Solving the Longest Substring Without Repeating Characters Problem in Java | Time Complexity: O(n)

Solving the Longest Substring Without Repeating Characters Problem in Java | Time Complexity: O(n)

Problem Definition

Given a string, find the length of the longest substring without repeating characters.

Input: A single string s.

Output: An integer representing the length of the longest substring without repeating characters.

Constraints:

Example:

    Input: "abcabcbb"
    Output: 3
    Explanation: The answer is "abc", with the length of 3.
    

Understanding the Problem

The core challenge of this problem is to identify the longest substring within a given string that does not contain any repeating characters. This problem is significant in various applications such as data validation, parsing, and text processing where unique sequences are required.

Potential pitfalls include misunderstanding the definition of a substring (which must be contiguous) and not efficiently handling the search for repeating characters, leading to suboptimal solutions.

Approach

To solve this problem, we can use a sliding window technique. The idea is to use two pointers to represent the current window of characters being considered and a set to track the characters within this window.

Here is a step-by-step approach:

This approach ensures that each character is processed at most twice, resulting in a linear time complexity of O(n).

Algorithm

Here is a detailed breakdown of the algorithm:

  1. Initialize left and right pointers to 0, and an empty set charSet.
  2. Iterate with the right pointer over the string.
  3. If the character at right is not in charSet, add it to the set and update the maximum length.
  4. If the character is in the set, remove the character at left from the set and move left to the right.
  5. Repeat until the right pointer reaches the end of the string.

Code Implementation

public class Solution {
    public int lengthOfLongestSubstring(String s) {
        // Initialize the set to store unique characters
        Set<Character> charSet = new HashSet<>();
        int left = 0, right = 0, maxLength = 0;

        // Iterate over the string with the right pointer
        while (right < s.length()) {
            // If the character is not in the set, add it and update maxLength
            if (!charSet.contains(s.charAt(right))) {
                charSet.add(s.charAt(right));
                maxLength = Math.max(maxLength, right - left + 1);
                right++;
            } else {
                // If the character is in the set, remove the leftmost character and move left pointer
                charSet.remove(s.charAt(left));
                left++;
            }
        }
        return maxLength;
    }
}

Complexity Analysis

The time complexity of this approach is O(n) because each character is processed at most twice (once by the right pointer and once by the left pointer). The space complexity is O(min(n, m)), where n is the length of the string and m is the size of the character set (which is constant for English letters, digits, symbols, and spaces).

Edge Cases

Consider the following edge cases:

These cases can be tested to ensure the robustness of the solution.

Testing

To test the solution comprehensively, consider the following test cases:

JUnit or any other testing framework can be used to automate these tests.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Conclusion

In this blog post, we discussed how to solve the problem of finding the longest substring without repeating characters using a sliding window technique. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is crucial for improving your algorithmic thinking and coding skills.

Additional Resources

For further reading and practice, consider the following resources: