➡️ Interviewer: Hello, Sideon. Can you please introduce yourself and your experience with Java programming?
👨🦰 Sideon: Hi, my name is Sideon and I have been programming with Java for about 3 years now.
➡️ Interviewer: That's great to hear. The problem we'll be discussing today is the "3. Longest Substring Without Repeating Characters" problem on LeetCode. Can you explain the problem statement in your own words?
👨🦰 Sideon: Sure. The problem is asking us to find the length of the longest substring in a given string that doesn't have any repeating characters.
➡️ Interviewer: That's right. Before we get into the solution, can you tell me what you would consider to be a "substring"?
Sideon: Sure. A substring is a contiguous sequence of characters within a string.
➡️ Interviewer: Excellent. Now let's get into the solution. Can you walk me through your thought process on how you would approach this problem?
👨🦰 Sideon: Sure. One way to approach this problem is to use a sliding window technique where we use two pointers, left and right, to traverse through the string. We would keep track of the characters seen so far in the window using a hash set. We would move the right pointer until we see a repeating character and then move the left pointer until we remove the repeating character from the window. We would keep track of the maximum window size seen so far.
➡️ Interviewer: That's a good start. Can you explain how the sliding window technique works and how it is used to solve this problem?
👨🦰 Sideon: Sure. The sliding window technique is used to solve problems where we need to find a substring that satisfies a given condition. In this problem, we need to find the longest substring without repeating characters. We use two pointers, left and right, to traverse through the string. We also use a hash set to keep track of the characters seen so far in the window. We move the right pointer until we see a repeating character and then move the left pointer until we remove the repeating character from the window. We would keep track of the maximum window size seen so far.
➡️ Interviewer: Good explanation. Now, can you walk me through a brute force approach to this problem?
👨🦰 Sideon: Sure. A brute force approach would be to check all possible substrings of the given string and check if each substring has unique characters. We would then return the length of the longest substring with unique characters.
➡️ Interviewer: That's correct. Can you explain the time and space complexity of this brute force approach?
👨🦰 Sideon: The time complexity of this brute force approach would be O(n^3) where n is the length of the string since we would have to check all possible substrings. The space complexity would be O(min(n, m)) where m is the size of the character set since we would need to store the characters seen so far in the hash set.
➡️ Interviewer: Good job. Now let's move on to the optimized approach. Can you explain how you optimized the brute force approach?
👨🦰 Sideon: Sure. Instead of checking all possible substrings, we can use the sliding window technique to check the longest substring with unique characters. We use two pointers, left and right, to traverse through the string. We also use a hash set to keep track of the characters seen so far in the window. We move the right pointer until we see a repeating character and then move the left pointer until we remove the repeating character from the window. We would keep track of the maximum window size seen so far.
➡️ Interviewer: Good job. Can you explain the time and space complexity of this optimized approach?
👨🦰 Sideon: The time complexity of this optimized approach would be O(n) where n is the length of the string since we would be traversing through the string only once. The space complexity would be O(min(n, m)) where m is the size of the character set since we would need to store the characters seen so far in the hash set.
➡️ Interviewer: Great explanation. Can you walk me through an example of how this optimized approach would work?
👨🦰 Sideon: Sure. Let's take the string "abcabcbb" as an example. We would start with an empty window and move the right pointer until we see a repeating character, which is 'a' at index 3. We would then move the left pointer until we remove the repeating character from the window, which is the first 'a' at index 0. Our window would then be "bc" and we would update the maximum window size seen so far to 2. We would continue this process until we reach the end of the string and return the maximum window size seen so far, which is 3.
➡️ Interviewer: Great example. Can you think of any edge cases or corner cases that we should consider for this problem?
👨🦰 Sideon: One edge case to consider is when the string is empty or has only one character, in which case the answer would be the length of the string. Another edge case is when all characters in the string are unique, in which case the answer would be the length of the string. A corner case to consider is when the length of the string is very large and the character set is very small, in which case the space complexity of the hash set would be a concern.
➡️ Interviewer: Good thinking. Now let's move on to the implementation. Can you write the optimized solution in Java?
👨🦰 Sideon: Sure. Here's the implementation:
public int lengthOfLongestSubstring(String s) {
int n = s.length();
Set<Character> set = new HashSet<>();
int maxLen = 0, i = 0, j = 0;
while (i < n && j < n) {
if (!set.contains(s.charAt(j))) {
set.add(s.charAt(j++));
maxLen = Math.max(maxLen, j - i);
} else {
set.remove(s.charAt(i++));
}
}
return maxLen;
}
➡️ Interviewer: Great job, Sideon. Does this implementation follow all the coding principles and standards? Is the implementation correct with all test and edge cases covered?
👨🦰 Sideon: Yes, I believe so. I have tested this implementation with various test cases including edge cases and it seems to be working correctly.
➡️ Interviewer: Excellent. Based on our discussion, I think you have a good understanding of the problem and the solution. One thing I would suggest is to be more careful with your variable naming and formatting to make the code more readable. Other than that, great job!
👨🦰 Sideon: Thank you for the feedback, I will make sure to improve my variable naming and formatting in the future.
➡️ Interviewer: Great, glad to hear that. One final question, Sideon. Can you think of any other possible solutions to this problem?
👨🦰 Sideon: Yes, I can think of one other possible solution. We can use a sliding window approach where we would maintain a window of non-repeating characters and move the window to the right until we see a repeating character. Once we see a repeating character, we would move the left pointer to the right until we remove the repeating character from the window, and then continue moving the right pointer to the right until we see another repeating character. We would continue this process until we reach the end of the string and return the maximum window size seen so far.
➡️ Interviewer: That's a good approach as well. However, I think the optimized approach we discussed earlier would have better time complexity since it only goes through the string once, whereas the sliding window approach would have to move the left pointer multiple times for each repeating character.
👨🦰 Sideon: That's a good point, and I agree that the optimized approach would be more efficient. Thank you for pointing that out.
➡️ Interviewer: You're welcome, Sideon. Overall, I think you did a great job in solving this problem and explaining your thought process. Thank you for your time and good luck with your future interviews!
👨🦰 Sideon: Thank you so much for your help, Interviewer. I really appreciate the opportunity to practice my problem-solving skills with you. Have a great day!
Lets Pace Up, Keep an eye out for similar Leetcode Interview Simulations on various problems, Though we kicked off with an easy one today, always check the comments.