Leetcode #799: Champagne Tower
In this guide, we solve Leetcode #799 Champagne Tower in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Dynamic Programming
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
class Solution:
def champagneTower(self, poured: int, query_row: int, query_glass: int) -> float:
f = [[0] * 101 for _ in range(101)]
f[0][0] = poured
for i in range(query_row + 1):
for j in range(i + 1):
if f[i][j] > 1:
half = (f[i][j] - 1) / 2
f[i][j] = 1
f[i + 1][j] += half
f[i + 1][j + 1] += half
return f[query_row][query_glass]
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.