Today I have done leetcode's August LeetCoding Challenge with cpp.

August LeetCoding Challenge 8

Description

Soup Servings

You have two soups, A and B, each starting with n mL. On every turn, one of the following four serving operations is chosen at random, each with probability 0.25 independent of all previous turns:

• pour 100 mL from type A and 0 mL from type B
• pour 75 mL from type A and 25 mL from type B
• pour 50 mL from type A and 50 mL from type B
• pour 25 mL from type A and 75 mL from type B

Note:

• There is no operation that pours 0 mL from A and 100 mL from B.
• The amounts from A and B are poured simultaneously during the turn.
• If an operation asks you to pour more than you have left of a soup, pour all that remains of that soup.

The process stops immediately after any turn in which one of the soups is used up.

Return the probability that A is used up before B, plus half the probability that both soups are used up in the same turn. Answers within 10-5 of the actual answer will be accepted.

Example 1:

Input: n = 50
Output: 0.62500
Explanation: 
If we perform either of the first two serving operations, soup A will become empty first.
If we perform the third operation, A and B will become empty at the same time.
If we perform 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
Explanation: 
If we perform the first serving operation, soup A will become empty first.
If we perform the second serving operations, A will become empty on performing operation [1, 2, 3], and both A and B become empty on performing operation 4.
If we perform the third operation, A will become empty on performing operation [1, 2], and both A and B become empty on performing operation 3.
If we perform the fourth operation, A will become empty on performing operation 1, and both A and B become empty on performing operation 2.
So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.71875.

Constraints:

• 0 <= n <= 109

Solution

vector<vector<double>> dp(193, vector<double>(193, -1));
class Solution {
  double solve(int a, int b) {
    if(a <= 0 && b <= 0) {
      return 0.5;
    } else if(a <= 0) {
      return 1.0;
    } else if(b <= 0) {
      return 0;
    } else if(dp[a][b] >= 0) {
      return dp[a][b];
    }
    dp[a][b] = 0.25 * (solve(a - 4, b) + solve(a - 3, b - 1) + solve(a - 2, b - 2) + solve(a - 1, b - 3));
    return dp[a][b];
  }
public:
  double soupServings(int n) {
    if(n > 4800) return 1;
    n += 24;
    n /= 25;
    return solve(n, n);
  }
};

// Accepted
// 43/43 cases passed (1 ms)
// Your runtime beats 84.83 % of cpp submissions
// Your memory usage beats 92.17 % of cpp submissions (8.5 MB)