375 Guess Number Higher or Lower II *****imp & hard

Dynamic Programming

We are playing the Guess Game. The game is as follows:

I pick a number from 1 to n. You have to guess which number I picked.

Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.

However, when you guess a particular number x, and you guess wrong, you pay $x. You win the game when you guess the number I picked.

Example:

n = 10, I pick 8.

First round:  You guess 5, I tell you that it's higher. You pay $5.
Second round: You guess 7, I tell you that it's higher. You pay $7.
Third round:  You guess 9, I tell you that it's lower. You pay $9.
Game over. 8 is the number I picked.
You end up paying $5 + $7 + $9 = $21.

Given a particular n ≥ 1, find out how much money you need to have to guarantee a win.

dp[i][j] : i - j 肯定赢所需多少钱

最大最小值问题：i - j 中，任意猜一个数x，获胜所花的钱为 x + max(helper(i ,x-1), helper(x+1 ,j))

time O(n ^ 3)

space O(n ^ 2)

dfs + dp

class Solution {
    int[][] dp;
    public int getMoneyAmount(int n) {
        dp = new int[n+1][n+1];
        return helper(1, n);
    }
    public int helper(int i, int j) {
        if (i >= j) {
            return 0;
        }
        if (dp[i][j] != 0) {
            return dp[i][j];
        }
        int res = Integer.MAX_VALUE;
        for(int x = i; x <= j; x++) {
            res = Math.min(res, x + Math.max(helper(i, x-1), helper(x + 1, j)));
        }
        dp[i][j] = res;
        return res;
    }
}

class Solution {
    public int getMoneyAmount(int n) {
        int[][] dp = new int[n + 1][n + 1];
        for (int i = n - 1; i > 0; i--) {
            for (int j = i + 1; j <= n; j++) {
                dp[i][j] = Integer.MAX_VALUE;
                for (int x = i; x < j; x++) {
                    dp[i][j] = Math.min(dp[i][j], x + Math.max(dp[i][x - 1], dp[x + 1][j]));
                }
            }
        }
        return dp[1][n];
    }
}