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 <= 1090 <= query_glass <= query_row < 100In this approach, we simulate the pouring process using a 2D array, where each element represents a glass and stores the amount of champagne in it. We iterate over each row, calculating the overflow and distributing it equally to the glasses directly below the current glass.
In this implementation, we use a 2D array 'tower' to track champagne levels in glasses. We initially pour all champagne into the top glass and simulate champagne overflow row by row.
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Time Complexity: O(n^2) where n is the number of rows traversed. Space Complexity: O(n^2) for the 2D array used to simulate the champagne distribution.
This approach optimizes the space complexity by using only a single array to store the champagne volumes, representing only two rows at a time.
Instead of using a 2D array, we use a single array to minimize space usage. This array tracks only the current row and is updated row-by-row in place, calculating overflows accordingly.
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Time Complexity: O(n^2). Space Complexity: O(n) due to the use of the single array.
| Approach | Complexity |
|---|---|
| Simulation Approach Using a Matrix | Time Complexity: O(n^2) where n is the number of rows traversed. Space Complexity: O(n^2) for the 2D array used to simulate the champagne distribution. |
| Space Optimized Approach | Time Complexity: O(n^2). Space Complexity: O(n) due to the use of the single array. |
Champagne Tower - Leetcode 799 - Python • NeetCodeIO • 13,578 views views
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