New 21 Game in C++

Suppose Rima plays the following game, that is loosely based on the card game "21". So Rima starts with 0 points, and draws numbers while she has less than K points. Now, during each draw, she gains an integer number of points randomly from the range [1, W], where W is given, and that is an integer. Now each draw is independent and the outcomes have equal probabilities. Rima stops drawing numbers when she gets K or more points. We have to find the probability that she has N or less points?

So if N = 6, K is 1 and W is 10, then the answer will be 0.6, as Rima gets a single card, then stops, In 6 out of 10 probabilities, She is at or below N = 6 points.

To solve this, we will follow these steps −

• If k is 0, or N >= K + W, then return 1
• make an array dp of size N + 1, set dp[0] := 1
• set wsum := 1.0, ret := 0.0
• for i in range 1 to N
  • dp[i] := wsum / W
  • if i < K, then wsum := wsum + dp[i], otherwise ret := ret + dp[i]
  • if i – W >= 0, then wsum := wsum – dp[i - W]
• return ret

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
   double new21Game(int N, int K, int W) {
      if(K == 0 || N >= K + W) return 1.0;
      vector <double> dp (N + 1);
      dp[0] = 1;
      double Wsum = 1.0;
      double ret = 0.0;
      for(int i = 1; i <= N; i++){
         dp[i] = Wsum / W;
         if(i < K){
            Wsum += dp[i];
         }else ret += dp[i];
         if(i - W >= 0) Wsum -= dp[i - W];
      }
      return ret;
   }
};
main(){
   Solution ob;
   cout << (ob.new21Game(6, 1, 10));
}

Input

6
1
10

Output

0.6