Introduction

In this blog post, we will explore how to solve the LeetCode problem "808" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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, and 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 an equal probability 0.25. If the remaining volume of soup is not enough to complete the operation, we will serve as much as possible. We stop once we no longer have some quantity of both types of soup. Note that we do not have an 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. Answers within 10-5 of the actual answer will be accepted. Example 1: Input: n = 50 Output: 0.62500 Explanation: If we choose the first two operations, A will become empty first. For the third operation, A and B will become empty at the same time. For the fourth operation, B will become empty first. So the total probability of A becoming empty first plus half the probability that A and B become empty at the same time, is 0.25 * (1 + 1 + 0.5 + 0) = 0.625. Example 2: Input: n = 100 Output: 0.71875 Constraints: 0 <= n <= 109

Explanation

Here's the breakdown of the problem and the solution:

• Dynamic Programming: Use dynamic programming to calculate the probability of A becoming empty first (or simultaneously with B). The state represents the remaining amounts of soup A and soup B.
• Memoization: Store the calculated probabilities in a memoization table (cache) to avoid redundant calculations, which significantly optimizes the solution for overlapping subproblems.

• Base Cases & Recursion: Define appropriate base cases for when either A or B (or both) are empty. Recursively calculate probabilities by considering all four possible operations and their respective probabilities (0.25).

• Runtime & Storage Complexity: O(N^2), O(N^2)

Code

    def soupServings(n: int) -> float:
    """
    Calculates the probability that soup A will be empty first, plus half the probability
    that A and B become empty at the same time.
    """

    if n > 4800:
        return 1.0  # Optimization for large n (approaches 1)

    n = (n + 24) // 25  # Scale down to reduce memory usage
    dp = {}  # Memoization cache

    def solve(a, b):
        if (a, b) in dp:
            return dp[(a, b)]

        if a <= 0 and b <= 0:
            return 0.5  # Both empty at the same time
        if a <= 0:
            return 1.0  # A empty first
        if b <= 0:
            return 0.0  # B empty first

        prob = 0.25 * (solve(a - 4, b) +
                        solve(a - 3, b - 1) +
                        solve(a - 2, b - 2) +
                        solve(a - 1, b - 3))

        dp[(a, b)] = prob
        return prob

    return solve(n, n)