375. Guess Number Higher or Lower II

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Problem

We are playing the Guessing Game. The game will work as follows:

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:

Solution

class Solution {
    int[][] matrix;

    public int getMoneyAmount(int n) {
        matrix = new int[n + 1][n + 1];
        return get(1, n);
    }

    private int get(int min, int max) {
        if (max - min < 3) {
            return max - min <= 0 ? 0 : max - 1;
        }
        if (matrix[min][max] != 0) {
            return matrix[min][max];
        }
        int select = max - 3;
        int minRes = Integer.MAX_VALUE;
        int res;
        int end = min + ((max - min) >> 1) - 1;
        int cnt = 0;
        while (true) {
            res = select + Math.max(get(min, select - 1), get(select + 1, max));
            if (res > minRes) {
                cnt++;
                if (cnt >= 3) {
                    break;
                }
            }
            if (res < minRes) {
                minRes = res;
            }
            select--;
            if (select <= end) {
                break;
            }
        }
        matrix[min][max] = minRes;
        return minRes;
    }
}

Explain:

nope.

Complexity: