#375

Medium

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

Guess Number Higher or Lower II

Minimize the maximum cost of guessing a number in a dynamic guessing game using optimal strategies.

Problem Statement

The Guessing Game works as follows: you are given a number range from 1 to n. Your task is to guess a number, and if it is not correct, you will have to pay the number you guessed. Your goal is to minimize the maximum amount of money you could lose in the worst-case scenario, regardless of which number is chosen.

For a given n, return the minimum amount of money needed to guarantee a win, ensuring that you use the best possible strategy for minimizing potential losses in the worst-case scenario.

Examples

Example 1

Input: n = 10

Output: 16

The winning strategy is as follows:

The range is [1,10]. Guess 7.
If this is my number, your total is $0. Otherwise, you pay$ 7.
If my number is higher, the range is [8,10]. Guess 9.
If this is my number, your total is $7. Otherwise, you pay$ 9.
If my number is higher, it must be 10. Guess 10. Your total is $7 +$ 9 = $16.
If my number is lower, it must be 8. Guess 8. Your total is $7 +$ 9 = $16.
If my number is lower, the range is [1,6]. Guess 3.
If this is my number, your total is $7. Otherwise, you pay$ 3.
If my number is higher, the range is [4,6]. Guess 5.
If this is my number, your total is $7 +$ 3 = $10. Otherwise, you pay$ 5.
If my number is higher, it must be 6. Guess 6. Your total is $7 +$ 3 + $5 =$ 15.
If my number is lower, it must be 4. Guess 4. Your total is $7 +$ 3 + $5 =$ 15.
If my number is lower, the range is [1,2]. Guess 1.
If this is my number, your total is $7 +$ 3 = $10. Otherwise, you pay$ 1.
If my number is higher, it must be 2. Guess 2. Your total is $7 +$ 3 + $1 =$ 11. The worst case in all these scenarios is that you pay $16. Hence, you only need$ 16 to guarantee a win.

Example 2

Input: n = 1

Output: 0

There is only one possible number, so you can guess 1 and not have to pay anything.

Example 3

Input: n = 2

Output: 1

There are two possible numbers, 1 and 2.

Guess 1.
If this is my number, your total is $0. Otherwise, you pay$ 1.
If my number is higher, it must be 2. Guess 2. Your total is $1. The worst case is that you pay$ 1.

Constraints

1 <= n <= 200

Solution Approach

Dynamic Programming Approach

We can solve this problem using dynamic programming, where we store the minimal cost required to guarantee a win for each subproblem range [i, j]. The cost for each range can be calculated by considering the cost of guessing a number k, then determining the cost of the two subranges [i, k-1] and [k+1, j]. The idea is to choose the k that minimizes the worst-case cost for each range.

State Transition Dynamic Programming

The key to this problem is the state transition between subproblems. The transition depends on minimizing the maximum loss, which means calculating the cost of the best guess and updating the state for each possible guess within the current range. This results in a decision tree where each state leads to smaller subproblems.

Optimal Substructure

The problem has an optimal substructure, meaning the optimal solution for a range can be constructed from optimal solutions to smaller subranges. By solving smaller subproblems first and building up, we can find the solution for the entire range. This approach ensures we minimize the maximum possible cost of guessing at each step.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time and space complexity depend on the final implementation of the dynamic programming approach. The algorithm typically involves iterating through each pair of start and end points in the range [1, n], making the time complexity approximately O(n^2). Space complexity is also O(n^2) due to the need to store intermediate results for all subranges.

What Interviewers Usually Probe

Can the candidate explain how the dynamic programming approach minimizes the worst-case cost?
Does the candidate understand how to construct subproblems for a given range and transition between them?
Can the candidate identify the optimal substructure and explain how it applies to this problem?

Common Pitfalls or Variants

Common pitfalls

Choosing the wrong number to guess within a range, leading to an inefficient solution with higher costs.
Failing to recognize the optimal substructure and trying to brute force the problem instead of using dynamic programming.
Not properly accounting for all possible subranges and making incorrect assumptions about the problem's state transitions.

Follow-up variants

What if the range starts at a number other than 1?
What if the cost function changes and isn't directly tied to the guess number?
How does this problem change if you're allowed to make multiple guesses instead of just one?

FAQ

What is the optimal strategy for the Guess Number Higher or Lower II problem?

The optimal strategy is to minimize the maximum cost by choosing guesses that break the range into smaller subranges, ensuring the worst-case cost is as low as possible.

What dynamic programming technique is used in this problem?

State transition dynamic programming is used to calculate the minimum cost required for each subrange and build the solution iteratively.

How do you calculate the cost of a guess in this problem?

The cost of a guess is calculated by considering the worst-case scenario for both the higher and lower subranges after making a guess.

What is the time complexity of solving this problem?

The time complexity is O(n^2) due to iterating over all possible subranges and computing the minimal costs for each range.

How can GhostInterview assist in solving Guess Number Higher or Lower II?

GhostInterview helps by guiding you through the problem's dynamic programming approach, ensuring a clear understanding of state transitions and the optimal solution process.

terminal

Solution

Solution 1: Dynamic Programming

We define $f[i][j]$ as the minimum cost required to guess any number in the interval $[i, j]$. Initially, $f[i][i] = 0$ because there is no cost to guess the only number, and for $i > j$, we also have $f[i][j] = 0$. The answer is $f[1][n]$.

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class Solution:
    def getMoneyAmount(self, n: int) -> int:
        f = [[0] * (n + 1) for _ in range(n + 1)]
        for i in range(n - 1, 0, -1):
            for j in range(i + 1, n + 1):
                f[i][j] = j + f[i][j - 1]
                for k in range(i, j):
                    f[i][j] = min(f[i][j], max(f[i][k - 1], f[k + 1][j]) + k)
        return f[1][n]

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