💼 Interview Question: - Longest Substring Without Repeating Characters

  Python implementation of the "Longest Substring Without Repeating Characters" problem using the sliding window technique:

def longest_substring(s):

    char_index_map = {}  # Stores the index of characters

    left = 0  # Left boundary of the sliding window

    max_length = 0  # Maximum length found

    for right in range(len(s)):

        if s[right] in char_index_map:

            left = max(left, char_index_map[s[right]] + 1)  # Move left boundary

        char_index_map[s[right]] = right  # Update index of character

        max_length = max(max_length, right - left + 1)  # Update max length

    return max_length

# Example usage:

s = "abcabcbb"

print(longest_substring(s))  # Output: 3

This algorithm efficiently finds the longest substring without repeating characters in O(n) time using a sliding window approach. The char_index_map dictionary keeps track of the last index of each character to avoid unnecessary repetitions.

🧠 : 2nd example:

Prompt

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

def length_of_longest_substring(s: str) -> int:

Input: s = "abcabcbb"  

Output: 3  

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

✅ What It's Testing:

1. Sliding Window Technique – Efficient string traversal.

2. Hashing/Data Structures – Using sets or dictionaries to track characters.

3. Edge Case Thinking – Empty strings, all identical characters, etc.

4. Time Complexity Awareness – Naive solutions may be O(n²); optimal is O(n).

def length_of_longest_substring(s: str) -> int:

    char_index = {}

    left = 0

    max_length = 0

    for right in range(len(s)):

        if s[right] in char_index and char_index[s[right]] >= left:

            left = char_index[s[right]] + 1

        char_index[s[right]] = right

        max_length = max(max_length, right - left + 1)

    return max_length

📊 Time and Space Complexity:

• Time Complexity: O(n), where n is the length of the string.

• Space Complexity: O(min(n, m)), where m is the character set size.