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808. Soup Servings

Description

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

Solutions

In this problem, since each operation is a multiple of $25$, we can consider every $25ml$ of soup as one unit. This reduces the data scale to $\left \lceil \frac{n}{25} \right \rceil$.

We design a function $dfs(i, j)$, which represents the probability result when there are $i$ units of soup $A$ and $j$ units of soup $B$ remaining.

When $i \leq 0$ and $j \leq 0$, it means both soups are finished, and we should return $0.5$. When $i \leq 0$, it means soup $A$ is finished first, and we should return $1$. When $j \leq 0$, it means soup $B$ is finished first, and we should return $0$.

Next, for each operation, we have four choices:

  • Take $4$ units from soup $A$ and $0$ units from soup $B$;
  • Take $3$ units from soup $A$ and $1$ unit from soup $B$;
  • Take $2$ units from soup $A$ and $2$ units from soup $B$;
  • Take $1$ unit from soup $A$ and $3$ units from soup $B$.

Each choice has a probability of $0.25$, so we can derive:

$$ dfs(i, j) = 0.25 \times (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3)) $$

We use memoization to store the results of the function.

Additionally, we find that when $n=4800$, the result is $0.999994994426$, and the required precision is $10^{-5}$. As $n$ increases, the result gets closer to $1$. Therefore, when $n \gt 4800$, we can directly return $1$.

The time complexity is $O(C^2)$, and the space complexity is $O(C^2)$. In this problem, $C=200$.

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class Solution:
    def soupServings(self, n: int) -> float:
        @cache
        def dfs(i: int, j: int) -> float:
            if i <= 0 and j <= 0:
                return 0.5
            if i <= 0:
                return 1
            if j <= 0:
                return 0
            return 0.25 * (
                dfs(i - 4, j)
                + dfs(i - 3, j - 1)
                + dfs(i - 2, j - 2)
                + dfs(i - 1, j - 3)
            )

        return 1 if n > 4800 else dfs((n + 24) // 25, (n + 24) // 25)

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class Solution {
    private double[][] f = new double[200][200];

    public double soupServings(int n) {
        return n > 4800 ? 1 : dfs((n + 24) / 25, (n + 24) / 25);
    }

    private double dfs(int i, int j) {
        if (i <= 0 && j <= 0) {
            return 0.5;
        }
        if (i <= 0) {
            return 1.0;
        }
        if (j <= 0) {
            return 0;
        }
        if (f[i][j] > 0) {
            return f[i][j];
        }
        double ans
            = 0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
        f[i][j] = ans;
        return ans;
    }
}

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class Solution {
public:
    double soupServings(int n) {
        double f[200][200] = {0.0};
        function<double(int, int)> dfs = [&](int i, int j) -> double {
            if (i <= 0 && j <= 0) return 0.5;
            if (i <= 0) return 1;
            if (j <= 0) return 0;
            if (f[i][j] > 0) return f[i][j];
            double ans = 0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
            f[i][j] = ans;
            return ans;
        };
        return n > 4800 ? 1 : dfs((n + 24) / 25, (n + 24) / 25);
    }
};

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func soupServings(n int) float64 {
    if n > 4800 {
        return 1
    }
    f := [200][200]float64{}
    var dfs func(i, j int) float64
    dfs = func(i, j int) float64 {
        if i <= 0 && j <= 0 {
            return 0.5
        }
        if i <= 0 {
            return 1.0
        }
        if j <= 0 {
            return 0
        }
        if f[i][j] > 0 {
            return f[i][j]
        }
        ans := 0.25 * (dfs(i-4, j) + dfs(i-3, j-1) + dfs(i-2, j-2) + dfs(i-1, j-3))
        f[i][j] = ans
        return ans
    }
    return dfs((n+24)/25, (n+24)/25)
}

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function soupServings(n: number): number {
    const f = new Array(200).fill(0).map(() => new Array(200).fill(-1));
    const dfs = (i: number, j: number): number => {
        if (i <= 0 && j <= 0) {
            return 0.5;
        }
        if (i <= 0) {
            return 1;
        }
        if (j <= 0) {
            return 0;
        }
        if (f[i][j] !== -1) {
            return f[i][j];
        }
        f[i][j] =
            0.25 * (dfs(i - 4, j) + dfs(i - 3, j - 1) + dfs(i - 2, j - 2) + dfs(i - 1, j - 3));
        return f[i][j];
    };
    return n >= 4800 ? 1 : dfs(Math.ceil(n / 25), Math.ceil(n / 25));
}

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