837. New 21 Game

Description

Alice plays the following game, loosely based on the card game "21".

Alice starts with 0 points and draws numbers while she has less than k points. During each draw, she gains an integer number of points randomly from the range [1, maxPts], where maxPts is an integer. Each draw is independent and the outcomes have equal probabilities.

Alice stops drawing numbers when she gets k or more points.

Return the probability that Alice has n or fewer points.

Answers within 10^-5 of the actual answer are considered accepted.

Example 1:

Input: n = 10, k = 1, maxPts = 10
Output: 1.00000
Explanation: Alice gets a single card, then stops.

Example 2:

Input: n = 6, k = 1, maxPts = 10
Output: 0.60000
Explanation: Alice gets a single card, then stops.
In 6 out of 10 possibilities, she is at or below 6 points.

Example 3:

Input: n = 21, k = 17, maxPts = 10
Output: 0.73278

Constraints:

0 <= k <= n <= 10⁴
1 <= maxPts <= 10⁴

Solutions

Solution 1: Memoized Search

We design a function $dfs(i)$, which represents the probability that when the current score is $i$, the final score does not exceed $n$ when we stop drawing numbers. The answer is $dfs(0)$.

The calculation method of function $dfs(i)$ is as follows:

If $i \ge k$, then we stop drawing numbers. If $i \le n$, return $1$, otherwise return $0$;
Otherwise, we can draw the next number $j$ in the range $[1,..\textit{maxPts}]$, then $dfs(i) = \frac{1}{maxPts} \sum_{j=1}^{maxPts} dfs(i+j)$.

Here we can use memoized search to accelerate the calculation.

The time complexity of the above method is $O(k \times \textit{maxPts})$, which will exceed the time limit, so we need to optimize it.

When $i \lt k$, the following equation holds:

\[\begin{aligned} dfs(i) &= (dfs(i + 1) + dfs(i + 2) + \cdots + dfs(i + \textit{maxPts})) / \textit{maxPts} & (1) \end{aligned}\]

When $i \lt k - 1$, the following equation holds:

\[\begin{aligned} dfs(i+1) &= (dfs(i + 2) + dfs(i + 3) + \cdots + dfs(i + \textit{maxPts} + 1)) / \textit{maxPts} & (2) \end{aligned}\]

Therefore, when $i \lt k-1$, we subtract equation $(2)$ from equation $(1)$ to get:

\[\begin{aligned} dfs(i) - dfs(i+1) &= (dfs(i + 1) - dfs(i + \textit{maxPts} + 1)) / \textit{maxPts} \end{aligned}\]

That is:

\[\begin{aligned} dfs(i) &= dfs(i + 1) + (dfs(i + 1) - dfs(i + \textit{maxPts} + 1)) / \textit{maxPts} \end{aligned}\]

If $i=k-1$, we have:

\[\begin{aligned} dfs(i) &= dfs(k - 1) = (dfs(k) + dfs(k + 1) + \cdots + dfs(k + \textit{maxPts} - 1)) / \textit{maxPts} & (3) \end{aligned}\]

We assume there are $i$ numbers not exceeding $n$, then $k+i-1 \leq n$, and since $i\leq \textit{maxPts}$, we have $i \leq \min(n-k+1, \textit{maxPts})$, so equation $(3)$ can be written as:

\[\begin{aligned} dfs(k-1) &= \min(n-k+1, \textit{maxPts}) / \textit{maxPts} \end{aligned}\]

In summary, we have the following state transition equation:

\[\begin{aligned} dfs(i) &= \begin{cases} 1, & i \geq k, i \leq n \\ 0, & i \geq k, i \gt n \\ \min(n-k+1, \textit{maxPts}) / \textit{maxPts}, & i = k - 1 \\ dfs(i + 1) + (dfs(i + 1) - dfs(i + \textit{maxPts} + 1)) / \textit{maxPts}, & i < k - 1 \end{cases} \end{aligned}\]

Time complexity $O(k + \textit{maxPts})$, space complexity $O(k + \textit{maxPts})$. Where $k$ is the maximum score.

Solution 2: Dynamic Programming

We can convert the memoized search in Solution 1 into dynamic programming.

Define $f[i]$ to represent the probability that when the current score is $i$, the final score does not exceed $n$ when we stop drawing numbers. The answer is $f[0]$.

When $k \leq i \leq \min(n, k + \textit{maxPts} - 1)$, we have $f[i] = 1$.

When $i = k - 1$, we have $f[i] = \min(n-k+1, \textit{maxPts}) / \textit{maxPts}$.

When $i \lt k - 1$, we have $f[i] = f[i + 1] + (f[i + 1] - f[i + \textit{maxPts} + 1]) / \textit{maxPts}$.

Time complexity $O(k + \textit{maxPts})$, space complexity $O(k + \textit{maxPts})$. Where $k$ is the maximum score.

class Solution {
    private double[] f;
    private int n, k, maxPts;

    public double new21Game(int n, int k, int maxPts) {
        f = new double[k];
        this.n = n;
        this.k = k;
        this.maxPts = maxPts;
        return dfs(0);
    }

    private double dfs(int i) {
        if (i >= k) {
            return i <= n ? 1 : 0;
        }
        if (i == k - 1) {
            return Math.min(n - k + 1, maxPts) * 1.0 / maxPts;
        }
        if (f[i] != 0) {
            return f[i];
        }
        return f[i] = dfs(i + 1) + (dfs(i + 1) - dfs(i + maxPts + 1)) / maxPts;
    }
}


// Solution 2
class Solution {
    public double new21Game(int n, int k, int maxPts) {
        if (k == 0) {
            return 1.0;
        }
        double[] f = new double[k + maxPts];
        for (int i = k; i < Math.min(n + 1, k + maxPts); ++i) {
            f[i] = 1;
        }
        f[k - 1] = Math.min(n - k + 1, maxPts) * 1.0 / maxPts;
        for (int i = k - 2; i >= 0; --i) {
            f[i] = f[i + 1] + (f[i + 1] - f[i + maxPts + 1]) / maxPts;
        }
        return f[0];
    }
}

class Solution {
public:
    double new21Game(int n, int k, int maxPts) {
        vector<double> f(k);
        function<double(int)> dfs = [&](int i) -> double {
            if (i >= k) {
                return i <= n ? 1 : 0;
            }
            if (i == k - 1) {
                return min(n - k + 1, maxPts) * 1.0 / maxPts;
            }
            if (f[i]) {
                return f[i];
            }
            return f[i] = dfs(i + 1) + (dfs(i + 1) - dfs(i + maxPts + 1)) / maxPts;
        };
        return dfs(0);
    }
};


// Solution 2
class Solution {
public:
    double new21Game(int n, int k, int maxPts) {
        if (k == 0) {
            return 1.0;
        }
        double f[k + maxPts];
        memset(f, 0, sizeof(f));
        for (int i = k; i < min(n + 1, k + maxPts); ++i) {
            f[i] = 1;
        }
        f[k - 1] = min(n - k + 1, maxPts) * 1.0 / maxPts;
        for (int i = k - 2; i >= 0; --i) {
            f[i] = f[i + 1] + (f[i + 1] - f[i + maxPts + 1]) / maxPts;
        }
        return f[0];
    }
};

class Solution:
    def new21Game(self, n: int, k: int, maxPts: int) -> float:
        @cache
        def dfs(i: int) -> float:
            if i >= k:
                return int(i <= n)
            if i == k - 1:
                return min(n - k + 1, maxPts) / maxPts
            return dfs(i + 1) + (dfs(i + 1) - dfs(i + maxPts + 1)) / maxPts

        return dfs(0)


# Solution 2
class Solution:
    def new21Game(self, n: int, k: int, maxPts: int) -> float:
        f = [0] * (k + maxPts)
        for i in range(k, min(n + 1, k + maxPts)):
            f[i] = 1
        f[k - 1] = min(n - k + 1, maxPts) / maxPts
        for i in range(k - 2, -1, -1):
            f[i] = f[i + 1] + (f[i + 1] - f[i + maxPts + 1]) / maxPts
        return f[0]

func new21Game(n int, k int, maxPts int) float64 {
	f := make([]float64, k)
	var dfs func(int) float64
	dfs = func(i int) float64 {
		if i >= k {
			if i <= n {
				return 1
			}
			return 0
		}
		if i == k-1 {
			return float64(min(n-k+1, maxPts)) / float64(maxPts)
		}
		if f[i] > 0 {
			return f[i]
		}
		f[i] = dfs(i+1) + (dfs(i+1)-dfs(i+maxPts+1))/float64(maxPts)
		return f[i]
	}
	return dfs(0)
}


// Solution 2
func new21Game(n int, k int, maxPts int) float64 {
	if k == 0 {
		return 1
	}
	f := make([]float64, k+maxPts)
	for i := k; i < min(n+1, k+maxPts); i++ {
		f[i] = 1
	}
	f[k-1] = float64(min(n-k+1, maxPts)) / float64(maxPts)
	for i := k - 2; i >= 0; i-- {
		f[i] = f[i+1] + (f[i+1]-f[i+maxPts+1])/float64(maxPts)
	}
	return f[0]
}

function new21Game(n: number, k: number, maxPts: number): number {
    const f = new Array(k).fill(0);
    const dfs = (i: number): number => {
        if (i >= k) {
            return i <= n ? 1 : 0;
        }
        if (i === k - 1) {
            return Math.min(n - k + 1, maxPts) / maxPts;
        }
        if (f[i] !== 0) {
            return f[i];
        }
        return (f[i] = dfs(i + 1) + (dfs(i + 1) - dfs(i + maxPts + 1)) / maxPts);
    };
    return dfs(0);
}


// Solution 2
function new21Game(n: number, k: number, maxPts: number): number {
    if (k === 0) {
        return 1;
    }
    const f: number[] = Array(k + maxPts).fill(0);
    for (let i = k; i < Math.min(n + 1, k + maxPts); ++i) {
        f[i] = 1;
    }
    f[k - 1] = Math.min(n - k + 1, maxPts) / maxPts;
    for (let i = k - 2; i >= 0; --i) {
        f[i] = f[i + 1] + (f[i + 1] - f[i + maxPts + 1]) / maxPts;
    }
    return f[0];
}

837 - New 21 Game

837. New 21 Game

Description

Solutions

Solution 1: Memoized Search

Solution 2: Dynamic Programming

All Problems

All Solutions

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837. New 21 Game

Description

Solutions

Solution 1: Memoized Search

Solution 2: Dynamic Programming

All Problems

All Solutions