799. Champagne Tower LeetCode Solution

In this guide, you will get 799. Champagne Tower LeetCode Solution with the best time and space complexity. The solution to Champagne Tower problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Champagne Tower solution in C++
Champagne Tower solution in Java
Champagne Tower solution in Python
Additional Resources

799. Champagne Tower LeetCode Solution image

Problem Statement of 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 <= 109
0 <= query_glass <= query_row < 100

Complexity Analysis

Time Complexity: O(|\texttt{query_row}|^2)
Space Complexity: O(n^2)

799. Champagne Tower LeetCode Solution in C++

class Solution {
 public:
  double champagneTower(int poured, int query_row, int query_glass) {
    vector<vector<double>> dp(query_row + 1, vector<double>(query_row + 1));
    dp[0][0] = poured;

    for (int i = 0; i < query_row; ++i)
      for (int j = 0; j <= i; ++j)
        if (dp[i][j] > 1) {
          dp[i + 1][j] += (dp[i][j] - 1) / 2.0;
          dp[i + 1][j + 1] += (dp[i][j] - 1) / 2.0;
        }

    return min(1.0, dp[query_row][query_glass]);
  }
};
/* code provided by PROGIEZ */

799. Champagne Tower LeetCode Solution in Java

class Solution {
  public double champagneTower(int poured, int query_row, int query_glass) {
    double[][] dp = new double[query_row + 1][query_row + 1];
    dp[0][0] = poured;

    for (int i = 0; i < query_row; ++i)
      for (int j = 0; j <= i; ++j)
        if (dp[i][j] > 1) {
          dp[i + 1][j] += (dp[i][j] - 1) / 2.0;
          dp[i + 1][j + 1] += (dp[i][j] - 1) / 2.0;
        }

    return Math.min(1.0, dp[query_row][query_glass]);
  }
}
// code provided by PROGIEZ

799. Champagne Tower LeetCode Solution in Python

class Solution {
 public:
  double champagneTower(int poured, int query_row, int query_glass) {
    vector<double> dp(query_row + 1);
    dp[0] = poured;

    for (int i = 0; i < query_row; ++i) {
      vector<double> newDp(query_row + 1);
      for (int j = 0; j <= i; ++j)
        if (dp[j] > 1) {
          newDp[j] += (dp[j] - 1) / 2.0;
          newDp[j + 1] += (dp[j] - 1) / 2.0;
        }
      dp = std::move(newDp);
    }

    return min(1.0, dp[query_glass]);
  }
};
 # code by PROGIEZ