# Solution For Soup Servings

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]
``````

## Step by Step Implementation For 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 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
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 equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can.  We stop once we no longer have some quantity of both types of soup.

Note that we do not have the 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.

class Solution {
public double soupServings(int N) {

}
}```
```def soupServings(N):

# if N is 0 or N is greater than or equal to 4800, then we know the answer is 1.0
if N >= 4800:
return 1.0

# if N is less than 0, then we know the answer is 0.0
elif N <= 0:
return 0.0

# we will use a dictionary to keep track of the answers to subproblems
# the key will be the amount of soup remaining, and the value will be the probability of getting a non-zero amount of soup
dp = {}

# we know that the probability of getting a non-zero amount of soup when there is no soup remaining is 0.0
dp[0] = 0.0

# we will fill in the dictionary using dynamic programming
# we will start by considering the subproblems where there is only 1 or 2 soup remaining
# we can calculate the probability of these subproblems by hand

# if there is only 1 soup remaining, the probability of getting a non-zero amount of soup is 1/2
dp[1] = 0.5

# if there is only 2 soup remaining, the probability of getting a non-zero amount of soup is 3/4
dp[2] = 0.75

# we can fill in the rest of the dictionary by considering all possible moves from each subproblem
# there are two possible moves: we can take 1/4 cup of soup, or we can take 1/2 cup of soup

# if we take 1/4 cup of soup, then we are left with 3/4 * N soup
# if we take 1/2 cup of soup, then we are left with 1/2 * N soup

# we can calculate the probability of getting a non-zero amount of soup from each of these subproblems
# and then take the maximum of these probabilities, since we are trying to maximize the probability of getting a non-zero amount of soup

# we can fill in the dictionary using this approach
for i in range(3, N+1):
dp[i] = max(0.25 * dp[3*i//4], 0.5 * dp[i//2])

# the answer to the original problem is the probability of getting a non-zero amount of soup when there is N soup remaining
return dp[N]```
```/**
* @param {number} N
* @return {number}
*/
var soupServings = function(N) {
// if N is greater than or equal to 4800, then return 1 as per the problem statement
if (N >= 4800) {
return 1.0;
}
// create an array to store the results of already calculated values
// this is to avoid recalculating the same values again and again
let dp = new Array(N + 1).fill(0);
// fill the first column of the array with 1.0 as per the problem statement
// this is the base case when N = 0
for (let i = 0; i <= N; i++) {
dp[i][0] = 1.0;
}
// fill the first row of the array with 0.5 as per the problem statement
// this is the base case when N = 1
for (let j = 0; j <= N; j++) {
dp[0][j] = 0.5;
}
// start from the second row and column and fill the array according to the recursive formula
for (let i = 1; i <= N; i++) {
for (let j = 1; j <= N; j++) {
dp[i][j] = 0.25 * (dp[Math.max(0, i - 4)][j] + dp[Math.max(0, i - 3)][j - 1] + dp[Math.max(0, i - 2)][j - 2] + dp[Math.max(0, i - 1)][j - 3]);
}
}
// return the result stored in the last cell of the array
return dp[N][N];
};```
```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 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
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 equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as we can.  We stop once we no longer have some quantity of both types of soup.

Note that we do not have the 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.

double soupServings(int N) {
if (N >= 4800) return 1.0;
int dp[N + 1][N + 1] = {};
for (int s = 0; s <= 2 * N; s++) {
for (int i = 0; i <= N; i++) {
int j = s - i;
if (j < 0 || j > N) continue;
double& ans = dp[i][j];
if (i == 0) ans = 1.0;
else if (j == 0) ans = 0.0;
else if (i == j) ans = 0.5;
else ans = 0.25 * (dp[max(0, i - 100)][j] + dp[max(0, i - 75)][max(0, j - 25)] + dp[max(0, i - 50)][max(0, j - 50)] + dp[max(0, i - 25)][max(0, j - 75)]);
}
}
return dp[N][N];
}```
```You and your friend decide to go for a meal at a restaurant and you agree to share the cost. The total cost of the meal is \$x and the fraction you have to pay is \$y, i.e. pay the cost \$xy.

Your friend asks that the bill be split in a way that minimises the largest amount you pay. That is, you should pay as little as possible for the first part of the bill, and as much as possible for the second part.

Complete the function that takes as input the cost of the meal x and the fraction you have to pay y. The function should return the minimum amount you have to pay.

public static double soupServings(double x, double y)
{