class Solution:
    '''
    Time Complexity: O(R^2)
    Space Complexity: O(R)
    where, R is query_row
    NOTE: You can optimize this further by only calculating first half of the row, since it's symmetric, 
    but this is good enough for the constraints given in the problem, and it's simple to understand and implement, good job!
    '''
    def champagneTower(self, poured: int, query_row: int, query_glass: int) -> float:
        curRow = [poured]
        nextRow = [0.0]*2

        for _ in range(query_row):
            for j in range(len(curRow)):
                excess = curRow[j]-1
                if excess > 0:
                    nextRow[j] += excess/2
                    nextRow[j+1] += excess/2
            curRow = nextRow
            nextRow = [0.0] * (len(curRow) + 1)
        
        return 1.0 if curRow[query_glass] > 1 else curRow[query_glass]