# Solving Leetcode Interviews in Seconds with AI: Guess Number Higher or Lower II


	# 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](https://www.chatmagic.app), 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
	```python
	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]
	```