【LEETCODE】375- Guess Number Higher or Lower II [Python]

题目：

https://leetcode.com/problems/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.Secondround:Youguess7,Itellyouthatit′shigher.Youpay 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.

题意：

For any given n, to find the minimum money you have to spend to guarantee a win.

分析：

Step 1: The structure

Since the target number is still now known, we have to find the minimum cost for all the target through 1 to n.

When you guess a number k, I will give you the response, there are 3 kinds, ‘k’s lower than/higher than/equal to target’. And I prefer to let you spend more and more money, so the target I chosen will fall in the part where the cost is the maximum.

In total, if I want to get the minimum cost for period 1 to n, I will guess every k through 1 to n, for each k, choose the max part from the right and left side of k, then find the k to minimize the cost.

In this process, some parts will be calculated for duplicate times, so we can use Dynamic Programming to store the results.

Step 2: A recursive solution

Let dp to be a matrix, and dp[i, j] stores the minimum cost for this period from i to j. Then we got the formula, k is the number you guess, i and j means the start and end of the period.

dp[i, j] = min( max( dp[i, k-1], dp[k+1, j] ) + k )

The final result we want to receive is f(1, n).

If i==j, dp[i, i] = 0.

If j==i+1, dp[i, i+1] = i.

Step 3: Computing the optimal costs

class Solution(object):
    def getMoneyAmount(self, n):
        """ :type n: int :rtype: int """
        dp = [ [0]*(n+1) for i in range(n+1) ]

        def solve(left, right):

            if left<right and dp[left][right]==0:

                dp[left][right] = min( max( solve(left, k-1), solve(k+1, right) )+k for k in range(left, right))

            return dp[left][right]       

        return solve(1, n)