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
The problem describes a champagne tower where champagne is poured into the top glass, and excess champagne flows equally to the glasses below. The goal is to determine how full a specific glass is after a certain amount of champagne is poured.
Both solutions presented utilize dynamic programming to efficiently solve this problem. Let's break down each approach:
This solution uses a 2D array f
to store the amount of champagne in each glass. f[i][j]
represents the amount of champagne in the j-th glass of the i-th row (0-indexed).
Algorithm:
Initialization: A 101x101 array f
is initialized with all zeros. The top glass (f[0][0]) is initialized with the amount of champagne poured.
Iteration: The code iterates through each row (i
) up to query_row
. For each row, it iterates through each glass (j
) in that row.
Excess Champagne Distribution: If a glass f[i][j]
has more than 1 unit of champagne, the excess (f[i][j] - 1
) is split equally between the two glasses below it (f[i+1][j]
and f[i+1][j+1]
). The amount in the current glass is then set to 1 (full).
Result: After iterating through all rows, the function returns the amount of champagne in the specified glass f[query_row][query_glass]
.
Time and Space Complexity:
f
.This solution improves upon Solution 1 by using a single one-dimensional array to store the champagne amounts. It avoids unnecessary computations by only considering the current row and the next row.
Algorithm:
Initialization: A single array f
is initialized, containing only the initial amount of champagne poured.
Iteration: The code iterates through each row (i
) up to query_row
. A new array g
(of size i+1
) is created to store the champagne amounts for the next row.
Excess Champagne Distribution: Similar to Solution 1, if a glass f[j]
has more than 1 unit, the excess is split equally between the two glasses below in g
.
Update: Array f
is updated with the new champagne levels from g
.
Result: After the loop, the function returns the minimum of 1 (a glass can't be more than full) and the amount in the specified glass in f
.
Time and Space Complexity:
Conclusion:
Both solutions correctly solve the problem, but Solution 2 is more efficient in terms of space complexity. The choice between them depends on whether the memory optimization is crucial given the problem constraints. For this particular problem with a maximum row of 100, the difference might not be significant, but for larger towers, Solution 2's advantage becomes more apparent.