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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.
Time Complexity: O(n^2) where n is the number of rows traversed. Space Complexity: O(n^2) for the 2D array used to simulate the champagne distribution.
1def champagneTower(poured, query_row, query_glass):
2 dp = [[0] * 101 for _ in range(101)]
3 dp[0][0] = poured
4 for i in range(100):
5 for j in range(i + 1):
6 if dp[i][j] >= 1:
7 overflow = (dp[i][j] - 1) / 2.0
8 dp[i + 1][j] += overflow
9 dp[i + 1][j + 1] += overflow
10 return min(1.0, dp[query_row][query_glass])
11
12print(f"{champagneTower(2, 1, 1):.5f}")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.
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.