(res, j - i + ``1``);`` ` `            ``// Update last index of j.``            ``lastIndex[str.charAt(j)] = j;``        ``}``        ``return` `res;``    ``}`` ` `    ``/* Driver program to test above function */``    ``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);``    ``}``}``// This code is contributed by Sumit Ghosh`

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)``  ``{`` ` `    ``// Creating a set to store the last positions of occurrence``    ``HashMap seen = ``new` `HashMap<>();  ``    ``int` `maximum_length = ``0``;`` ` `    ``// starting the initial point of window to index 0``    ``int` `start = ``0``;`` ` `    ``for``(``int` `end = ``0``; end < s.length(); end++)``    ``{``      ``// Checking if we have already seen the element or not``      ``if``(seen.containsKey(s.charAt(end)))``      ``{``        ``// If we have seen the number, move the start pointer``        ``// to position after the last occurrence``        ``start = Math.max(start, seen.get(s.charAt(end)) + ``1``);``      ``}`` ` `      ``// Updating the last seen value of the character``      ``seen.put(s.charAt(end), end);``      ``maximum_length = Math.max(maximum_length, end-start + ``1``);``    ``}``    ``return` `maximum_length;``  ``}`` ` `  ``// Driver code``  ``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);``  ``}``}`` ` `// This code is contributed by rutvik_56.`

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!

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