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375. Guess Number Higher or Lower II

Description

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

  1. I pick a number between 1 and n.
  2. You guess a number.
  3. If you guess the right number, you win the game.
  4. 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.
  5. 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

Solutions

  • class Solution {
    public int getMoneyAmount(int n) {
        int[][] dp = new int[n + 10][n + 10];
        for (int l = 2; l <= n; ++l) {
            for (int i = 1; i + l - 1 <= n; ++i) {
                int j = i + l - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k <= j; ++k) {
                    int t = Math.max(dp[i][k - 1], dp[k + 1][j]) + k;
                    dp[i][j] = Math.min(dp[i][j], t);
                }
            }
        }
        return dp[1][n];
    }
}

  • class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n + 10, vector<int>(n + 10));
        for (int l = 2; l <= n; ++l) {
            for (int i = 1; i + l - 1 <= n; ++i) {
                int j = i + l - 1;
                dp[i][j] = INT_MAX;
                for (int k = i; k <= j; ++k) {
                    int t = max(dp[i][k - 1], dp[k + 1][j]) + k;
                    dp[i][j] = min(dp[i][j], t);
                }
            }
        }
        return dp[1][n];
    }
};

  • class Solution:
    def getMoneyAmount(self, n: int) -> int:
        dp = [[0] * (n + 10) for _ in range(n + 10)]
        for l in range(2, n + 1):
            for i in range(1, n - l + 2):
                j = i + l - 1
                dp[i][j] = inf
                for k in range(i, j + 1):
                    t = max(dp[i][k - 1], dp[k + 1][j]) + k
                    dp[i][j] = min(dp[i][j], t)
        return dp[1][n]


  • func getMoneyAmount(n int) int {
	dp := make([][]int, n+10)
	for i := 0; i < len(dp); i++ {
		dp[i] = make([]int, n+10)
	}
	for l := 2; l <= n; l++ {
		for i := 1; i+l-1 <= n; i++ {
			j := i + l - 1
			dp[i][j] = math.MaxInt32
			for k := i; k <= j; k++ {
				t := max(dp[i][k-1], dp[k+1][j]) + k
				dp[i][j] = min(dp[i][j], t)
			}
		}
	}
	return dp[1][n]
}

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