onenoughtone

Solution Approach

Solution Overview

There are 3 main approaches to solve this problem:

Dynamic Programming with Memoization - Complex approach
Iterative Dynamic Programming - Complex approach
Mathematical Insight - Complex approach

Approach 1: Dynamic Programming with Memoization

The key insight for solving this problem is to use dynamic programming with memoization to calculate the probability for different amounts of soup A and soup B.

Let's define a function P(a, b) as the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time, given that we have a ml of soup A and b ml of soup B.

The base cases are:

If a ≤ 0 and b ≤ 0, then P(a, b) = 0.5 (both soups become empty at the same time).
If a ≤ 0 and b > 0, then P(a, b) = 1 (soup A becomes empty first).
If a > 0 and b ≤ 0, then P(a, b) = 0 (soup B becomes empty first).

For the recursive case, we consider all four operations with equal probability:

P(a, b) = 0.25 * (P(a-100, b) + P(a-75, b-25) + P(a-50, b-50) + P(a-25, b-75))

To avoid redundant calculations, we can use memoization to store the results of subproblems that we've already solved.

Additionally, we can optimize the solution by observing that for large values of n, the probability approaches 1. This is because soup A is consumed at a faster rate than soup B in the given operations. Specifically, for n > 4800, the probability is very close to 1 (within the required precision of 10^-5), so we can return 1 directly.

Another optimization is to scale down the problem by dividing n by 25 and working with smaller units. This simplifies the calculations and reduces the number of recursive calls.

Algorithm:

If n {'>'} 4800, return 1 (the probability approaches 1 for large n).
Scale down the problem by dividing n by 25 and rounding up to the nearest integer.
Define a recursive function P(a, b) that calculates the probability for given amounts of soup A and soup B.
Base cases:
a. If a ≤ 0 and b ≤ 0, return 0.5 (both soups become empty at the same time).
b. If a ≤ 0 and b > 0, return 1 (soup A becomes empty first).
c. If a > 0 and b ≤ 0, return 0 (soup B becomes empty first).
Recursive case:
a. P(a, b) = 0.25 * (P(a-4, b) + P(a-3, b-1) + P(a-2, b-2) + P(a-1, b-3)).
Use memoization to avoid redundant calculations.
Return P(n, n), which represents the probability for the initial state.

Time Complexity:

O(n^2)

In the worst case, we need to calculate P(a, b) for all pairs (a, b) where 0 ≤ a, b ≤ n. However, due to the optimization for large n, the actual time complexity is much better for large inputs.

Space Complexity:

O(n^2)

We use a memoization table to store the results of subproblems, which requires O(n^2) space.

Approach 2: Iterative Dynamic Programming

We can also solve this problem using an iterative dynamic programming approach, which avoids the overhead of recursion.

Let's define dp[a][b] as the probability that soup A will be empty first, plus half the probability that A and B become empty at the same time, given that we have a ml of soup A and b ml of soup B.

The base cases are the same as in the recursive approach:

If a ≤ 0 and b ≤ 0, then dp[a][b] = 0.5.
If a ≤ 0 and b > 0, then dp[a][b] = 1.
If a > 0 and b ≤ 0, then dp[a][b] = 0.

We can fill the DP table iteratively, starting from smaller values of a and b and moving to larger values:

dp[a][b] = 0.25 * (dp[a-4][b] + dp[a-3][b-1] + dp[a-2][b-2] + dp[a-1][b-3])

The final answer is dp[n][n].

This approach has the same time and space complexity as the recursive approach with memoization, but it avoids the overhead of recursion and may be more efficient in practice.

Algorithm:

If n {'>'} 4800, return 1 (the probability approaches 1 for large n).
Scale down the problem by dividing n by 25 and rounding up to the nearest integer.
Initialize a 2D DP array dp of size (n+1) × (n+1).
Set the base cases:
a. dp[0][0] = 0.5 (both soups become empty at the same time).
b. For a = 0 and b > 0, dp[a][b] = 1 (soup A becomes empty first).
c. For a > 0 and b = 0, dp[a][b] = 0 (soup B becomes empty first).
Fill the DP array iteratively:
a. For a from 1 to n:
i. For b from 1 to n:
1. dp[a][b] = 0.25 * (dp[max(0, a-4)][b] + dp[max(0, a-3)][max(0, b-1)] + dp[max(0, a-2)][max(0, b-2)] + dp[max(0, a-1)][max(0, b-3)]).
Return dp[n][n], which represents the probability for the initial state.

Time Complexity:

O(n^2)

We need to fill in the DP table of size (n+1) × (n+1), which requires O(n^2) time.

Space Complexity:

O(n^2)

We use a 2D DP array of size (n+1) × (n+1) to store the results, which requires O(n^2) space.

Approach 3: Mathematical Insight

For this problem, there's an interesting mathematical insight: as n increases, the probability that soup A will be empty first approaches 1.

This is because, on average, soup A is consumed at a faster rate than soup B in the given operations:

Operation 1: 100 ml of A, 0 ml of B (ratio: 100:0)
Operation 2: 75 ml of A, 25 ml of B (ratio: 75:25)
Operation 3: 50 ml of A, 50 ml of B (ratio: 50:50)
Operation 4: 25 ml of A, 75 ml of B (ratio: 25:75)

The average consumption per operation is:

Soup A: (100 + 75 + 50 + 25) / 4 = 250 / 4 = 62.5 ml
Soup B: (0 + 25 + 50 + 75) / 4 = 150 / 4 = 37.5 ml

Since soup A is consumed at a faster rate (62.5 ml vs 37.5 ml per operation on average), for large values of n, soup A is more likely to be empty first.

Through empirical testing, it has been found that for n > 4800, the probability is very close to 1 (within the required precision of 10^-5), so we can return 1 directly for such cases.

This insight allows us to optimize our solution significantly for large inputs.

Algorithm:

If n {'>'} 4800, return 1 (the probability approaches 1 for large n).
Otherwise, use either the recursive or iterative dynamic programming approach to calculate the probability.

Time Complexity:

O(1) for n {'>'} 4800, O(n^2) otherwise

For large values of n, we return 1 directly, which takes constant time. For smaller values, we use dynamic programming, which takes O(n^2) time.

Space Complexity:

O(1) for n {'>'} 4800, O(n^2) otherwise

For large values of n, we don't need any additional space. For smaller values, we use a memoization table or DP array, which requires O(n^2) space.

Pseudocode

solution.pseudo

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function soupServings(n):
    if n {'>'} 4800:
        return 1.0
 
    n = ceil(n / 25)  // Scale down the problem
    memo = new map or 2D array
 
    function P(a, b):
        if a <= 0 and b <= 0:
            return 0.5
        if a <= 0:
            return 1.0
        if b <= 0:
            return 0.0
 
        if (a, b) in memo:
            return memo[(a, b)]
 
        result = 0.25 * (P(a-4, b) + P(a-3, b-1) + P(a-2, b-2) + P(a-1, b-3))
        memo[(a, b)] = result
        return result
 
    return P(n, n)

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