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We can use a dynamic programming approach with memoization to store the probability results for known quantities of soup A and B. We'll treat the problem states as a grid where each cell holds the probability of that state leading to soup A becoming empty first. The key to optimizing this solution is to memoize intermediate results to avoid redundant calculations by storing probabilities in a 2D array or dictionary.
Time Complexity: O(n^2) due to memoization.
Space Complexity: O(n^2) for the memoization table.
1public class SoupServings {
2 public double soupServings(int N) {
3 if (N > 4800) return 1.0;
4 return calculate((N + 24) / 25, (N + 24) / 25, new HashMap<>());
5 }
6 private double calculate(int a, int b, Map<String, Double> memo) {
7 if (a <= 0 && b <= 0) return 0.5;
8 if (a <= 0) return 1.0;
9 if (b <= 0) return 0.0;
10 String key = a + "," + b;
11 if (memo.containsKey(key)) return memo.get(key);
12 double result = 0.25 * (calculate(a - 4, b, memo) +
13 calculate(a - 3, b - 1, memo) +
14 calculate(a - 2, b - 2, memo) +
15 calculate(a - 1, b - 3, memo));
16 memo.put(key, result);
17 return result;
18 }
19}This Java solution uses a recursive function with memoization stored in a HashMap. Similar base cases are checked and results stored with a concatenated string key to identify states. Reduction by integer division by 25 ensures manageable state space.
An alternative is solving the problem iteratively using a DP table to store results for each state. We build up from small values to larger values of soup A and B, filling out the DP table based on possible outcomes until we reach the desired amounts.
Time Complexity: O(n^2) as we fill up each cell of an N-by-N table.
Space Complexity: O(n^2) for maintaining the DP table.
1#include <iostream>
2#include <vector>
3using namespace std;
4
5double soupServings(int N) {
6 if (N > 4800) return 1.0;
N = (N + 24) / 25;
vector<vector<double>> dp(N + 1, vector<double>(N + 1, 0));
for (int i = 0; i <= N; ++i) {
for (int j = 0; j <= N; ++j) {
if (i == 0 && j == 0) {
dp[i][j] = 0.5;
} else if (i == 0) {
dp[i][j] = 1;
} else if (j == 0) {
dp[i][j] = 0;
} else {
dp[i][j] = 0.25 * (getDP(dp, i - 4, j) + getDP(dp, i - 3, j - 1) + getDP(dp, i - 2, j - 2) + getDP(dp, i - 1, j - 3));
}
}
}
return dp[N][N];
}
double getDP(const vector<vector<double>>& dp, int i, int j) {
if (i <= 0 && j <= 0) {
return 0.5;
}
if (i <= 0) {
return 1;
}
if (j <= 0) {
return 0;
}
return dp[i][j];
}This solution implements the iterative DP approach in C++. We initialize a 2D vector to hold probabilities and populate it from known base cases upward. The getDP function simplifies access and calculations for the DP table.