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.)

Examples

Example 1

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Input: poured = 1, query_row = 1, query_glass = 1
Output: 0.00000
Explanation: We poured 1 cup of champagne 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

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Input: poured = 2, query_row = 1, query_glass = 1
Output: 0.50000
Explanation: We poured 2 cups of champagne 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 champagne.

Example 3

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Input: poured = 100000009, query_row = 33, query_glass = 17
Output: 1.00000

Solution

Method 1 – Dynamic Programming Simulation

Intuition: The problem can be visualized as simulating the flow of champagne through a triangular array (pyramid) of glasses. Each glass can hold at most 1 cup; any overflow is split equally between the two glasses below it. By tracking the amount in each glass row by row, we can determine the fullness of any glass.

Approach:

  1. Create a 2D array dp where dp[i][j] represents the amount of champagne in the glass at row i and position j.
  2. Pour all champagne into the top glass (dp[0][0]).
  3. For each row from 0 to query_row - 1, iterate through each glass:
  • If a glass has more than 1 cup, calculate the overflow.
  • Split the overflow equally to the two glasses below (dp[i+1][j] and dp[i+1][j+1]).
  1. The answer is the minimum of 1 and the amount in dp[query_row][query_glass] (since a glass can’t be more than full).

Code

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class Solution {
public:
   double champagneTower(int poured, int row, int glass) {
      vector<vector<double>> dp(row + 2, vector<double>(row + 2, 0.0));
      dp[0][0] = poured;
      for (int i = 0; i <= row; ++i) {
        for (int j = 0; j <= i; ++j) {
           double overflow = max(0.0, (dp[i][j] - 1.0) / 2.0);
           dp[i+1][j] += overflow;
           dp[i+1][j+1] += overflow;
        }
      }
      return min(1.0, dp[row][glass]);
   }
};
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func champagneTower(poured int, row int, glass int) float64 {
   dp := make([][]float64, row+2)
   for i := range dp {
      dp[i] = make([]float64, row+2)
   }
   dp[0][0] = float64(poured)
   for i := 0; i <= row; i++ {
      for j := 0; j <= i; j++ {
        overflow := (dp[i][j] - 1.0) / 2.0
        if overflow > 0 {
           dp[i+1][j] += overflow
           dp[i+1][j+1] += overflow
        }
      }
   }
   if dp[row][glass] > 1.0 {
      return 1.0
   }
   return dp[row][glass]
}
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class Solution {
   public double champagneTower(int poured, int row, int glass) {
      double[][] dp = new double[row + 2][row + 2];
      dp[0][0] = poured;
      for (int i = 0; i <= row; i++) {
        for (int j = 0; j <= i; j++) {
           double overflow = Math.max(0, (dp[i][j] - 1) / 2.0);
           dp[i + 1][j] += overflow;
           dp[i + 1][j + 1] += overflow;
        }
      }
      return Math.min(1, dp[row][glass]);
   }
}
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class Solution {
   fun champagneTower(poured: Int, row: Int, glass: Int): Double {
      val dp = Array(row + 2) { DoubleArray(row + 2) }
      dp[0][0] = poured.toDouble()
      for (i in 0..row) {
        for (j in 0..i) {
           val overflow = ((dp[i][j] - 1.0) / 2.0).coerceAtLeast(0.0)
           dp[i + 1][j] += overflow
           dp[i + 1][j + 1] += overflow
        }
      }
      return dp[row][glass].coerceAtMost(1.0)
   }
}
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class Solution:
   def champagneTower(self, poured: int, row: int, glass: int) -> float:
      dp: list[list[float]] = [[0.0] * (row + 2) for _ in range(row + 2)]
      dp[0][0] = float(poured)
      for i in range(row + 1):
        for j in range(i + 1):
           overflow: float = max(0.0, (dp[i][j] - 1.0) / 2.0)
           dp[i + 1][j] += overflow
           dp[i + 1][j + 1] += overflow
      return min(1.0, dp[row][glass])
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impl Solution {
   pub fn champagne_tower(poured: i32, row: i32, glass: i32) -> f64 {
      let n = row as usize + 2;
      let mut dp = vec![vec![0.0; n]; n];
      dp[0][0] = poured as f64;
      for i in 0..=row as usize {
        for j in 0..=i {
           let overflow = ((dp[i][j] - 1.0) / 2.0).max(0.0);
           dp[i + 1][j] += overflow;
           dp[i + 1][j + 1] += overflow;
        }
      }
      dp[row as usize][glass as usize].min(1.0)
   }
}

Complexity

  • ⏰ Time complexity: O(row^2) — We fill each glass up to the target row.
  • 🧺 Space complexity: O(row^2) — We store the amount in each glass up to the target row.