CSES Coin Piles Explained | 2 Conditions = Full Solution скачать в хорошем качестве

CSES Coin Piles Explained | 2 Conditions = Full Solution

In this video, we solve the Coin Piles problem from CSES by breaking it down into two simple and sufficient conditions. Instead of simulating moves, we directly analyze the math behind the problem: 1️⃣ (a + b) % 3 == 0 Each move removes exactly 3 coins in total, so the sum must be divisible by 3. 2️⃣ max(a, b) ≤ 2 × min(a, b) The larger pile can never be more than twice the smaller one, otherwise it’s impossible to balance the moves. I explain why these conditions are necessary and sufficient, and how they cover all edge cases efficiently in O(1) time per test case. Perfect for understanding the logic, not just memorizing the solution. Topics Covered: Observing patterns instead of brute force Why modulo 3 matters Intuition behind the a ≤ 2b constraint Clean and optimal solution approach Tags / Hashtags: #CSES #CoinPiles #CompetitiveProgramming #MathLogic #DSA #ProblemSolving #O1Solution