Move from brute-force thinking to an efficient approach using dynamic programming strategy.
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 < 100Problem summary: 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
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Dynamic Programming
1 1 1
2 1 1
100000009 33 17
number-of-ways-to-build-house-of-cards)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #799: Champagne Tower
class Solution {
public double champagneTower(int poured, int query_row, int query_glass) {
double[][] f = new double[101][101];
f[0][0] = poured;
for (int i = 0; i <= query_row; ++i) {
for (int j = 0; j <= i; ++j) {
if (f[i][j] > 1) {
double half = (f[i][j] - 1) / 2.0;
f[i][j] = 1;
f[i + 1][j] += half;
f[i + 1][j + 1] += half;
}
}
}
return f[query_row][query_glass];
}
}
// Accepted solution for LeetCode #799: Champagne Tower
func champagneTower(poured int, query_row int, query_glass int) float64 {
f := [101][101]float64{}
f[0][0] = float64(poured)
for i := 0; i <= query_row; i++ {
for j := 0; j <= i; j++ {
if f[i][j] > 1 {
half := (f[i][j] - 1) / 2.0
f[i][j] = 1
f[i+1][j] += half
f[i+1][j+1] += half
}
}
}
return f[query_row][query_glass]
}
# Accepted solution for LeetCode #799: Champagne Tower
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]
// Accepted solution for LeetCode #799: Champagne Tower
impl Solution {
pub fn champagne_tower(poured: i32, query_row: i32, query_glass: i32) -> f64 {
let mut f = vec![vec![0.0; 101]; 101];
f[0][0] = poured as f64;
for i in 0..=query_row as usize {
for j in 0..=i {
if f[i][j] > 1.0 {
let half = (f[i][j] - 1.0) / 2.0;
f[i][j] = 1.0;
f[i + 1][j] += half;
f[i + 1][j + 1] += half;
}
}
}
f[query_row as usize][query_glass as usize]
}
}
// Accepted solution for LeetCode #799: Champagne Tower
function champagneTower(poured: number, query_row: number, query_glass: number): number {
const f: number[][] = Array.from({ length: 101 }, () => Array(101).fill(0));
f[0][0] = poured;
for (let i = 0; i <= query_row; ++i) {
for (let j = 0; j <= i; ++j) {
if (f[i][j] > 1) {
const half = (f[i][j] - 1) / 2.0;
f[i][j] = 1;
f[i + 1][j] += half;
f[i + 1][j + 1] += half;
}
}
}
return f[query_row][query_glass];
}
Use this to step through a reusable interview workflow for this problem.
Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.
Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.
Review these before coding to avoid predictable interview regressions.
Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.
Usually fails on: Correctness breaks on cases that differ only in hidden state.
Fix: Define state so each unique subproblem maps to one DP cell.