[leetcode #799] Champagne Tower

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 <= 10⁹
・ 0 <= query_glass <= query_row < 100

Idea

샴페인 타워를 만들었을 때 query_row 번째 행과 query_glass 열에 있는 잔에 얼마만큼의 샴페인이 찼는지 확인하는 문제다.

double 데이터 형을 저장하는 2d 배열을 만든다. 해당 배열에 데이터를 채울 때 위 샴페인이 1보다 큰 샴페인을 가졌는지 확인한다. 샴페인이 1보다 크다면 아래로 1/2만큼 흐르므로 해당 샴페인을 아랫잔에 더한다. 양 끝의 잔은 왼쪽 또는 오른쪽 위의 샴페인 잔만 확인하고, 나머지 경우는 왼쪽 오른쪽 모두 확인한다.

계산이 끝난 뒤 주어진 query_row, query_glass에 해당하는 값을 넘겨주면 된다. 이 때 값이 1보다 크면 샴페인이 다 찼다는 의미이므로 1을 그대로 리턴한다.

Solution

class Solution {
    public double champagneTower(int poured, int query_row, int query_glass) {
        double[][] glasses = new double[100][100];
        glasses[0][0] = poured;

        for (int i=1; i <= query_row; i++) {
            for (int j=0; j <= i; j++) {
                if (j == 0 && glasses[i-1][j] > 1) {
                    glasses[i][j] = (glasses[i-1][j] - 1) / 2;
                } else if (j == i && glasses[i-1][j-1] > 1) {
                    glasses[i][j] = (glasses[i-1][j-1] - 1) / 2;
                } else if (j > 0 && j < i){
                    if (glasses[i-1][j-1] > 1)
                        glasses[i][j] += (glasses[i-1][j-1] - 1) / 2;
                    if (glasses[i-1][j] > 1)
                        glasses[i][j] += (glasses[i-1][j] - 1) / 2;
                }
            }
        }

        return Math.min(1, glasses[query_row][query_glass]);
    }
}

Reference

https://leetcode.com/problems/champagne-tower/