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Interval I

Description

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.

Question Link

https://leetcode.com/problems/guess-number-higher-or-lower-ii/description/

Example I:

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.

Questions

• LeetCode - 375. Guess Number Higher or Lower II

Transform Function

for(int len = 1; len <= n; len++) {
    for(int i = 1; i + len <= n + 1; i++) {
        int global_min = Integer.MAX_VALUE;
        for(int k = i + 1; k < j; k++) {
            int temp = k + Math.max(dp[i][k - 1], dp[k + 1][j]);
            global_min = Math.min(global_min, temp);
        }
        dp[i][j] = global_min;
    }
}

tips: len works from len >= 3, because i ~ (k - 1) and (k + 1) ~ j needs at least 3 interval length

Solution

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