LeetCode 799 | Champagne Tower | DP Simulation Explained | Java | Medium скачать в хорошем качестве

LeetCode 799 | Champagne Tower | DP Simulation Explained | Java | Medium

LeetCode 799 – Champagne Tower (Java) Today's LeetCode Daily 🚀 We stack glasses in a pyramid. Each glass holds 1 cup. If a glass gets more than 1 cup, the extra champagne is split equally to the left and right glasses below. Goal: Return how full the glass at (query_row, query_glass) is. Approach Used: Dynamic Programming (Simulation) Step 1: Create a 2D DP array of size 101 x 101. Step 2: Pour the champagne into dp[0][0]. Step 3: For each glass: If dp[i][j] is greater than 1, overflow = (dp[i][j] - 1) divided by 2 Add overflow to: dp[i + 1][j] dp[i + 1][j + 1] Step 4: Return minimum of 1 and dp[query_row][query_glass]. Key Idea: Simulate overflow row by row. Time Complexity: O(row squared) Space Complexity: O(100 squared) Problem Link: https://leetcode.com/problems/champag... Java Solution (GitHub): https://github.com/Amandf/LeetCode-So... More daily Java solutions. Clean code. No mic. Pure logic. #leetcode #leetcodedailychallenge #dynamicprogramming #java #dsa #CodingInterview #MediumProblem