Introduction

In this blog post, we will explore how to solve the LeetCode problem "799" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

We stack glasses in a pyramid, where the first row has 1 glass, the second row has 2 glasses, and so on until the 100th row. Each glass holds one cup of champagne. Then, some champagne is poured into the first glass at the top. When the topmost glass is full, any excess liquid poured will fall equally to the glass immediately to the left and right of it. When those glasses become full, any excess champagne will fall equally to the left and right of those glasses, and so on. (A glass at the bottom row has its excess champagne fall on the floor.) For example, after one cup of champagne is poured, the top most glass is full. After two cups of champagne are poured, the two glasses on the second row are half full. After three cups of champagne are poured, those two cups become full - there are 3 full glasses total now. After four cups of champagne are poured, the third row has the middle glass half full, and the two outside glasses are a quarter full, as pictured below. Now after pouring some non-negative integer cups of champagne, return how full the jth glass in the ith row is (both i and j are 0-indexed.) Example 1: Input: poured = 1, query_row = 1, query_glass = 1 Output: 0.00000 Explanation: We poured 1 cup of champange to the top glass of the tower (which is indexed as (0, 0)). There will be no excess liquid so all the glasses under the top glass will remain empty. Example 2: Input: poured = 2, query_row = 1, query_glass = 1 Output: 0.50000 Explanation: We poured 2 cups of champange to the top glass of the tower (which is indexed as (0, 0)). There is one cup of excess liquid. The glass indexed as (1, 0) and the glass indexed as (1, 1) will share the excess liquid equally, and each will get half cup of champange. Example 3: Input: poured = 100000009, query_row = 33, query_glass = 17 Output: 1.00000 Constraints: 0 <= poured <= 109 0 <= query_glass <= query_row < 100

Explanation

Here's the solution to the champagne tower problem:

• Simulate the Pouring: The core idea is to simulate the pouring process row by row. We keep track of the amount of champagne in each glass.
• Excess Distribution: When a glass overflows (contains more than 1 cup), the excess is split equally between the two glasses below it.

• Query Result: Finally, we return the amount of champagne in the requested glass, making sure it doesn't exceed 1 (full).

• Runtime Complexity: O(R^2), where R is the number of rows.

• Storage Complexity: O(R^2)

Code

    def champagneTower(poured: int, query_row: int, query_glass: int) -> float:
    """
    Simulates the pouring of champagne into a tower of glasses and returns how full a specified glass is.

    Args:
        poured: The total amount of champagne poured into the top glass.
        query_row: The row index of the glass to query (0-indexed).
        query_glass: The glass index within the row to query (0-indexed).

    Returns:
        The amount of champagne in the specified glass, capped at 1.0.
    """

    tower = [[0.0] * (i + 1) for i in range(100)]  # Create the tower (up to 100 rows)
    tower[0][0] = float(poured)

    for i in range(query_row + 1):
        for j in range(i + 1):
            if tower[i][j] > 1:
                excess = tower[i][j] - 1
                tower[i][j] = 1  # Glass is full
                if i + 1 < 100:
                    tower[i + 1][j] += excess / 2.0
                    tower[i + 1][j + 1] += excess / 2.0

    return min(1.0, tower[query_row][query_glass])