algoadvance

Leetcode 808. Soup Servings

Problem Statement

There are two types of soup: type A and type B. Initially, we have N ml of soup A and N ml of soup B. We serve soup according to the following probability operations:

We stop serving the soup when either type of soup becomes empty or both become empty. The operation of serving soup obeys this hierarchy:

  1. If both A and B become empty simultaneously, we get result 0.5.
  2. If only soup A becomes empty while soup B has some remaining, we get result 1.
  3. If only soup B becomes empty while soup A has some remaining, we get result 0.

Return the probability that soup A will be empty first, plus half the probability that A and B will become empty at the same time.

Clarifying Questions

Before we proceed, let’s clarify a few details:

  1. What is the maximum value for N?
  2. Should we consider N to be large, possibly requiring optimization techniques?

Assuming the above information is standard:

Strategy

  1. Base Cases:
    • If N ≤ 0, return 0.5, since N = 0 implies both soups are empty.
  2. Dynamic Programming with Memoization:
    • Use a map to store intermediate results.
    • Key: the tuple (A, B) representing current amounts of soups A and B.
    • Value: the probability of serving soup A first or both soups together from state (A, B).
  3. Recursive Function:
    • Define a function probability(A, B) which returns the desired probability for given amounts A and B.
    • Use the recursive relation considering the four service operations and their associated probabilities.
    • Memoize results to avoid redundant calculations.

Time Complexity

Given the memoization approach, the complexity is O(N^2) in terms of states (A, B).

Code

Let’s implement this solution in C++:

#include <map>
#include <utility>

using namespace std;

class Solution {
public:
    map<pair<int, int>, double> memo;

    double probability(int A, int B) {
        if (A <= 0 && B <= 0) return 0.5;
        if (A <= 0) return 1.0;
        if (B <= 0) return 0.0;
        
        if (memo.find({A, B}) != memo.end()) {
            return memo[{A, B}];
        }
        
        double result = 0.25 * (probability(A - 100, B) +
                                probability(A - 75, B - 25) +
                                probability(A - 50, B - 50) +
                                probability(A - 25, B - 75));
        
        memo[{A, B}] = result;
        return result;
    }

    double soupServings(int N) {
        if (N > 4800) return 1.0; // Fast return for large N based on empirical observations
        return probability(N, N);
    }
};

Explanation of the Code

  1. Base Cases: Handle the cases when either or both soups reach zero.
  2. Memoization: Use a map memo to store and reuse results.
  3. Recursive Function: The function probability calculates the probability based on recursive states.
  4. Optimization for Large N: If N is larger than a threshold (4800), return 1.0 directly as the probability asymptotically approaches 1.0.

Cut your prep time in half and DOMINATE your interview with AlgoAdvance AI