375. Guess Number Higher or Lower II

Posted 2021-02-09 ruruozhenhao

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.

 

Approach #1: DP. [C++]

class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
        return getMoneyAmount(dp, 1, n);
    }
    
private:
    int getMoneyAmount(vector<vector<int>>& dp, int s, int e) {
        if (s >= e) return 0;
        if (dp[s][e] != 0) return dp[s][e];
        
        int res = INT_MAX;
        for (int i = s; i <= e; ++i) {
            int tmp = i + max(getMoneyAmount(dp, s, i-1), getMoneyAmount(dp, i+1, e));
            res = min(res, tmp);
        }
        
        dp[s][e] = res;
        
        return res;
    }
};

 

Analysis:

dp[i][j] : the index from i to j how much money you have to pay.

res = i + max(fuc(), fuc());

the reasion why we use max instead of min is that we want to find the minimum money which we have to pay. If we use min in this place it will return 0 alway.

 

Approach #2: bottom to up. [C++]

    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n+1, vector<int>(n+1, 0));
        for (int i = 2; i <= n; ++i) {
            for (int j = i-1; j > 0; --j) {
                int globalMin = INT_MAX;
                for (int k = j+1; k < i; ++k) {
                    int localMin = k + max(dp[j][k-1], dp[k+1][i]);
                    globalMin = min(globalMin, localMin);
                }
                dp[j][i] = j+1 == i ? j : globalMin;
            }
        }
        return dp[1][n];
    }

　　

Analysis:

If j+1 == i, dp[j][i] represent two adjacent numbers, so we return the little one as the value of dp[j][i]. why not is the bigger one? I think it is realted to we solve this problem using bottom to up method. so we should get the little one‘s value firstly.