Скачать или смотреть Day 3: Find First Unique Character in String | logic building |

*🔥 LOGIC BUILDING SERIES DAY 3: CRACKING THE UNIQUE CHARACTER PUZZLE! 🔥*

Welcome back, Codee Fam! It's Day 3 of our epic Logic Building journey, and today, we're tackling the *BIGGEST* interview question that every developer MUST master: *Find the First Non-Repeating Character in a String.*

Don't let this problem scare you! Your best friend, Codee, is here to turn this complex logic into simple, Bakchodi-style fun. We'll show you the most efficient and clean way to solve this using the powerful *Frequency Mapping* technique.

*🚀 WHAT WE COVER IN THIS DEEP DIVE:*

*1. The Two-Pass Algorithm Explained (O(n) Efficiency):*
Why do we need to loop through the string twice*? We break down the **Two-Pass Strategy* step-by-step. The first loop handles the counting (The Census), and the second loop handles the finding (The Search). This ensures $O(n)$ time complexity, which is crucial for big tech interviews!

*2. The Frequency Map Magic:*
Learn how to use a simple JavaScript object (`{}`) as a highly efficient *Hash Map* to store character frequencies. We demystify the smart short-hand: `frequency[ch] = (frequency[ch] || 0) + 1`. This line is pure genius!

*3. Case-Insensitivity Fix (The Hidden Bug):*
We show you how to prevent a major error by using **`.toLowerCase()`**. This small fix ensures your code treats 'P' and 'p' as the same character, making your solution robust and interview-ready.

*4. The Final Search:*
Understand why we must iterate from the start of the string again in the second loop to guarantee we return the *first* non-repeating character, maintaining the original string's order.

Get ready to level up your JavaScript skills and conquer this common logic challenge. Hit that like button if you’re ready to become a *Codee* legend!

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*👉 NEXT UP:* Day 4 will be even bigger! Subscribe and press the notification bell so you don't miss it!

*💬 Let me know in the comments: What is the next complex logic problem you want Codee to break down?*