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Leetcode 808. Soup Servings

Problem Statement

The problem is taken from LeetCode (#808): “Soup Servings”.

There are two types of soup: type A and type B. Initially, we have N ml of soup. There are four kinds of operations:

  1. Serve 100 ml of soup A and 0 ml of soup B
  2. Serve 75 ml of soup A and 25 ml of soup B
  3. Serve 50 ml of soup A and 50 ml of soup B
  4. Serve 25 ml of soup A and 75 ml of soup B

When we serve some soup, we consume the specified number of milliliters from soup A and soup B. After soup has been served, we consider the following events later:

Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time.

Clarifying Questions

  1. What should be the behavior if N is very large?
  2. Is there any limit to the size of N for which the precision should be accounted?

Assumption: For large values of N, the result is very close to 1.

Strategy

  1. Use memoization with recursion to avoid redundant calculations.
  2. Adjust serving operations and base cases to keep track of the probabilities.
  3. Use a dynamic programming approach to store intermediate results.

Code

import java.util.HashMap;
import java.util.Map;

public class SoupServings {
    private Map<String, Double> memo = new HashMap<>();

    public double soupServings(int N) {
        if (N > 4800) 
            return 1.0;
        
        return serve(N, N);
    }
    
    private double serve(int A, int B) {
        if (A <= 0 && B <= 0)
            return 0.5;
        if (A <= 0)
            return 1.0;
        if (B <= 0)
            return 0.0;
        
        String key = A + "," + B;
        if (memo.containsKey(key))
            return memo.get(key);
    
        double probability = 0.25 * (
            serve(A - 100, B) +
            serve(A - 75, B - 25) +
            serve(A - 50, B - 50) +
            serve(A - 25, B - 75)
        );
        
        memo.put(key, probability);
        return probability;
    }

    public static void main(String[] args) {
        SoupServings solution = new SoupServings();
        System.out.println(solution.soupServings(50));  // Example input
        System.out.println(solution.soupServings(100)); // Example input
    }
}

Strategy Explanation

  1. Memoization: The serve function uses a HashMap to memorize previously computed values to avoid redundant calculations and retain the intermediate results.
  2. Base Cases:
    • When both A and B are exhausted simultaneously, return 0.5.
    • When only A is exhausted, return 1.0.
    • When only B is exhausted, return 0.0.
  3. Operations: The four kinds of operations are recursively applied, and the resultant probability is averaged by multiplying by 0.25.
  4. Large N optimization: For large values (greater than 4800), the result is approximated to 1.0 due to diminishing probability differences.

Time Complexity

The time complexity is O(N^2) due to memoization, where N is the soup quantity in the worst case, noting that N in this context will be reduced by certain thresholds, making the number of effective calls intractable. This is efficient given memoized results are reused.

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