Introduction

In this blog post, we will explore how to solve the LeetCode problem "375" 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

We are playing the Guessing Game. The game will work as follows: I pick a number between 1 and n. You guess a number. If you guess the right number, you win the game. If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing. Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game. Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick. Example 1: Input: n = 10 Output: 16 Explanation: The winning strategy is as follows: - The range is [1,10]. Guess 7. - If this is my number, your total is $0. Otherwise, you pay $7. - If my number is higher, the range is [8,10]. Guess 9. - If this is my number, your total is $7. Otherwise, you pay $9. - If my number is higher, it must be 10. Guess 10. Your total is $7 + $9 = $16. - If my number is lower, it must be 8. Guess 8. Your total is $7 + $9 = $16. - If my number is lower, the range is [1,6]. Guess 3. - If this is my number, your total is $7. Otherwise, you pay $3. - If my number is higher, the range is [4,6]. Guess 5. - If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $5. - If my number is higher, it must be 6. Guess 6. Your total is $7 + $3 + $5 = $15. - If my number is lower, it must be 4. Guess 4. Your total is $7 + $3 + $5 = $15. - If my number is lower, the range is [1,2]. Guess 1. - If this is my number, your total is $7 + $3 = $10. Otherwise, you pay $1. - If my number is higher, it must be 2. Guess 2. Your total is $7 + $3 + $1 = $11. The worst case in all these scenarios is that you pay $16. Hence, you only need $16 to guarantee a win. Example 2: Input: n = 1 Output: 0 Explanation: There is only one possible number, so you can guess 1 and not have to pay anything. Example 3: Input: n = 2 Output: 1 Explanation: There are two possible numbers, 1 and 2. - Guess 1. - If this is my number, your total is $0. Otherwise, you pay $1. - If my number is higher, it must be 2. Guess 2. Your total is $1. The worst case is that you pay $1. Constraints: 1 <= n <= 200

Explanation

Here's the approach, analysis, and code:

• Dynamic Programming: Use dynamic programming to solve this problem. dp[i][j] represents the minimum cost to guarantee a win when the range is from i to j.
• Optimal Substructure: The optimal cost dp[i][j] can be found by trying each number k in the range [i, j] as the first guess. The cost would be k + max(dp[i][k-1], dp[k+1][j]). We choose the k that minimizes this cost.

• Base Cases: dp[i][i] = 0 (if the range has only one number, no cost is needed) and dp[i][i+1] = i (if there are two numbers i and i+1, we pick i. If it's wrong, we pay i and know that i+1 is the answer.).

• Time Complexity: O(n^3), Space Complexity: O(n^2)

Code

    def getMoneyAmount(n: int) -> int:
    """
    Calculates the minimum amount of money needed to guarantee a win in the Guessing Game.

    Args:
        n: The upper bound of the range of numbers to guess (1 to n).

    Returns:
        The minimum amount of money needed.
    """

    dp = [[0] * (n + 1) for _ in range(n + 1)]

    for length in range(2, n + 1):
        for i in range(1, n - length + 2):
            j = i + length - 1
            dp[i][j] = float('inf')
            for k in range(i, j):
                dp[i][j] = min(dp[i][j], k + max(dp[i][k - 1] if i < k else 0, dp[k + 1][j] if k < j else 0))

    return dp[1][n]