Sponsored
Sponsored
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.
1def soupServings(N):
2 if N > 4800:
3 return 1
4
5 memo = {}
6
7 def dp(a, b):
8 if a <= 0 and b <= 0:
9 return 0.5
10 if a <= 0:
11 return 1
12 if b <= 0:
13 return 0
14 if (a, b) in memo:
15 return memo[(a, b)]
16
17 memo[(a, b)] = 0.25 * (dp(a - 100, b) + dp(a - 75, b - 25) + dp(a - 50, b - 50) + dp(a - 25, b - 75))
18 return memo[(a, b)]
19
20 N = (N + 24) // 25
21 return dp(N, N)In this Python solution, we use a recursive function with memoization. If the amount of soup A or B is less than or equal to zero, we return the base probabilities. The recursive step calculates the probability for each serving option and stores the result in a memo dictionary to avoid recalculating. For large n, greater than 4800, the probability approaches 1, so we return 1 directly.
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.