## Similar Problems

Similar Problems not available

# Soup Servings

## Companies:

LeetCode:  Soup Servings Leetcode Solution

Difficulty: Unknown

Topics:

The Soup Servings problem on LeetCode is a probability problem that requires us to calculate the probability of A getting his desired volume of soup before B. Here's a detailed solution.

Problem Description:

There are two types of soup, A and B. Initially, we have N ml of each type of soup. There are four kinds of actions you can perform:

• Serve 100 ml of soup A and 0 ml of soup B
• Serve 75 ml of soup A and 25 ml of soup B
• Serve 50 ml of soup A and 50 ml of soup B
• Serve 25 ml of soup A and 75 ml of soup B

These actions can be performed in any order, any number of times. Your task is to determine the probability that soup A will be finished first, assuming that each serving is independent of the others.

Solution:

To solve this problem, we can use a dynamic programming approach. We can create a table dp, where dp[i][j] represents the probability that A will be empty when there are i ml of soup A left and j ml of soup B left.

We start by initializing dp[N][N] = 1, since the probability that A will be empty when there are N ml of soup A and N ml of soup B left is 1. We also initialize dp[0][0] = 0.5, since the probability that A and B will be empty at the same time is 0.5.

For each remaining i and j, we can fill out the dp table in the following way:

• dp[i][j] = 0.25(dp[i-4][j] + dp[i-3][j-1] + dp[i-2][j-2] + dp[i-1][j-3])

This equation represents the probabilities of all possible serving actions. For example, the first term dp[i-4][j] represents the probability of serving 100 ml of soup A and 0 ml of soup B. Similarly, the second term dp[i-3][j-1] represents the probability of serving 75 ml of soup A and 25 ml of soup B.

We can stop filling out the dp table when dp[0][i] or dp[1][i] or dp[2][i] or dp[3][i] is greater than 0.999, since the probability that A will be empty before B is at least 0.999.

Finally, we can return dp[a][b], where a = ceil(N/100) and b = ceil(N/100), since the sizes of the serving actions are in multiples of 25 ml.

Time Complexity:

The time complexity of this solution is O(N^2), since we need to fill out the entire dp table.

Space Complexity:

The space complexity of this solution is also O(N^2), since we need to store the entire dp table.

Here's the full code for the solution in Python:

class Solution: def soupServings(self, N: int) -> float: if N > 5000: return 1.0

``````    def ceildiv(a, b):
return -(-a // b)

N = ceildiv(N, 25)

dp = [[0.0] * (N+1) for _ in range(N+1)]
dp[N][N] = 1.0
dp[0][0] = 0.5

for i in range(1, N+1):
dp[0][i] = 1.0
dp[i][0] = 0.0

for i in range(1, N+1):
for j in range(1, N+1):
dp[i][j] = 0.25 * (dp[max(i-4,0)][j] +
dp[max(i-3,0)][max(j-1,0)] +
dp[max(i-2,0)][max(j-2,0)] +
dp[max(i-1,0)][max(j-3,0)])

return dp[N][N]
``````

## Solution Implementation

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