Champagne Tower solution and explanation
Google
9 more videos on this problem
Champagne Tower - Leetcode 799 - PythonNeetCodeIO · 13:48 · 17K views
Champagne Tower | Recursion | Memoization | GOOGLE | Leetcode - 799codestorywithMIK · 29:28 · 13K views
Champagne Tower | Live Coding with Explanation | Leetcode - 799Algorithms Made Easy · 11:29 · 8K views
Champagne Tower | Leetcode 799 | Matrix | Day-4Ayushi Sharma · 27:53 · 4K views
Champagne Tower | Leetcode 799 | Easy DP | Live coding session 🔥🔥Coding Decoded · 12:40 · 3K views
Champagne Tower | LeetCode 799 | Pascal’s Triangle Simulation & Dynamic ProgrammingSanyam IIT Guwahati · 22:20 · 826 views
Champagne Tower | LeetCode 799 | October LeetCode Challenge | Day 26Learn Code Repeat · 12:17 · 650 views
Daily Leetcode #2146: Feb 14, 2026 - 799. Champagne TowerProgramming Live with Larry · 13:11 · 570 views
799. Champagne Tower | Leetcode Daily - PythonLeetcode Daily · 8:09 · 506 views
Champagne Tower | LeetCode 799 | Coders CampCoders Camp · 11:57 · 461 views
The problem
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
Approach overview
Problem Overview: You pour poured cups of champagne into the top glass of a triangular tower. Each glass holds exactly 1 cup. Any overflow splits evenly to the two glasses below it. Given query_row and query_glass, compute how full that specific glass is after the champagne settles.
Approach 1: Simulation Using a Matrix (Time: O(n^2), Space: O(n^2))
The most direct way is to simulate the tower row by row using a 2D matrix. Create a matrix where tower[r][c] stores how much champagne reaches that glass. Start by placing poured in the top glass. Iterate through each row and check if a glass exceeds capacity (greater than 1). Any extra amount (value - 1) is split evenly and added to the two glasses below: (r+1, c) and (r+1, c+1). Continue propagating overflow until reaching query_row.
This approach mirrors the physical process of champagne flowing down the pyramid. The matrix ensures every overflow is tracked correctly. Because the tower up to row n contains roughly n^2/2 glasses, the simulation runs in O(n^2) time and uses O(n^2) space. This technique resembles classic dynamic programming where each state depends on the row above.
Approach 2: Space Optimized DP (Time: O(n^2), Space: O(n))
The observation: each row only depends on the previous row. Instead of storing the entire triangle, maintain a single array representing the current row. Start with an array where the first element equals poured. For every row, iterate from right to left and compute overflow contributions. If dp[i] > 1, the excess (dp[i] - 1) / 2 flows to dp[i] and dp[i+1] in the next iteration.
Processing from right to left prevents overwriting values needed for calculations. After processing query_row iterations, the target glass is simply min(1, dp[query_glass]). The logic still evaluates every glass up to the target row, so the time complexity remains O(n^2), but memory drops to O(n). This pattern is common in dynamic programming problems where transitions only depend on the previous layer of a triangular structure.
The state transition is straightforward: overflow from (r, c) contributes equally to (r+1, c) and (r+1, c+1). Because the tower structure resembles Pascal’s triangle, the DP flow is easy to reason about once you visualize the champagne distribution.
Recommended for interviews: Interviewers typically expect the dynamic programming simulation. Start with the matrix-based simulation to show you understand how overflow propagates through the tower. Then mention the space optimization that compresses rows into a single array. Demonstrating both versions shows strong reasoning about DP state transitions and memory optimization, which is common in problems involving triangular arrays or layered simulations.
Approach 1: Simulation Approach Using a Matrix
In 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.
def champagneTower(poured, query_row, query_glass):
dp = [[0] * 101 for _ in range(101)]
dp[0][0] = poured
for i in range(100):
for j in range(i + 1):
if dp[i][j] >= 1:
overflow = (dp[i][j] - 1) / 2.0
dp[i + 1][j] += overflow
dp[i + 1][j + 1] += overflow
return min(1.0, dp[query_row][query_glass])
print(f"{champagneTower(2, 1, 1):.5f}")import java.util.Arrays;
public class ChampagneTower {
public static double champagneTower(int poured, int query_row, int query_glass) {
double[][] dp = new double[101][101];
dp[0][0] = poured;
for (int i = 0; i < 100; i++) {
for (int j = 0; j <= i; j++) {
if (dp[i][j] >= 1) {
double overflow = (dp[i][j] - 1) / 2.0;
dp[i + 1][j] += overflow;
dp[i + 1][j + 1] += overflow;
}
}
}
return Math.min(1.0, dp[query_row][query_glass]);
}
public static void main(String[] args) {
System.out.printf("%.5f\n", champagneTower(2, 1, 1));
}
}#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
double champagneTower(int poured, int query_row, int query_glass) {
vector<vector<double>> dp(101, vector<double>(101, 0.0));
dp[0][0] = poured;
for (int i = 0; i < 100; ++i) {
for (int j = 0; j <= i; ++j) {
if (dp[i][j] >= 1) {
double overflow = (dp[i][j] - 1) / 2.0;
dp[i + 1][j] += overflow;
dp[i + 1][j + 1] += overflow;
}
}
}
return min(1.0, dp[query_row][query_glass]);
}
int main() {
cout << fixed << setprecision(5) << champagneTower(2, 1, 1) << endl;
return 0;
}function champagneTower(poured, query_row, query_glass) {
let tower = Array.from({ length: 101 }, () => Array(101).fill(0));
tower[0][0] = poured;
for (let i = 0; i < 100; i++) {
for (let j = 0; j <= i; j++) {
if (tower[i][j] >= 1) {
let overflow = (tower[i][j] - 1) / 2;
tower[i + 1][j] += overflow;
tower[i + 1][j + 1] += overflow;
}
}
}
return Math.min(1.0, tower[query_row][query_glass]);
}
console.log(champagneTower(2, 1, 1).toFixed(5));#include <stdio.h>
#include <string.h>
#define MAX_ROWS 100
double champagneTower(int poured, int query_row, int query_glass) {
double tower[MAX_ROWS][MAX_ROWS];
memset(tower, 0, sizeof(tower));
tower[0][0] = poured;
for (int i = 0; i <= query_row; ++i) {
for (int j = 0; j <= i; ++j) {
double overflow = (tower[i][j] - 1.0) / 2.0;
if (overflow > 0) {
tower[i+1][j] += overflow;
tower[i+1][j+1] += overflow;
}
}
}
return fmin(1.0, tower[query_row][query_glass]);
}
int main() {
printf("%.5f\n", champagneTower(2, 1, 1));
return 0;
}using System;
public class Solution {
public double ChampagneTower(int poured, int query_row, int query_glass) {
double[,] tower = new double[101, 101];
tower[0, 0] = poured;
for (int i = 0; i < 100; i++) {
for (int j = 0; j <= i; j++) {
if (tower[i, j] >= 1) {
double overflow = (tower[i, j] - 1) / 2.0;
tower[i + 1, j] += overflow;
tower[i + 1, j + 1] += overflow;
}
}
}
return Math.Min(1.0, tower[query_row, query_glass]);
}
public static void Main(string[] args) {
Solution s = new Solution();
Console.WriteLine("{0:F5}", s.ChampagneTower(2, 1, 1));
}
}We use a 2D list 'dp' to maintain the amount of champagne. For each glass, calculate the overflow and distribute it equally between the two glasses beneath it.
Time Complexity: O(n^2). Space Complexity: O(n^2).
Approach 2: Space Optimized Approach
This approach optimizes the space complexity by using only a single array to store the champagne volumes, representing only two rows at a time.
def champagneTower(poured, query_row, query_glass):
dp = [0] * 101
dp[0] = poured
for i in range(query_row):
for j in range(i, -1, -1):
if dp[j] > 1:
overflow = (dp[j] - 1.0) / 2.0
dp[j + 1] += overflow
dp[j] = 1.0
return min(1.0, dp[query_glass])
print(f"{champagneTower(2, 1, 1):.5f}")import java.util.Arrays;
public class ChampagneTower {
public static double champagneTower(int poured, int query_row, int query_glass) {
double[] dp = new double[101];
dp[0] = poured;
for (int i = 0; i < query_row; ++i) {
for (int j = i; j >= 0; --j) {
double overflow = (dp[j] - 1.0) / 2.0;
if (overflow > 0) {
dp[j + 1] += overflow;
dp[j] = 1.0;
}
}
}
return Math.min(1.0, dp[query_glass]);
}
public static void main(String[] args) {
System.out.printf("%.5f\n", champagneTower(2, 1, 1));
}
}#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
double champagneTower(int poured, int query_row, int query_glass) {
vector<double> dp(101, 0.0);
dp[0] = poured;
for (int i = 0; i < query_row; ++i) {
for (int j = i; j >= 0; --j) {
double overflow = (dp[j] - 1) / 2.0;
if (overflow > 0) {
dp[j + 1] += overflow;
dp[j] = 1.0;
}
}
}
return min(1.0, dp[query_glass]);
}
int main() {
cout << fixed << setprecision(5) << champagneTower(2, 1, 1) << endl;
return 0;
}function champagneTower(poured, query_row, query_glass) {
let dp = new Array(101).fill(0);
dp[0] = poured;
for (let i = 0; i < query_row; i++) {
for (let j = i; j >= 0; j--) {
if (dp[j] > 1) {
let overflow = (dp[j] - 1.0) / 2.0;
dp[j + 1] += overflow;
dp[j] = 1.0;
}
}
}
return Math.min(1.0, dp[query_glass]);
}
console.log(champagneTower(2, 1, 1).toFixed(5));#include <stdio.h>
#include <string.h>
#define MAX_COLS 100
double champagneTower(int poured, int query_row, int query_glass) {
double tower[MAX_COLS];
memset(tower, 0, sizeof(tower));
tower[0] = poured;
for (int i = 0; i < query_row; ++i) {
for (int j = i; j >= 0; --j) {
if (tower[j] > 1) {
double overflow = (tower[j] - 1) / 2.0;
tower[j + 1] += overflow;
tower[j] = 1.0;
}
}
}
return fmin(1.0, tower[query_glass]);
}
int main() {
printf("%.5f\n", champagneTower(2, 1, 1));
return 0;
}using System;
public class Solution {
public double ChampagneTower(int poured, int query_row, int query_glass) {
double[] dp = new double[101];
dp[0] = poured;
for (int i = 0; i < query_row; i++) {
for (int j = i; j >= 0; j--) {
double overflow = (dp[j] - 1.0) / 2.0;
if (overflow > 0) {
dp[j + 1] += overflow;
dp[j] = 1.0;
}
}
}
return Math.Min(1.0, dp[query_glass]);
}
public static void Main(string[] args) {
Solution s = new Solution();
Console.WriteLine("{0:F5}", s.ChampagneTower(2, 1, 1));
}
}We use a linear array 'dp' to hold the champagne amounts for each glass in the current row, updating in place to manage the overflow process effectively as we advance through the rows.
Time Complexity: O(n^2). Space Complexity: O(n).
Approach 3: Simulation
We directly simulate the process of pouring champagne.
Define a 2D array f, where f[i][j] represents the amount of champagne in the j-th glass of the i-th layer. Initially, f[0][0] = poured.
For each layer, if the amount of champagne in the current glass f[i][j] is greater than 1, the champagne will flow to two glasses in the next layer. The amount flowing in is \frac{f[i][j]-1}{2}, which is the amount of champagne in the current glass minus 1 and then divided by 2. Then the amount of champagne in the current glass is updated to 1.
After the simulation ends, return f[query_row][query_glass].
The time complexity is O(n^2), and the space complexity is O(n^2), where n is the number of layers, i.e., query_row.
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];
}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]
}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]
}
}Approach 4: Simulation (Space Optimization)
Since the amount of champagne in each layer only depends on the amount in the previous layer, we can use a rolling array approach to optimize space complexity, converting the 2D array to a 1D array.
The time complexity is O(n^2), and the space complexity is O(n), where n is the number of layers, i.e., query_row.
function champagneTower(poured: number, query_row: number, query_glass: number): number {
let f: number[] = [poured];
for (let i = 1; i <= query_row; ++i) {
const g: number[] = new Array(i + 1).fill(0);
for (let j = 0; j < i; ++j) {
if (f[j] > 1) {
const half = (f[j] - 1) / 2.0;
g[j] += half;
g[j + 1] += half;
}
}
f = g;
}
return Math.min(1, f[query_glass]);
}func champagneTower(poured int, query_row int, query_glass int) float64 {
f := []float64{float64(poured)}
for i := 1; i <= query_row; i++ {
g := make([]float64, i+1)
for j, v := range f {
if v > 1 {
half := (v - 1) / 2.0
g[j] += half
g[j+1] += half
}
}
f = g
}
return math.Min(1, f[query_glass])
}impl Solution {
pub fn champagne_tower(poured: i32, query_row: i32, query_glass: i32) -> f64 {
let mut f = vec![poured as f64];
for i in 1..=query_row {
let mut g = vec![0.0; (i + 1) as usize];
for j in 0..i as usize {
if f[j] > 1.0 {
let half = (f[j] - 1.0) / 2.0;
g[j] += half;
g[j + 1] += half;
}
}
f = g;
}
f[query_glass as usize].min(1.0)
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Simulation Approach Using a Matrix | Time Complexity: O(n^2). Space Complexity: O(n^2). |
| Space Optimized Approach | Time Complexity: O(n^2). Space Complexity: O(n). |
| Simulation | – |
| Simulation (Space Optimization) | – |
Complexity analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Simulation Using a Matrix | O(n^2) | O(n^2) | Best for understanding the overflow process clearly and visualizing the entire tower |
| Space Optimized Dynamic Programming | O(n^2) | O(n) | Preferred when memory usage matters or when implementing a concise interview solution |
Frequently asked questions
Is Champagne Tower easy or hard?
Champagne Tower is considered a medium difficulty problem on LeetCode. The logic is straightforward once you model the overflow correctly, but recognizing the dynamic programming pattern and optimizing space can be tricky for beginners.
How to solve Champagne Tower in O(n)?
The problem cannot be solved in O(n) time because every row up to query_row must be evaluated. However, you can reduce the space complexity to O(n) by storing only the current row in a 1D array and updating it from right to left while propagating overflow.
What is the best approach for Champagne Tower?
The standard solution uses dynamic programming to simulate how overflow spreads through the tower. Each glass passes its extra champagne equally to the two glasses below. This row-by-row simulation runs in O(n^2) time and can be implemented with either a 2D matrix or a space‑optimized 1D array.
What data structure is used in Champagne Tower?
Most solutions use either a 2D array representing the champagne tower or a 1D array for the space‑optimized dynamic programming version. The structure behaves like a triangular DP table where each state distributes overflow to two children.
What is the time complexity of Champagne Tower?
The time complexity is O(n^2), where n is the target row. The algorithm processes every glass up to that row because overflow from each glass may affect the two glasses below it. Even with space optimization, the number of computations remains quadratic.
Champagne Tower Python or Java solution approach?
Python and Java implementations follow the same dynamic programming idea. Use an array to track how much champagne reaches each glass, compute overflow as (value − 1) / 2, and distribute it to the next row. The final answer is min(1, amount_in_target_glass).
Is Champagne Tower asked at Google, Amazon, or Meta?
Champagne Tower appears in interview preparation lists for companies such as Amazon and Google. It tests dynamic programming intuition, simulation skills, and the ability to optimize space when working with triangular structures.