LeetCode #375 — MEDIUM

Guess Number Higher or Lower II

Move from brute-force thinking to an efficient approach using math strategy.

Solve on LeetCode
The Problem

Problem Statement

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
Patterns Used

Roadmap

  1. Brute Force Baseline
  2. Core Insight
  3. Algorithm Walkthrough
  4. Edge Cases
  5. Full Annotated Code
  6. Interactive Study Demo
  7. Complexity Analysis
Step 01

Brute Force Baseline

Problem summary: We are playing the Guessing Game. The game will work as follows: I pick a number between 1 and n. You guess a number. If you guess the right number, you win the game. 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. 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math · Dynamic Programming

Example 1

10

Example 2

1

Example 3

2

Related Problems

  • Flip Game II (flip-game-ii)
  • Guess Number Higher or Lower (guess-number-higher-or-lower)
  • Can I Win (can-i-win)
  • Find K Closest Elements (find-k-closest-elements)
Step 02

Core Insight

What unlocks the optimal approach

  • The best strategy to play the game is to minimize the maximum loss you could possibly face. Another strategy is to minimize the expected loss. Here, we are interested in the <b>first</b> scenario.
  • Take a small example (n = 3). What do you end up paying in the worst case?
  • Check out <a href="https://en.wikipedia.org/wiki/Minimax">this article</a> if you're still stuck.
  • The purely recursive implementation of minimax would be worthless for even a small n. You MUST use dynamic programming.
  • As a follow-up, how would you modify your code to solve the problem of minimizing the expected loss, instead of the worst-case loss?
Interview move: turn each hint into an invariant you can check after every iteration/recursion step.
Step 03

Algorithm Walkthrough

Iteration Checklist

  1. Define state (indices, window, stack, map, DP cell, or recursion frame).
  2. Apply one transition step and update the invariant.
  3. Record answer candidate when condition is met.
  4. Continue until all input is consumed.
Use the first example testcase as your mental trace to verify each transition.
Step 04

Edge Cases

Minimum Input
Single element / shortest valid input
Validate boundary behavior before entering the main loop or recursion.
Duplicates & Repeats
Repeated values / repeated states
Decide whether duplicates should be merged, skipped, or counted explicitly.
Extreme Constraints
Upper-end input sizes
Re-check complexity target against constraints to avoid time-limit issues.
Invalid / Corner Shape
Empty collections, zeros, or disconnected structures
Handle special-case structure before the core algorithm path.
Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #375: Guess Number Higher or Lower II
class Solution {
    public int getMoneyAmount(int n) {
        int[][] f = new int[n + 1][n + 1];
        for (int i = n - 1; i > 0; --i) {
            for (int j = i + 1; j <= n; ++j) {
                f[i][j] = j + f[i][j - 1];
                for (int k = i; k < j; ++k) {
                    f[i][j] = Math.min(f[i][j], Math.max(f[i][k - 1], f[k + 1][j]) + k);
                }
            }
        }
        return f[1][n];
    }
}
Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Press Step or Run All to begin.
Step 07

Complexity Analysis

Time
O(n × m)
Space
O(n × m)

Approach Breakdown

RECURSIVE
O(2ⁿ) time
O(n) space

Pure recursion explores every possible choice at each step. With two choices per state (take or skip), the decision tree has 2ⁿ leaves. The recursion stack uses O(n) space. Many subproblems are recomputed exponentially many times.

DYNAMIC PROGRAMMING
O(n × m) time
O(n × m) space

Each cell in the DP table is computed exactly once from previously solved subproblems. The table dimensions determine both time and space. Look for the state variables — each unique combination of state values is one cell. Often a rolling array can reduce space by one dimension.

Shortcut: Count your DP state dimensions → that’s your time. Can you drop one? That’s your space optimization.
Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.

State misses one required dimension

Wrong move: An incomplete state merges distinct subproblems and caches incorrect answers.

Usually fails on: Correctness breaks on cases that differ only in hidden state.

Fix: Define state so each unique subproblem maps to one DP cell.