Algorithm
Problem Name: 799. Champagne Tower
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 < 100
Code Examples
#1 Code Example with C++ Programming
Code -
C++ Programming
class Solution {
public:
double champagneTower(int poured, int query_row, int query_glass) {
vector<double>tower((1 + 100) * 100 / 2);
tower[0] = poured;
for(int i = 0; i < 99; i++){
for(int j = 0; j < = i; j++){
int idx = index(i, j);
if(tower[idx] > 1){
double remain = tower[idx] - 1;
tower[idx] = 1;
tower[index(i + 1, j)] += remain / 2;
tower[index(i + 1, j + 1)] += remain / 2;
}
}
}
return tower[index(query_row, query_glass)];
}
int index(int row, int col){
return row * (row + 1) / 2 + col;
}
};
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Output
#2 Code Example with Java Programming
Code -
Java Programming
class Solution {
public double champagneTower(int poured, int query_row, int query_glass) {
double[][] glassCapacity = new double[query_row + 1][query_glass + 2];
glassCapacity[0][0] = poured;
for (int currRow = 0; currRow < query_row; currRow++) {
for (int currCol = 0; currCol < = query_glass; currCol++) {
double currCapacity = Math.max(glassCapacity[currRow][currCol] - 1.0, 0);
glassCapacity[currRow + 1][currCol] += currCapacity / 2.0;
glassCapacity[currRow + 1][currCol + 1] += currCapacity / 2.0;
}
}
return Math.min(glassCapacity[query_row][query_glass], 1.0);
}
}
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#3 Code Example with Javascript Programming
Code -
Javascript Programming
const champagneTower = function(poured, query_row, query_glass) {
let curRow = [poured]
for(let i = 0; i < = query_row; i++) {
const nxtRow = Array(i + 2).fill(0)
for(let j = 0; j < = i; j++) {
if(curRow[j] > 1) {
nxtRow[j] += (curRow[j] - 1) / 2
nxtRow[j + 1] += (curRow[j] - 1) / 2
curRow[j] = 1
}
}
if(i !== query_row) curRow = nxtRow
}
return curRow[query_glass]
};
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#4 Code Example with Python Programming
Code -
Python Programming
class Solution:
def champagneTower(self, poured, query_row, query_glass):
"""
:type poured: int
:type query_row: int
:type query_glass: int
:rtype: float
"""
glasses=[[poured if i==0 and j==0 else 0 for i in range(j+1)] for j in range(query_row+1)]
for i in range(1,len(glasses)):
for j in range(len(glasses[i])):
if j-1>=0 and glasses[i-1][j-1]>1: glasses[i][j]+=(glasses[i-1][j-1]-1)/2
if j<=i-1 and glasses[i-1][j]>1: glasses[i][j]+=(glasses[i-1][j]-1)/2
return glasses[query_row][query_glass] if glasses[query_row][query_glass]<=1 else 1
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Output