Question

375. Guess Number Higher or Lower II

中文文档

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

Solution 1

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]

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];
  }
};

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]
}