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375. Guess Number Higher or Lower II
Description
We are playing the Guessing Game. The game will work as follows:
 I pick a number between
1
andn
.  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 payx
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+l1 <= n; i++ { j := i + l  1 dp[i][j] = math.MaxInt32 for k := i; k <= j; k++ { t := max(dp[i][k1], dp[k+1][j]) + k dp[i][j] = min(dp[i][j], t) } } } return dp[1][n] }