Dynamic Programming - 375. Guess Number Higher or Lower II

375. Guess Number Higher or Lower II

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.

思路：

这一题是374题的升级题目，374只需要用一个二分查找就能在不超时的情况下猜中数字，而这题是在猜数字的情况下，猜错了就得付钱，求的是保证能赢花出去的钱，做法是使用动态规划来做，在1-n个数里面，我们任意猜一个数(设为i)，保证获胜所花的钱应该为 i + max(w(1 ,i-1), w(i+1 ,n))，这里w(x,y))表示猜范围在(x,y)的数保证能赢应花的钱，则我们依次遍历 1-n作为猜的数，求出其中的最小值即为答案。

代码： go :

func getMoneyAmount(n int) int {
    var dp = make([][]int, n+1)
    for i := range dp {
        dp[i] = make([]int, n+1)
    }
    
    for j := 2; j <= n; j++ {
        for i := j-1; i > 0; i-- {
            globalMin :=math.MaxInt32
            for k := i+1; k < j; k++ {
                localMax := k + max(dp[i][k-1], dp[k+1][j])
                globalMin = min(globalMin, localMax)
            }
            if i+1 == j {
                dp[i][j] = i 
            } else {
                dp[i][j] = globalMin
            }
        }
    }
    
    return dp[1][n];
}

// recursive dp solution
func getMoneyAmount(n int) int {

    var dp = make([][]int, n+1)
    for i := range dp {
        dp[i] = make([]int, n+1)
    }
    
    return dfs(1, n, &dp)
}


func dfs(start, end int, dp *[][]int) int {
    if start >= end {
        return 0
    }
    if (*dp)[start][end] != 0 {
        return (*dp)[start][end]
    }
    
    res := math.MaxInt32
    for i := start; i <= end; i++ {
        res = min(res, i + max(dfs(start, i - 1, dp), dfs(i + 1, end, dp)))
    }
    
    (*dp)[start][end] = res
    
    return res;
}

func min(i, j int) int {
    if i < j {
        return i
    }
    return j
}

func max(i, j int) int {
    if i > j {
        return i
    }
    return j
}