Problem Description We're playing a number guessing game with the following rules:

• I pick a number between 1 and n.
• You guess which number I picked.
• Each time you guess wrong, I tell you whether my number is higher or lower.
• When you guess a number x and it's wrong, you pay $x.
• You win when you guess the correct number. Given n ≥ 1, determine the minimum amount of money you need to guarantee a win.

Example For n = 10, if I pick 8:

• First guess: 5 (higher) → pay $5
• Second guess: 7 (higher) → pay $7
• Third guess: 9 (lower) → pay $9 You win with 8. Total payment: $5 + $7 + $9 = $21.

Dynamic Programming Approach This problem can be solved using dynamic programming. The key insight is to break down the problem into smaller subproblems and build up the solution.

Understanding the Problem For any range [i, j], we need to find the minimum cost to guarantee a win. The optimal strategy involves choosing a pivot number k (where i ≤ k ≤ j) and considering the worst-case scenario:

• If the target number is less than k, we need to solve the subproblem [i, k-1]
• If the target number is greater than k, we need to solve the subproblem [k+1, j] The cost for choosing k is k + max(cost[i, k-1], cost[k+1, j]). We want to minimize this cost over all possible k.

DP Table Construction We'll use a 2D table dp where dp[i][j] represents the minimum cost to guarantee a win for the range [i, j].

For small ranges:

• When i == j: dp[i][j] = 0 (no cost needed as we've found the number)
• When j == i+1: dp[i][j] = i (only one guess needed)

For larger ranges, we fill the table by increasing the range size:

for length from 2 to n:
  for i from 1 to n-length+1:
    j = i + length - 1
    dp[i][j] = infinity
    for k from i to j:
      cost = k + max(dp[i][k-1], dp[k+1][j])
      dp[i][j] = min(dp[i][j], cost)

Implementatino Here's the Python implementation using dynamic programming:

def get_minimum_money(n):
    """
    Calculate the minimum amount of money needed to guarantee a win in the guessing game.
    
    Args:
        n (int): The upper bound of the number range (1 to n)
    
    Returns:
        int: The minimum amount of money required to guarantee a win
    """
    # Create a DP table initialized with zeros
    dp = [[0] * (n + 2) for _ in range(n + 2)]
    
    # Fill the table for increasing range sizes
    for length in range(2, n + 1):
        for start in range(1, n - length + 2):
            end = start + length - 1
            dp[start][end] = float('inf')
            
            # Try all possible pivot points
            for pivot in range(start, end + 1):
                # Calculate cost for current pivot
                cost = pivot + max(dp[start][pivot - 1], dp[pivot + 1][end])
                dp[start][end] = min(dp[start][end], cost)
    
    return dp[1][n]

Optimization The above solution has a time complexity of O(n³) which can be optimized. We can observe that the optimal pivot tends to be around the middle of the range, so we can limit our search to a smaller window around the middle:

def get_minimum_money_optimized(n):
    """
    Optimized version of the guessing game solution with reduced time complexity.
    
    Args:
        n (int): The upper bound of the number range (1 to n)
    
    Returns:
        int: The minimum amount of money required to guarantee a win
    """
    dp = [[0] * (n + 2) for _ in range(n + 2)]
    
    for length in range(2, n + 1):
        for start in range(1, n - length + 2):
            end = start + length - 1
            dp[start][end] = float('inf')
            
            # Only check pivots in a smaller window around the middle
            left = max(start, (start + end) // 2 - 1)
            right = min(end, (start + end) // 2 + 2)
            
            for pivot in range(left, right + 1):
                cost = pivot + max(dp[start][pivot - 1], dp[pivot + 1][end])
                dp[start][end] = min(dp[start][end], cost)
    
    return dp[1][n]

Complexity Analysis

• Time Complexity: O(n³) for the basic solution, O(n²) for the optimized version
• Space Complexity: O(n²) for storing the DP table

Example Walkthrough For n = 4, the DP table would be filled as follows:

The answer is dp[1][4] = 4, meaning we need at least $4 to guarantee a win for n = 4.