Length of the longest substring without repeating characters in JavaThe task of discovering the length of the longest substring without repeating characters is a crucial challenge in algorithmic programming. The problem involves identifying the maximum length of a continuous section in a provided string where each character occurs only once. Addressing this challenge in the Java programming landscape necessitates adeptly utilizing string manipulation tools and algorithmic paradigms. The problem is practical in optimizing computational processes, particularly in data compression and string processing scenarios. Example 1: Input: "pqrstuvwxy" Output: 10 Explanation: The entire string "pqrstuvwxy" contains no repeating characters, constituting the longest substring with a length of 10. Example 2: Input: "abacabadabacaba" Output: 4 Explanation: The distinct substrings without repeating characters are "abac" and "bacab," with a length of 4. Example 3: Input: "xyzxyzabcd" Output: 6 Explanation: The longest substrings without repeating characters are "xyzabc" and "yzabcd," both with lengths of 6. Example 4: Input: "mississippi" Output: 4 Explanation: The longest substring without repeating characters is "issi," with a length of 4. Approach: Using Sliding Window in O(n3) time.The Sliding Window approach, used to find the longest substring without repeated characters in O(n^3) time, involves moving a window through the text to check for substrings. Algorithm:Step 1: Set n as the length of the input string s and the maxLength to 0. Step 2: Iterate over all possible starting indices of the substring using a loop with index i from 0 to n.
Step 3: The allUnique Method iterates through the substring using two nested loops, comparing each character with every other character. Step 4: If a repeated character is found it will return false; otherwise it will return true. Step 5: If the substring is unique (allUnique returns true), calculate the length of the substring (j  i) and update maxLength with the maximum of its current value and the calculated length. Step 6: After both loops are complete, return the final value of maxLength as the length of the longest substring without repeating characters. Step 7: Check if all characters in a given substring (from index start to end  1) are unique.
Step 8: Display the result of the code. Implementation:Filename: LongestSubstring1.java Output: Length of the longest substring without repeating characters: 7 Time Complexity: The time complexity of the provided code is O(n^3), where 'n' is the length of the input string. Auxiliary Space: The auxiliary space complexity is O(1) or constant space. Additionally, the helper method allUnique utilizes only a few local variables, and its space complexity remains constant regardless of the input size. Approach: Using Sliding Window in O(n2) time:The Sliding Window approach, employed to find the length of the longest substring without repeating characters in O(n^2) time, efficiently adjusts a window over the input and is suitable for moderately sized inputs. Algorithm:Step 1: Import Necessary Packages: Import java.io.* for input/output functionality and the Import java.util.*. Step 2: Define Method lengthOfLongestSubstring and also the Input of the String s and the Output of the Integer (Maximum length of the substring without repeating characters) Step 3: Now Initialize Variables. Set n as the length of the input string s and Set maxLength to 0. Step 4: Begin a loop with index i ranging from 0 to n  1.
Step 5: Create a boolean array visited of size 256 to keep track of visited characters using ASCII values. Step 6: For each character at index j:
Step 7: If the substring is unique (no repeating characters), calculate the length of the substring (j  i + 1). Step 8: Update maxLength with the maximum of its current value and the calculated length. Implementation:Filename: LongestSubstring2.java Output: Length of the longest substring without repeating characters: 7 Time Complexity: The time complexity of the provided code is O(n^2), where 'n' is the length of the input string. It is because the code utilizes nested loops. The outer loop runs 'n' times, and for each iteration of the outer loop, the inner loop runs at most 'n' times. Auxiliary Space: The auxiliary space complexity is O(1) or constant space. Approach: Using Binary Search Top of FormThe Binary Search approach is like a smart way of finding things in an organized list. When looking for the length of the longest substring without repeating characters, it helps search for the best length more quickly. AlgortihmStep 1: Initialize the variable in the code. Set n as the length of the input string s and also the variables low to 1, high to n, and result to 0. Step 2: While low is less than or equal to high, do the following:
Step 3: Initialize an empty set set to keep track of characters in the current window.
Step 4: While the character at the current end violates uniqueness (present in the set):
Step 5: Add the current character to the set.
Step 6: If a substring of length mid is possible without repeating characters (isValid returns true), do the following:
Step 7: If a substring of length mid is not possible, do the following:
Step 8: After the binary search concludes, return the result's final value as the longest substring's length without repeating characters. Implementation:Filename: BinarySearchLongestSubstring.java Output: Length of the longest substring without repeating characters: 7 Time Complexity: The code has a time complexity of O(n log n), where "n" is the length of the input string. Auxiliary Space: The auxiliary space complexity is O(n), where "n" is the length of the input string. Approach: Using Sliding WindowBy using the Sliding Window approach, we can efficiently explore different substrings to find the longest one without any repeating characters. Algorithm:Step 1: Initialize the length of the input string is 0, return 0 (no characters to process).
Step 2: Initialize the maximum length (maxLength) to 0.
Step 3: Initialize two pointers, leftPointer and rightPointer, both set to 0.
Step 4: If the character at rightPointer is already visited:
Step 5: The variable maxLength now holds the length of the longest substring without repeating characters. Step 6: Display the result. Implementation:Filename: SlidingWindowLongestSubstring.java Output: Given Input String: javatpoint Length of the Longest Substring without repeating characters: 7 Time Complexity: The time complexity of the provided code is O(N), where N is the length of the input string. Auxiliary Space Complexity: The auxiliary space complexity is O(1) since the extra space used is independent of the input size. Approach: Storing the last index of each characterThe Index Storage approach involves keeping track of the last index where each character appeared in the text. It helps in efficiently determining the length of the longest substring without any repeated characters. Algorithm:Step 1: Initialize n to the length of the input string s. Step 2: Create an array lastIndex of size 256 to store the last index of each character, initialized with 1 for all characters.
Step 3: Start iterating through the characters of the input string using a sliding window approach. Step 4: Set the end pointer to the initial position, starting from 0 and moving towards the end of the input string (n1). Step 5: For each character encountered at the current end index: Check if the Character is Already Present:
Step 6: If the character is already present in the current substring:
Step 7: Update the lastIndex of the current character to the end index.
Step 8: Update maxLength by comparing it with the length of the current substring: It ensures that maxLength always stores the length of the longest substring without repeating characters. Step 9: After the iteration completes, return the final value of maxLength as the length of the longest substring without repeating characters. Implementation:Filename: LastIndexLongestSubstring.java Output: Length of the longest substring without repeating characters: 7 Time Complexity: The time complexity of the provided code is O(n), where 'n' is the length of the input string. Auxiliary Space: The auxiliary space complexity is O(1).
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