LeetCode 808 - Soup Servings

The problem gives us two soups, A and B, each starting with exactly n milliliters. At every turn, one of four serving operations is chosen uniformly at random. Each operation removes different amounts from A and B simultaneously.

leetcodemediummathdynamic-programmingprobability-and-statistics

LeetCode Problem 808

Difficulty: 🟡 Medium
Topics: Math, Dynamic Programming, Probability and Statistics

Solution

Problem Understanding

The problem gives us two soups, A and B, each starting with exactly n milliliters. At every turn, one of four serving operations is chosen uniformly at random. Each operation removes different amounts from A and B simultaneously.

The four possible operations are:

  • Remove 100 mL from A and 0 mL from B
  • Remove 75 mL from A and 25 mL from B
  • Remove 50 mL from A and 50 mL from B
  • Remove 25 mL from A and 75 mL from B

Each operation has probability 0.25.

The process continues until at least one soup becomes empty. The required return value is:

  • the probability that A becomes empty before B
  • plus half the probability that both become empty during the same operation

In other words:

  • If A empties first, contribute 1
  • If both empty together, contribute 0.5
  • If B empties first, contribute 0

The final answer is the expected probability value across all possible random serving sequences.

A very important detail is that if an operation requests more soup than remains, we simply pour whatever is left. For example, if A has only 20 mL remaining and we attempt to remove 100 mL, A becomes exactly 0.

The constraint is extremely large:

0 <= n <= 10^9

A direct simulation over all possible states measured in individual milliliters would be completely infeasible. The solution must exploit mathematical structure and probability behavior to reduce the state space dramatically.

Several edge cases are important:

  • n = 0, both soups are already empty, so the answer should be 0.5
  • Very large n, where the probability converges extremely close to 1
  • States where both soups become empty simultaneously
  • Operations that attempt to remove more soup than remains

The problem guarantees that the answer only needs precision within 10^-5, which becomes crucial for the optimization.

Approaches

Brute Force Approach

A naive recursive solution would model every possible serving sequence explicitly.

From a state (a, b), we recursively try all four serving operations and average their probabilities. The recurrence naturally follows from the equal probability of each operation.

The recursive relation would be:

$$P(a,b)=\frac14\sum P(\text{next state})$$

with base cases:

  • If a <= 0 and b <= 0, return 0.5
  • If a <= 0, return 1
  • If b <= 0, return 0

This gives the correct answer because it exactly models the probabilistic process.

However, without optimization, the number of states grows exponentially. Even with memoization, using raw milliliter values still creates far too many states when n can be as large as 10^9.

Key Insight

The critical observation is that all serving amounts are multiples of 25.

The operations are:

  • (100, 0) → (4, 0)
  • (75, 25) → (3, 1)
  • (50, 50) → (2, 2)
  • (25, 75) → (1, 3)

after dividing by 25.

This means we can scale the entire problem down by converting the soups into units of 25 mL.

Instead of tracking exact milliliters, we track:

$$m = \left\lceil \frac{n}{25} \right\rceil$$

This reduces the state space dramatically.

An even more important insight is that as n becomes large, the probability approaches 1. This happens because every operation removes at least as much soup from A as from B, and on average A is depleted faster.

Empirically and mathematically, once n > 4800, the answer differs from 1 by less than 10^-5. Since the problem accepts that tolerance, we can directly return 1.0 for sufficiently large n.

This transforms the problem into a manageable memoized dynamic programming solution.

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Explores all serving sequences recursively
Optimal O(m²) O(m²) Memoized DP on scaled states

Here m = ceil(n / 25) and is effectively bounded by about 200 because of the large-n optimization.

Algorithm Walkthrough

  1. Handle the large n optimization first.

When n becomes sufficiently large, the answer converges extremely close to 1. Empirical analysis shows that for n > 4800, the error is already below 10^-5.

Therefore:

if n > 4800:
    return 1.0

This avoids unnecessary recursion and keeps the algorithm efficient. 2. Convert the problem into units of 25 mL.

Since every serving size is divisible by 25, we scale the state space:

$$m = \left\lceil \frac{n}{25} \right\rceil$$

In code:

m = (n + 24) // 25

This reduces the number of unique states dramatically. 3. Define the recursive DP function.

Let:

dp(a, b)

represent the probability that A empties first plus half the probability of a tie, starting with a units of A and b units of B. 4. Define the base cases carefully.

These are the stopping conditions:

  • If both soups are empty simultaneously:
return 0.5
  • If only A is empty:
return 1.0
  • If only B is empty:
return 0.0
  1. Recursively evaluate the four operations.

Each operation occurs with probability 0.25.

The transitions become:

  • (a-4, b)
  • (a-3, b-1)
  • (a-2, b-2)
  • (a-1, b-3)

The recurrence is:

$$dp(a,b)=\frac14\left( dp(a-4,b)+ dp(a-3,b-1)+ dp(a-2,b-2)+ dp(a-1,b-3) \right)$$ 6. Use memoization.

Many states repeat during recursion. Without caching, the recursion tree would explode exponentially.

A memoization cache stores previously computed states so each state is computed only once. 7. Return the final answer.

The final state is:

dp(m, m)

Why it works

The algorithm works because the recursive recurrence exactly models the probabilistic process described in the problem. Every state transition corresponds to one possible serving operation, and each operation contributes equally with probability 0.25.

The base cases correctly encode the scoring rules:

  • 1 when A empties first
  • 0.5 when both empty together
  • 0 when B empties first

Memoization guarantees each state is evaluated once, and scaling by 25 preserves correctness because every serving operation changes soup quantities only in multiples of 25.

Python Solution

class Solution:
    def soupServings(self, n: int) -> float:
        if n > 4800:
            return 1.0

        units = (n + 24) // 25
        memo = {}

        def dp(a: int, b: int) -> float:
            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)]

            probability = 0.25 * (
                dp(a - 4, b) +
                dp(a - 3, b - 1) +
                dp(a - 2, b - 2) +
                dp(a - 1, b - 3)
            )

            memo[(a, b)] = probability
            return probability

        return dp(units, units)

The implementation begins with the large-input optimization. For sufficiently large n, the probability is effectively 1, so we return immediately.

Next, the soup quantities are converted into 25 mL units. This compression is essential because it drastically reduces the number of states.

The recursive dp function computes the probability for a given state (a, b).

The base cases directly encode the problem requirements:

  • both empty → 0.5
  • only A empty → 1.0
  • only B empty → 0.0

The memo dictionary caches previously computed states. This prevents redundant recursive computation and transforms exponential recursion into polynomial complexity.

Finally, the recurrence averages the four equally likely serving operations and stores the result in the cache.

Go Solution

func soupServings(n int) float64 {
    if n > 4800 {
        return 1.0
    }

    units := (n + 24) / 25

    memo := make(map[[2]int]float64)

    var dp func(int, int) float64

    dp = func(a int, b int) float64 {
        if a <= 0 && b <= 0 {
            return 0.5
        }

        if a <= 0 {
            return 1.0
        }

        if b <= 0 {
            return 0.0
        }

        key := [2]int{a, b}

        if value, exists := memo[key]; exists {
            return value
        }

        probability := 0.25 * (
            dp(a-4, b) +
                dp(a-3, b-1) +
                dp(a-2, b-2) +
                dp(a-1, b-3),
        )

        memo[key] = probability
        return probability
    }

    return dp(units, units)
}

The Go implementation follows the same logic as the Python solution.

A Go map with a fixed-size array key is used for memoization:

map[[2]int]float64

This is an efficient and idiomatic way to memoize two-dimensional DP states.

Go does not support nested named functions directly, so the recursive closure is declared with:

var dp func(int, int) float64

before assignment.

Integer overflow is not a concern because the scaled state size remains very small after the optimization cutoff.

Worked Examples

Example 1

Input:

n = 50

Scaled units:

$$m = \lceil 50/25 \rceil = 2$$

Initial state:

dp(2, 2)

Possible operations:

Operation Next State Result
(4,0) (-2,2) A empty first → 1
(3,1) (-1,1) A empty first → 1
(2,2) (0,0) Both empty → 0.5
(1,3) (1,-1) B empty first → 0

Final probability:

$$0.25 \times (1 + 1 + 0.5 + 0)$$

$$= 0.625$$

Example 2

Input:

n = 100

Scaled units:

$$m = 4$$

Initial state:

dp(4,4)

The recursive evaluation expands into smaller states.

Transition Next State
(4,0) (0,4)
(3,1) (1,3)
(2,2) (2,2)
(1,3) (3,1)

Evaluating recursively:

State Probability
dp(0,4) 1
dp(1,3) 0.875
dp(2,2) 0.625
dp(3,1) 0.375

Final result:

$$0.25 \times (1 + 0.875 + 0.625 + 0.375)$$

$$= 0.71875$$

Complexity Analysis

Measure Complexity Explanation
Time O(m²) Each scaled state (a,b) is computed once
Space O(m²) Memoization stores all reachable states

Here:

$$m = \left\lceil \frac{n}{25} \right\rceil$$

However, because we return immediately for n > 4800, m is effectively bounded by about 192. This makes the solution extremely fast in practice.

The recursion explores only reachable states, and memoization ensures no state is recomputed.

Test Cases

solution = Solution()

assert abs(solution.soupServings(50) - 0.625) < 1e-5   # provided example 1
assert abs(solution.soupServings(100) - 0.71875) < 1e-5  # provided example 2

assert abs(solution.soupServings(0) - 0.5) < 1e-5  # both soups already empty

assert abs(solution.soupServings(25) - 0.625) < 1e-5  # smallest nonzero unit

assert solution.soupServings(5000) == 1.0  # large n optimization

assert solution.soupServings(4801) == 1.0  # threshold boundary

assert solution.soupServings(200) > solution.soupServings(100)  # probability increases

assert 0.0 <= solution.soupServings(300) <= 1.0  # valid probability range

assert solution.soupServings(1000) > 0.99  # convergence toward 1

assert abs(solution.soupServings(1) - 0.625) < 1e-5  # rounding up to one unit
Test Why
n = 50 Verifies provided example
n = 100 Verifies recursive behavior
n = 0 Tests simultaneous empty base case
n = 25 Tests smallest positive scaled state
n = 5000 Tests optimization shortcut
n = 4801 Tests threshold handling
n = 200 Ensures monotonic probability growth
n = 300 Ensures valid probability bounds
n = 1000 Verifies convergence behavior
n = 1 Tests ceil-based scaling

Edge Cases

One important edge case is n = 0. In this situation, both soups are already empty before any serving occurs. According to the problem statement, simultaneous emptying contributes 0.5 to the final probability. A naive implementation might incorrectly return 1 or 0. The implementation handles this correctly through the base case:

if a <= 0 and b <= 0:
    return 0.5

Another critical edge case occurs when a serving operation removes more soup than remains. For example, if only 10 mL of A remains and we attempt to remove 100 mL, the soup simply becomes empty. Implementations that assume exact subtraction without clamping can produce invalid negative-state logic. This solution avoids that issue naturally by allowing negative values and checking base cases using <= 0.

Very large inputs are another major edge case. Since n can be as large as 10^9, any solution attempting to build DP tables proportional to n would fail immediately. The implementation handles this using two optimizations:

  • scaling by 25
  • early return for n > 4800

Without this optimization, even memoization could still become unnecessarily expensive.

A subtle edge case involves rounding during scaling. Since operations occur in increments of 25 mL, values like n = 1 still behave like one full 25 mL unit. Using floor division would incorrectly map such cases to zero. The implementation correctly uses ceiling division:

(n + 24) // 25

which preserves the correct probabilistic behavior.