837

837. New 21 Game (Medium)

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, W], where W is an integer.  Each draw is independent and the outcomes have equal probabilities.

Alice stops drawing numbers when she gets K or more points.  What is the probability that she has N or less points?

Example 1:

Input: N = 10, K = 1, W = 10
Output: 1.00000
Explanation:  Alice gets a single card, then stops.

Example 2:

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

Example 3:

Input: N = 21, K = 17, W = 10
Output: 0.73278

Note:

1. 0 <= K <= N <= 10000
2. 1 <= W <= 10000
3. Answers will be accepted as correct if they are within 10^-5 of the correct answer.
4. The judging time limit has been reduced for this question.

Related Topics:
Dynamic Programming

Solution 1. Brute Force (TLE)

Let dp[i] be the possibility to land on node i.

For each node i in [0, K), try jump j steps to node i + j where j is in [1, W]. When landing on node i + j from node i, p[i + j] should increase by dp[i] / W.

// OJ: https://leetcode.com/problems/new-21-game/
// Author: github.com/lzl124631x
// Time: O(KW)
// Space: O(1)
// NOTE: This solution will get TLE
class Solution {
public:
    double new21Game(int N, int K, int W) {
        if (N >= K + W - 1) return 1;
        vector<double> dp(K + W);
        dp[0] = 1;
        for (int i = 0; i < K; ++i) {
            for (int j = 1; j <= W && i + j <= N; ++j) dp[i + j] += dp[i] / W;
        }
        return accumulate(begin(dp) + K, begin(dp) + N + 1, 0.0);
    }
};

Solution 2. DP

Let dp[i] be the probability of getting i points.

Assume K = 5, W = 3.

dp[1] = 1/W
dp[2] = 1/W + dp[1]/W
dp[3] = 1/W + dp[1]/W + dp[2]/W

dp[4] = dp[1]/W + dp[2]/W + dp[3]/W
dp[5] = dp[2]/W + dp[3]/W + dp[4]/W

dp[6] = dp[3]/W + dp[4]/W
dp[7] = dp[4]/W

dp[0] = 1
dp[1] =         dp[0]/W
dp[2] = dp[1] + dp[1]/W
dp[3] = dp[2] + dp[2]/W

dp[4] = dp[3] + dp[3]/W - dp[0]/W
dp[5] = dp[4] + dp[4]/W - dp[1]/W

dp[6] = dp[5]           - dp[2]/W
dp[7] = dp[6]           - dp[3]/W

So we have the formula:

dp[0] = 1
dp[i] = (i > 1 ? dp[i-1] : 0)
         + (i <= K ? dp[i-1]/W : 0)
         - (i-W-1 >= 0 ? dp[i-W-1]/W : 0)

The answer is sum( dp[i] | K <= i <= N ).

// OJ: https://leetcode.com/problems/new-21-game/
// Author: github.com/lzl124631x
// Time: O(min(N, K + W))
// Space: O(min(N, K + W))
class Solution {
public:
    double new21Game(int N, int K, int W) {
        if (!K || N >= K + W - 1) return 1;
        vector<double> dp(N + 1);
        dp[0] = 1;
        double ans = 0;
        for (int i = 1; i <= N; ++i) {
            if (i > 1) dp[i] += dp[i - 1];
            if (i <= K) dp[i] += dp[i - 1] / W;
            if (i > W) dp[i] -= dp[i - W - 1]/W;
            if (i >= K) ans += dp[i];
        }
        return ans;
    }
};

Solution 3. DP

// OJ: https://leetcode.com/problems/new-21-game/
// Author: github.com/lzl124631x
// Time: O(K + W)
// Space: O(K + W)
class Solution {
public:
    double new21Game(int N, int K, int W) {
        if (!K || N >= K + W - 1) return 1;
        vector<double> dp(K + W);
        for (int i = K; i < K + W && i <= N; ++i) dp[i] = 1;
        double sum = min(N - K + 1, W);
        for (int i = K - 1; i >= 0; --i) {
            dp[i] = sum / W;
            sum += dp[i] - dp[i + W];
        }
        return dp[0];
    }
}