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Guess Number Higher or Lower II

We are playing the Guessing Game. The game will work as follows:

  1. I pick a number between 1 and n.
  2. You guess a number.
  3. If you guess the right number, you win the game.
  4. If you guess the wrong number, then I will tell you whether the number I picked is higher or lower, and you will continue guessing.
  5. Every time you guess a wrong number x, you will pay x dollars. If you run out of money, you lose the game.

Given a particular n, return the minimum amount of money you need to guarantee a win regardless of what number I pick.

 

Example 1:

Input: n = 10
Output: 16
Explanation: 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
Explanation: There is only one possible number, so you can guess 1 and not have to pay anything.

Example 3:

Input: n = 2
Output: 1
Explanation: 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 Explanation: Guess Number Higher or Lower II

This problem can be solved efficiently using dynamic programming. The core idea is to break down the problem into smaller overlapping subproblems and store their solutions to avoid redundant calculations.

Approach:

  1. State Definition: We use a 2D array dp (or f in the code examples) where dp[i][j] represents the minimum amount of money needed to guarantee a win when the range of possible numbers is [i, j].

  2. Base Cases:

    • dp[i][i] = 0: If the range contains only one number, you've already guessed it, so the cost is 0.
    • dp[i][j] = 0 for i > j: This case is invalid as i cannot be greater than j in a valid range.

  3. Recursive Relation: For a range [i, j], we can choose any number k within the range (i ≤ k ≤ j) as our guess. If the chosen number k is the target number, the cost is 0. Otherwise, we incur a cost of k, and we need to recursively solve the problem for either the lower half ([i, k-1]) or the upper half ([k+1, j]), depending on whether the target number is less than or greater than k. The worst-case scenario (maximum cost) needs to be considered.

    The recursive relation is:

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

  4. Iteration: We iterate through the dp array, filling in the values using the recursive relation. We start from smaller ranges and work our way up to larger ranges, leveraging the previously computed values. This bottom-up approach is essential for dynamic programming's efficiency.

  5. Result: The final answer is dp[1][n], which represents the minimum cost to guarantee a win for the range [1, n].

Time and Space Complexity Analysis:

  • Time Complexity: O(n³). The three nested loops in the code iterate through the dp array, resulting in cubic time complexity.

  • Space Complexity: O(n²). We use a 2D array of size n x n to store the dp values.

Code Examples (with detailed explanations already provided in the previous response):

The code examples provided in Python, Java, C++, Go, and TypeScript demonstrate the dynamic programming approach effectively. They all follow the same logic, differing only in syntax. The comments within the code explain each step clearly. The key is the nested loop structure which implements the recursive relation described above.