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:
0 <= K <= N <= 100001 <= W <= 10000- Answers will be accepted as correct if they are within
10^-5of the correct answer. - The judging time limit has been reduced for this question.
Related Topics:
Dynamic Programming
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);
}
};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;
}
};// 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];
}
}