grid47 Exploring patterns and algorithms
Aug 18, 2024
6 min read
Solution to LeetCode 808: Soup Servings Problem
You have two types of soup: A and B, with an initial volume of n ml of each. You can serve soup using four possible operations, each with a 25% chance of being chosen randomly. When serving, the amounts of soups A and B used are specified by each operation. If there is insufficient soup for a full operation, serve as much as possible. The process stops when one of the soups runs out. Return the probability that soup A will be exhausted first, plus half the probability that both soups will be exhausted at the same time.
Problem
Approach
Steps
Complexity
Input: The input consists of a single integer n, which represents the initial amount (in ml) of both soup A and soup B.
Example: Input: n = 50
Constraints:
• 0 <= n <= 10^9
Output: Return the probability that soup A becomes empty first, plus half the probability that both soups become empty simultaneously. The answer must be accurate to 10^-5.
Example: Output: 0.62500
Constraints:
• The answer will be considered correct if the error is within 10^-5 of the actual answer.
Goal: The goal is to calculate the probability of different outcomes where either soup A or both soups become empty based on random choices of operations.
Steps:
• Represent the problem as a recursive dynamic programming problem, where each state tracks the remaining soup quantities of A and B.
• Use memoization to store the results of already computed states to avoid redundant calculations.
• Each recursive call simulates one of the four operations and computes the resulting probabilities, considering the constraints of the problem.
Goal: Ensure that the input is within the provided bounds and that the solution handles edge cases.
Steps:
• n is between 0 and 10^9.
Assumptions:
• We assume that the operation probabilities are equal (0.25 for each of the four operations).
• The process stops once either soup A or both soups are exhausted.
• Input: Input: n = 50
• Explanation: With 50 ml of soup A and B, there are four possible operations to serve the soup. Depending on the operation chosen, either soup A or both soups may be emptied. The probability of each outcome is calculated by simulating the operations and tracking when one or both soups become empty.
Approach: The approach leverages dynamic programming with memoization to efficiently calculate the probability of each possible outcome based on the operations performed. This avoids recalculating the same state multiple times.
Observations:
• The problem requires recursive thinking due to the multiple choices of operations and the changes in the soup quantities.
• Memoization can help optimize the recursive approach by storing the results of already calculated states and reusing them.
Steps:
• Define a recursive function that calculates the probability of A or both soups becoming empty, based on the remaining quantities of soup A and B.
• Use memoization to store the results of each state (remaining soup quantities of A and B).
• Handle base cases where one of the soups is empty and return the corresponding probability for those cases.
• Return the final probability after considering all operations.
Empty Inputs:
• If n = 0, both soups are already empty, so the result is 0.
Large Inputs:
• For very large values of n, the memoization approach ensures that we do not recompute the same states repeatedly.
Special Values:
• If n is very small (e.g., n = 1), the result depends on the available operations and how quickly one soup is exhausted.
Constraints:
• Ensure that the solution works efficiently for large values of n.
classSolution {
vector<vector<double>> mem;
public:double soupServings(int n) {
mem.resize(200, vector<double>(200, 0));
return n >4800?1:f((n +24)/25, (n+24)/25);
}
doublef(int a, int b) {
if(a <=0&& b <=0) return0.5;
if(a <=0) return1;
if(b <=0) return0;
if(mem[a][b] >0) return mem[a][b];
mem[a][b] =0.25* (f(a-4, b) + f(a-3, b-1) + f(a -2, b -2) + f(a-1, b-3));
return mem[a][b];
}
1 : Class Definition
classSolution {
This defines a class named 'Solution' where the problem-solving method will be implemented.
2 : Variable Declaration
vector<vector<double>> mem;
Declares a 2D vector 'mem' to store the computed probabilities of soup servings.
3 : Access Modifier
public:
This is the access modifier, making the following methods accessible from outside the class.
4 : Method Declaration
doublesoupServings(int n) {
This method calculates the probability of serving the soups until one runs out, given an initial amount 'n'.
5 : Vector Initialization
mem.resize(200, vector<double>(200, 0));
Resizes the 'mem' vector to store 200 rows and columns, initializing all values to 0.
6 : Conditional Return
return n >4800?1:f((n +24)/25, (n+24)/25);
Checks if 'n' is greater than 4800, returning 1 if true, otherwise calling the function 'f' with adjusted parameters.
7 : Method Declaration
doublef(int a, int b) {
This method is a helper recursive function that calculates the probability for each state of the soup servings.
8 : Base Case Check
if(a <=0&& b <=0) return0.5;
Checks if both soups are empty, returning a probability of 0.5 (equal chances of either soup running out first).
9 : Base Case Check
if(a <=0) return1;
If soup 'a' is empty, return a probability of 1 (soup 'b' runs out first).
10 : Base Case Check
if(b <=0) return0;
If soup 'b' is empty, return a probability of 0 (soup 'a' runs out first).
11 : Memoization Check
if(mem[a][b] >0) return mem[a][b];
If the probability for the current state has already been computed, return the stored value from 'mem'.
12 : Memoization Update
mem[a][b] =0.25* (f(a-4, b) + f(a-3, b-1) + f(a -2, b -2) + f(a-1, b-3));
Recursively computes the probability for the current state by considering all possible moves and updates the 'mem' vector with the result.
13 : Return Statement
return mem[a][b];
Returns the computed probability stored in 'mem' for the current state (a, b).
Best Case: O(1) for small n values where the result can be computed directly.
Average Case: O(n^2) due to the recursive nature of the solution with memoization for a grid of possible states.
Worst Case: O(n^2), where n is the maximum possible value.
Description: The time complexity depends on the number of unique states and the recursive depth of the function.
Best Case: O(1), for very small n values where minimal space is needed.
Worst Case: O(n^2), since the memoization stores results for each pair of soup quantities.
Description: The space complexity is determined by the memoization table, which stores results for each pair of remaining soup quantities.
