Algorithm
Problem Name: 808. Soup Servings
There are two types of soup: type A and type B. Initially, we have n ml of each type of soup. There are four kinds of operations:
- Serve
100ml of soup A and0ml of soup B, - Serve
75ml of soup A and25ml of soup B, - Serve
50ml of soup A and50ml of soup B, and - Serve
25ml of soup A and75ml of soup B.
When we serve some soup, we give it to someone, and we no longer have it. Each turn, we will choose from the four operations with an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup.
Note that we do not have an operation where all 100 ml's of soup B are used first.
Return the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time. Answers within 10-5 of the actual answer will be accepted.
Example 1:
Input: n = 50 Output: 0.62500 Explanation: If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For 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
Constraints:
0 <= n <= 109
Code Examples
#1 Code Example with Python Programming
Code -
Python Programming
class Solution:
def soupServings(self, N):
visited = {}
def dfs(a, b):
if (a, b) in visited: return visited[(a, b)]
elif a <= 0 or b <= 0: return (a < b and 1) or (a == b and 0.5) or (b < a and 0)
visited[(a, b)] = 0.25 * (dfs(a - 100, b) + dfs(a - 75, b - 25) + dfs(a - 50, b -50) + dfs(a - 25, b - 75))
return visited[(a, b)]
return N > 4800 and 1 or round(dfs(N, N), 5)
Input
Output