Longest Substring Without Repeating Characters | Leetcode скачать в хорошем качестве

Longest Substring Without Repeating Characters | 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: https://learninmaster.com/ 🐦 Twitter: https://x.com/KumarDhaneshwar 📸 Instagram: [YourInstagramHandle] 💼 LinkedIn:   / 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