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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.
1import java.util.Arrays;
2
3public class ChampagneTower {
4 public static double champagneTower(int poured, int query_row, int query_glass) {
5 double[][] dp = new double[101][101];
6 dp[0][0] = poured;
7 for (int i = 0; i < 100; i++) {
8 for (int j = 0; j <= i; j++) {
9 if (dp[i][j] >= 1) {
10 double overflow = (dp[i][j] - 1) / 2.0;
11 dp[i + 1][j] += overflow;
12 dp[i + 1][j + 1] += overflow;
13 }
14 }
15 }
16 return Math.min(1.0, dp[query_row][query_glass]);
17 }
18
19 public static void main(String[] args) {
20 System.out.printf("%.5f\n", champagneTower(2, 1, 1));
21 }
22}Using a 2D array 'dp', we simulate the champagne pouring process, distributing overflow champagne to subsequent rows. Each element stores the amount of champagne in a corresponding glass.
This approach optimizes the space complexity by using only a single array to store the champagne volumes, representing only two rows at a time.
Time Complexity: O(n^2). Space Complexity: O(n) due to the use of the single array.
1function champagneTower
Using an array 'dp', we track champagne amount in each current row glass, utilizing a space-efficient strategy to update and handle overflows across rows sequentially.