How do you solve Guess Number Higher or Lower II on LeetCode?

Guess Number Higher or Lower II (LeetCode #375) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n²) time complexity and O(n²) space complexity. Key insight: Minimizing the maximum cost is crucial to guarantee a win.

What is the brute force solution for Guess Number Higher or Lower II?

The brute force approach for Guess Number Higher or Lower II is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves trying every possible guess and calculating the maximum cost for each guess. This is simple but inefficient as it explores all combinations. The optimal Optimal Solution approach improves this to O(n²).

What algorithm or data structure is used to solve Guess Number Higher or Lower II?

Guess Number Higher or Lower II uses the following concepts: Math, Dynamic Programming, Game Theory. The recommended patterns to study are: Dynamic Programming, Game Theory.

What are the interview tips for Guess Number Higher or Lower II?

Always think about the worst-case scenario when dealing with guessing games. Explain your thought process clearly, especially how you handle different ranges. Practice similar dynamic programming problems to get comfortable with state transitions.

What are common mistakes when solving Guess Number Higher or Lower II?

Not considering all possible guesses and their consequences. Failing to use memoization or dynamic programming to optimize the solution.

#375

Guess Number Higher or Lower II

Medium

MathDynamic ProgrammingGame TheoryDynamic ProgrammingGame Theory

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n²)
Space	O(1)	O(n²)

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Intuition

Time O(n²)Space O(n²)

The optimal solution uses dynamic programming to minimize the maximum cost by storing results of subproblems. This avoids redundant calculations and speeds up the process significantly.

⚙️

Algorithm

3 steps

1Step 1: Create a 2D array dp where dp[i][j] represents the minimum cost to guarantee a win between numbers i and j.
2Step 2: Fill the dp array for ranges of increasing lengths, calculating the cost for each possible guess in that range.
3Step 3: For each guess, calculate the worst-case cost and update the dp table accordingly.

solution.py12 lines

1# Full working Python code
2class Solution:
3    def getMoneyAmount(self, n: int) -> int:
4        dp = [[0] * (n + 1) for _ in range(n + 1)]
5        for length in range(2, n + 1):
6            for start in range(1, n - length + 2):
7                end = start + length - 1
8                dp[start][end] = float('inf')
9                for x in range(start, end):
10                    cost = x + max(dp[start][x - 1], dp[x + 1][end])
11                    dp[start][end] = min(dp[start][end], cost)
12        return dp[1][n]

ℹ

Complexity note: The time complexity remains O(n²) due to the nested loops, but we now use O(n²) space to store results of subproblems, which allows us to avoid redundant calculations.

1Minimizing the maximum cost is crucial to guarantee a win.
2Dynamic programming helps break down the problem into manageable subproblems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.