One Pass Greedy Solution | Minimum Deletions to Make String Balanced | LeetCode 1653

Описание: In this video, we solve LeetCode 1653. Minimum Deletions to Make String Balanced using a simple counting observation.

The key idea is to understand what a balanced string means:
• All 'a' characters must come before any 'b'

So at every index, we decide:
• Delete 'b' characters seen on the left, or
• Delete remaining 'a' characters on the right

Instead of trying all deletions, we maintain:
1️⃣ Count of remaining 'a' characters
2️⃣ Count of 'b' characters seen so far

The minimum deletions at any point becomes:
count_b + count_a

This leads to a clean single-pass O(n) solution with constant space.

This problem is part of today’s LeetCode Daily Problem of the Day (POTD).

🔹 What you’ll learn:
• How to convert string problems into counting problems
• Why tracking left and right counts works
• How to minimize deletions greedily
• O(n) time and O(1) space solution

🔹 Topics:
• Strings
• Greedy
• Prefix Counting

Clear problem explanation, intuition-first walkthrough, and clean Java implementation.

Problem link - https://leetcode.com/problems/minimum...

Java Code - https://pastebin.com/79pR2GtJ
Python Code - https://pastebin.com/jRbq9cTH

Time complexity - O(N)
Space Complexity - O(1)

Page Marker Extension - https://chromewebstore.google.com/det...

Video Chapters -
00:00 - Problem Statement
05:11 - Approach
19:32 - Code
21:46 - Time and Space Complexity


Leetcode daily playlist -    • LeetCode Daily Problems  

LeetCode Easy POTD playlist -    • LeetCode Easy  

LeetCode Medium POTD playlist -    • LeetCode Medium  

LeetCode Hard POTD playlist -    • LeetCode Hard  

If you want to practice daily and improve your problem-solving skills, this is the place to follow along.

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