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375. Guess Number Higher or Lower II (Medium)

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.

Related Topics:
Dynamic Programming, Minimax

Similar Questions:

Solution 1. DP Top-down

Let dp[i][j] be the answer to the subproblem with number range [i, j].

dp[i][j] = min( k + max(dp[i][k-1], dp[k+1][j]) | i <= k <= j )
dp[i][j] = 0   where i >= j

// OJ: https://leetcode.com/problems/guess-number-higher-or-lower-ii/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
    int solve(vector<vector<int>> &dp, int i, int j) {
        if (i >= j) return 0;
        if (dp[i][j]) return dp[i][j];
        dp[i][j] = INT_MAX;
        for (int k = i; k <= j; ++k) dp[i][j] = min(dp[i][j], k + max(solve(dp, i, k - 1), solve(dp, k + 1, j)));
        return dp[i][j];
    }
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n + 1, vector<int>(n + 1));
        return solve(dp, 1, n);
    }
};

Solution 2. DP Bottom-up

// OJ: https://leetcode.com/problems/guess-number-higher-or-lower-ii/
// Author: github.com/lzl124631x
// Time: O(N^2)
// Space: O(N^2)
class Solution {
public:
    int getMoneyAmount(int n) {
        vector<vector<int>> dp(n + 1, vector<int>(n + 1));
        for (int i = n - 1; i >= 1; --i) {
            for (int j = i + 1; j <= n; ++j) {
                dp[i][j] = INT_MAX;
                for (int k = i; k <= j; ++k) {
                    dp[i][j] = min(dp[i][j], k + max(k == i ? 0 : dp[i][k - 1], k == j ? 0 : dp[k + 1][j]));
                }
            }
        }
        return dp[1][n];
    }
};