Champagne Tower Python Solution with Tests

Tested Python solution for LeetCode 799 with 14 pytest cases. Generate a practice environment with lcpy.

LeetCode 799, Medium. Topics: Dynamic Programming. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 14 parametrized pytest cases, and a playground notebook:

lcpy gen -n 799   # by problem number
lcpy gen -s champagne_tower   # by problem name

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Problem

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.)

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Examples

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.

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.

Input: poured = 100000009, query_row = 33, query_glass = 17
Output: 1.00000

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Constraints

• 0 <= poured <= 10^9
• 0 <= query_glass <= query_row < 100

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

class Solution:
    # Time: O(query_row^2)
    # Space: O(query_row)
    def champagne_tower(self, poured: int, query_row: int, query_glass: int) -> float:
        row = [float(poured)]
        for _ in range(query_row):
            nxt = [0.0] * (len(row) + 1)
            for i, amount in enumerate(row):
                excess = max(0.0, amount - 1.0) / 2.0
                nxt[i] += excess
                nxt[i + 1] += excess
            row = nxt
        return min(1.0, row[query_glass])

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Complexity

Time	Space
O(query_row^2)	O(query_row)

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Tags

NeetCode All.