Longest Substring Without Repeating Characters | Leetcode #youtube #leetcode

🚀 *Longest Substring Without Repeating Characters* is a must-know LeetCode problem for coding interviews! In this video, I'll break it down with a *step-by-step explanation**, including **brute force* and **optimized sliding window approach**.

🔥 *What you'll learn:*
✔️ Problem explanation with examples
✔️ Brute force approach (O(n²))
✔️ Optimal Sliding Window technique (O(n))
✔️ Dry run walkthrough
✔️ Coding interview tips!

💡 *Challenge:* Modify the function to return the actual substring instead of its length! Comment below with your solution.

💬 Have questions? Drop them in the comments!

🔔 *Subscribe for more coding tutorials:* [Your Channel Link]

📌 *Related Videos:*
[Similar Sliding Window Problems](#)
[Top LeetCode Questions](#)

"Hey everyone! Welcome back to the channel. If you're preparing for coding interviews, today's problem is a must-know – ‘Longest Substring Without Repeating Characters.’ This is a common question asked in FAANG interviews, and I’ll walk you through the best approach to solve it efficiently. Let’s get started!"

📌 Section 1: Problem Breakdown 📝 [On-screen text: "Problem Statement" + Example inputs and outputs]
🎙️ "Given a string s, we need to find the length of the longest substring without repeating characters. For example:

abcabcbb → The longest substring is "abc", so the output is 3.
bbbbb → The longest substring is "b", output 1.
pwwkew → "wke", output 3."
💡 "A substring is a contiguous sequence in a string, meaning characters should be consecutive."

📌 Section 2: Brute Force Approach (❌ Not Efficient) 🎙️ "Let's first consider a naive approach – checking all possible substrings and ensuring they don’t have repeating characters. However, this leads to a time complexity of O(n²), which is too slow for large inputs."

def length_of_longest_substring(s):
max_length = 0
for i in range(len(s)):
seen = set()
for j in range(i, len(s)):
if s[j] in seen:
break
seen.add(s[j])
max_length = max(max_length, j - i + 1)
return max_length

Problem: Nested loops make this inefficient.

📌 Section 3: Optimal Solution - Sliding Window (✅ Efficient) 🎙️ "A better approach is using the Sliding Window Technique with a HashSet. We use two pointers (left and right) to dynamically track the longest unique substring in O(n) time complexity."

💡 Algorithm Steps: 1️⃣ Initialize left pointer at 0, a set to track characters, and max_length = 0.
2️⃣ Move the right pointer forward, adding characters to the set.
3️⃣ If a duplicate is found, remove characters from the left until the set is unique again.
4️⃣ Keep track of the maximum length found.


def length_of_longest_substring(s):
char_set = set()
left = 0
max_length = 0

for right in range(len(s)):
while s[right] in char_set:
char_set.remove(s[left])
left += 1
char_set.add(s[right])
max_length = max(max_length, right - left + 1)

return max_length

Section 5: Summary & Closing 🎙️ "Here’s a quick recap:

Brute Force: O(n²) (Not efficient).
Sliding Window: O(n) (Best solution!).
Use cases: Text processing, search engines, and pattern recognition.
🎯 Challenge for You: Try modifying the function to return the substring itself instead of its length! Let me know in the comments how you did it."

👋 "If you found this helpful, don't forget to like, subscribe, and hit the 🔔 bell for more coding interview prep!"

🎯 *Don't forget to LIKE 👍 & SUBSCRIBE 🔔 for more coding tutorials!* 🚀

📢 *Follow for More Updates:*
🌐 Website: learninmaster.com/
🐦 Twitter: x.com/KumarDhaneshwar
📸 Instagram: [YourInstagramHandle]
💼 LinkedIn: www.linkedin.com/in/dhaneshwar-singh-b00abb83/

Longest Substring Without Repeating Characters | Leetcode 3 | Optimal Sliding Window Solution Explained
leetcode,coding interview,sliding window,python tutorial,programming,tech interview,leetcode solutions,software engineer,coding challenge,faang interview

#codinginterview #leetcode #python #slidingwindow #programming #faang
#youtubeshorts #youtube #youtuber #youtubevideo #trending #trend #trenfingshorts #trendingvideo #trendingreels #viralvideo #viralshort #viralreels #viralvideos