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Java-based LeetCode algorithm problem solutions, regularly updated

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package g0801_0900.s0808_soup_servings;

// #Medium #Dynamic_Programming #Math #Probability_and_Statistics
// #2022_03_23_Time_1_ms_(96.04%)_Space_41.7_MB_(78.22%)

/**
 * 808 - Soup Servings\.
 *
 * Medium
 *
 * There are two types of soup: **type A** and **type B**. Initially, we have `n` ml of each type 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** , and
 * 4.  Serve `25` ml of **soup A** and `75` ml of **soup B**.
 *
 * When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability `0.25`. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
 *
 * **Note** that we do not have an operation where all `100` ml's of **soup B** are used first.
 *
 * Return _the probability that **soup A** will be empty first, plus half the probability that **A** and **B** become empty at the same time_. Answers within 10^-5 of the actual answer will be accepted.
 *
 * **Example 1:**
 *
 * **Input:** n = 50
 *
 * **Output:** 0.62500
 *
 * **Explanation:** If we choose the first two operations, A will become empty first. 
 *
 * For the third operation, A and B will become empty at the same time. 
 *
 * For the fourth operation, B will become empty first. 
 *
 * So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 \* (1 + 1 + 0.5 + 0) = 0.625.
 *
 * **Example 2:**
 *
 * **Input:** n = 100
 *
 * **Output:** 0.71875
 *
 * **Constraints:**
 *
 * *   0 <= n <= 10⁹
**/
public class Solution {
    public double soupServings(int n) {
        return solve(n);
    }

    private double solve(int n) {
        n = n / 25 + (n % 25 > 0 ? 1 : 0);
        if (n >= 500) {
            return 1.0;
        }
        return find(n, n, new Double[n + 1][n + 1]);
    }

    private double find(int a, int b, Double[][] mem) {
        if (a <= 0 && b <= 0) {
            return 0.5;
        } else if (a <= 0) {
            return 1;
        } else if (b <= 0) {
            return 0;
        }
        double prob;
        if (mem[a][b] != null) {
            return mem[a][b];
        }
        prob = find(a - 4, b, mem);
        prob += find(a - 3, b - 1, mem);
        prob += find(a - 2, b - 2, mem);
        prob += find(a - 1, b - 3, mem);
        mem[a][b] = 0.25 * prob;
        return mem[a][b];
    }
}