826. Soup Servings¶

Difficulty: Medium

826. Soup Servings

Medium

You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns:

pour 100 mL from type A and 0 mL from type B
pour 75 mL from type A and 25 mL from type B
pour 50 mL from type A and 50 mL from type B
pour 25 mL from type A and 75 mL from type B

Note:

There is no operation that pours 0 mL from A and 100 mL from B.
The amounts from A and B are poured simultaneously during the turn.
If an operation asks you to pour more than you have left of a soup, pour all that remains of that soup.

The process stops immediately after any turn in which one of the soups is used up.

Return the probability that A is used up before B, plus half the probability that both soups are used up in the same turn. Answers within 10^-5 of the actual answer will be accepted.

Example 1:

Input: n = 50
Output: 0.62500
Explanation: 
If we perform either of the first two serving operations, soup A will become empty first.
If we perform the third operation, A and B will become empty at the same time.
If we perform 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
Explanation: 
If we perform the first serving operation, soup A will become empty first.
If we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4.
If we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3.
If we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.

Constraints:

0 <= n <= 10⁹

Solution¶

class Solution {
    private Map<Pair, Double> mp = new HashMap<>();

    private double recur(int a, int b) {
        if (a <= 0 && b <= 0) return 0.5;
        if (a <= 0 && b > 0) return 1;
        if (a > 0 && b <= 0) return 0;
        if (mp.containsKey(new Pair(a, b))) return mp.get(new Pair(a, b));

        double op1 = recur(a - 100, b);
        double op2 = recur(a - 75, b - 25);
        double op3 = recur(a - 50, b - 50);
        double op4 = recur(a - 25, b - 75);

        double result = 0.25 * (op1 + op2 + op3 + op4);
        mp.put(new Pair(a, b), result);
        return result;
    }

    public double soupServings(int n) {
        if (n >= 4800) return 1;
        double ans = recur(n, n);
        return ans;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here