Interview Problem Explained: Guess Number Higher or Lower II(LeetCode 375, Minimax & Dynamic Programming)

Guess Number Higher or Lower II: Problem Statement

You are playing a game where you must guess a number between 1 and n. Each time you make a guess and it’s incorrect, you must pay the value of your guess. After each guess, you are told whether the correct answer is higher or lower than your guess.

Your goal is to calculate the minimum total cost required to guarantee a win, regardless of what the number is.

This means, no matter what number you guess, you must assume the opponent (or the game's rules) will always steer the situation to maximize your cost. You need to find the optimal guessing strategy and its minimum guaranteed cost, even in the worst-case scenario.

Example (n = 3):

Your goal is to determine which number to guess first among 1, 2, or 3 to minimize your total guaranteed payment.

1. If you guess 2 first:

   • Pay $2

   • Case 1: Answer is 1 (lower) → range becomes [1][1], extra cost $0 (guess 1 next)

   • Case 2: Answer is 3 (higher) → range becomes [3][3], extra cost $0 (guess 3 next)

   • Opponent maximizes your cost: max($0, $0) = $0

   • Total worst-case cost: $2 + $0 = $2

2. If you guess 1 first:

   • Pay $1

   • If answer is 1: game ends, extra cost $0

   • If answer is higher (2 or 3): range [2,[3]

     • Guess 2: pay $2, if answer is 3, next guess is 3 ($0 more), total $2

     • Guess 3: pay $3, if answer is 2, next guess is 2 ($0 more), total $3

     • Minimum guaranteed cost for [2,[3] is $2

   • Opponent forces the higher cost, so extra cost is $2

   • Total worst-case cost: $1 + $2 = $3

3. If you guess 3 first: (symmetric to guessing 1)

   • Pay $3

   • If answer is lower (1 or 2): range [1,[2]

     • Guess 1: pay $1, if answer is 2, next guess is 2 ($0 more), total $1

   • Extra cost: $1

   • Total: $3 + $1 = $4

You must calculate the worst-case cost for each possible first guess (k = 1, 2, 3), and pick the minimum. Here, the answer is $2.

Problem Requirements

1. Input: Integer n (1 ≤ n ≤ 200)

2. Output: Return the minimum total cost required to guarantee a win, regardless of the number chosen.

Tip: This is LeetCode Problem 375.
LeetCode’s problem statements are often concise and may confuse learners. Codetree provides extra explanations and examples for deeper understanding.

Step 1. Naive Approach

Idea

The core of this problem is to find the minimum cost that guarantees a win, no matter what the answer is. You want to minimize your cost, but the opponent always acts to maximize it. This is the basis of the Minimax strategy: "I try to minimize my loss, while the opponent tries to maximize it."

The most natural way to solve this is with recursion. Define a function solve(start, end) that returns the minimum guaranteed cost to win for the range [start, end].

1. Which number should you guess?: Try every possible first guess k in [start, end].

2. Cost for guessing k:

   • Pay k

   • If k is correct: done, no extra cost

   • If not: you are told if the answer is lower ([start, k-1]) or higher ([k+1, end]). The opponent will always force you into the more expensive subrange: max(solve(start, k-1), solve(k+1, end))

3. Choose the optimal k:

   • For each k, compute k + max(solve(start, k-1), solve(k+1, end))

   • The answer for solve(start, end) is the minimum of these values over all k

Logic Breakdown

solve(start, end): Returns the minimum guaranteed cost to win for the range [start, end].

1. Base Case:

   • If start >= end: No numbers or only one number left. No cost needed, so return 0.

2. Recursive Step:

   • For each k in [start, end]:

     • Cost = k (guess) + worst-case extra cost: max(solve(start, k-1), solve(k+1, end))

   • Return the minimum cost among all possible k

• If k is not the correct answer:

  • If the answer is smaller than k, the remaining range becomes [start, k-1]. The additional cost for this subrange is solve(start, k-1).

  • If the answer is larger than k, the remaining range becomes [k+1, end]. The additional cost for this subrange is solve(k+1, end).

• Opponent’s role: The opponent will force the scenario with the higher additional cost. Thus, the extra cost after guessing k is max(solve(start, k-1), solve(k+1, end)).

• Total worst-case cost for guessing k: k + max(solve(start, k-1), solve(k+1, end)).

• Goal: Minimize this total cost across all possible k. solve(start, end) computes this value for every k in [start, end] and returns the minimum.

[mathjax] [latex] \text{solve}(\text{start}, \text{end}) = \min_{k=\text{start} \dots \text{end}} (k + \max(\text{solve}(\text{start}, k-1), \text{solve}(k+1, \text{end}))) [/latex]

Code Implementation

def solve(start, end):
    if start >= end:
        return 0

    min_cost_for_range = (end - start + 1) * end

    for k in range(start, end + 1):
        # Choose k as the current guess
        # Cost = k (current guess) + worst-case extra cost
        cost_for_this_guess = k + max(solve(start, k - 1), solve(k + 1, end))
        min_cost_for_range = min(min_cost_for_range, cost_for_this_guess)

    return min_cost_for_range

print(solve(1, N)) # Output the minimum cost for the range 1 ~ N

Time & Space Complexity Analysis

• Time Complexity: O(2^N). Each recursive call tries all possible k in [start, end], and for each, calls solve twice for the left and right subranges. This forms a binary tree of height n, leading to about 2^N calls. For N ≥ 200, this is infeasible.

• Space Complexity: O(N). The maximum recursion depth is proportional to N.

Step 2. Memoization

Idea

The problem with the naive recursive approach is that the same subproblem (start, end) is solved multiple times—overlapping subproblems. For example, solve(1, 10) may call solve(4, 6) multiple times through different paths.

The key idea to improve efficiency is dynamic programming (DP), specifically memoization. Store the result of each subproblem, and if it is needed again, return the stored value instead of recomputing.

Logic Breakdown

To apply memoization, we use a dictionary to store the results of the solve(start, end) function.

1. Create a memo dictionary where each key is a (start, end) pair, and the value is the precomputed minimum cost for that range.

2. When solve(start, end) is called:

   • If (start, end) exists in memo: Return the stored value immediately (no recomputation).

   • If(start, end)does not exist inmemo:

     • Compute the cost using the same logic as in Step 1 (naive recursion).

     • Store the computed result in memo[(start, end)] for future reuse.

This memoization technique is a form of dynamic programming (DP), specifically the Top-down approach, where previously computed results are cached and reused to eliminate redundant calculations.

Code Implementation

memo = {}
def solve(start, end):
    if start >= end:
        return 0

    if (start, end) in memo:
        return memo[(start, end)]

    min_cost_for_range = (end - start + 1) * end

    for k in range(start, end + 1):
        # Choose k as the current guess
        # Cost = k (current guess) + worst-case extra cost
        cost_for_this_guess = k + max(solve(start, k - 1), solve(k + 1, end))
        min_cost_for_range = min(min_cost_for_range, cost_for_this_guess)

    memo[(start, end)] = min_cost_for_range
    return min_cost_for_range

print(solve(1, N)) # Output the minimum cost for the range 1 ~ N

Time & Space Complexity

• Time Complexity:O(N^3)

  • There are O(N^2) subproblems (for each possible [start, end]), and for each, we try up to N possible guesses k.

• Space Complexity:O(N^2)

  • The memoization table stores up to N^2 results.

Step 3. Tabulation

Idea

If memoization is a top-down approach—starting from the main problem and solving smaller subproblems recursively while storing the results—tabulation is a bottom-up dynamic programming method. It begins by solving the smallest subproblems first and then builds up the solution to the original problem by filling in a DP table.

In this problem, we define dp[i][j] as the minimum cost to guarantee a win in the range [i, j]. This is conceptually the same as memo[(i, j)] used in the memoization approach. The recurrence relation also remains the same:

dp[i][j] = min_{k=i..j} (k + max(dp[i][k-1], dp[k+1][(with the rule that dp[x][y] = 0 if x ≥ y

Logic Breakdown

In the tabulation approach, the order in which the DP table is filled is critical. To compute dp[i][j], we need the values of smaller subproblems like dp[i][k - 1] and dp[k + 1][j]. That’s why we must fill the DP table in increasing order of interval length.

1. State Definition: dp[i][j] represents the minimum cost required to guarantee a win when guessing a number in the range from i to j.

2. Recurrence Relation: dp[i][j] = min(k + max(dp[i][k - 1], dp[k + 1][j])) for all valid values of k in [i, j]. For base cases where x ≥ y, we assume dp[x][y] = 0. For example, dp[i][i - 1] (an empty range) or dp[i][i] (a single number that can be guessed directly) both have a cost of 0.

3. Base Cases:

   • dp[i][i] = 0 since the cost of guessing a single remaining number is zero.

   • Typically, initializing the entire DP table with zero takes care of base cases like dp[i][i - 1] (empty range) and dp[i][i] (length-1 range) naturally.

4. DP Table Filling Order (Key to Interval DP):

   • Outer loop by interval length: Start with intervals of length L = 2 up to n. We skip L = 1 since dp[i][i] = 0 by default.

   • Start index loop: For each length L, iterate over the starting index i from 1 to n − L + 1.

   • End index calculation: Given i and L, the end index j is simply i + L − 1.

   • Split point (first guess k_pivot) loop: For each interval [i, j], iterate through all possible first guesses k_pivot in the range [i, j]. Use the recurrence: dp[i][j] = min(dp[i][j], k_pivot + max(dp[i][k_pivot - 1], dp[k_pivot + 1][j])) Note that both dp[i][k_pivot - 1] and dp[k_pivot + 1][j] represent smaller sub-intervals than dp[i][j], which is why they must already have been computed in earlier iterations of smaller L.

The final answer is stored in dp[1][n], which represents the minimum cost to guarantee a win in the range from 1 to n.

Code Implementation

# dp[i][j]: minimum cost to guarantee a win in range i to j
dp = [[0] * (N + 2) for _ in range(N + 2)]

for L in range(2, N + 1):  # Interval length
    for i in range(1, N - L + 2):  # Start index
        j = i + L - 1  # End index
        dp[i][j] = float('inf')
        for k_pivot in range(i, j + 1):
            cost_for_this_guess = k_pivot + max(dp[i][k_pivot-1], dp[k_pivot+1][j])
            dp[i][j] = min(dp[i][j], cost_for_this_guess)

print(dp[1][N])  # Final answer

Time & Space Complexity

• Time Complexity: O(N^3) We iterate over the segment length L from 2 to N (O(N)). For each length L, we iterate the starting point i from 1 to N - L + 1 (O(N)). For each segment [i, j], we try all possible pivot points k_pivot from i to j (up to O(N)). This results in three nested loops, leading to a total time complexity of O(N) × O(N) × O(N) = O(N^3).

• Space Complexity: O(N^2) We use a N x N DP table to store the minimum costs for all subranges, requiring O(N^2) space.

Frequently Asked Questions

Q: What is the Minimax algorithm?

A: Minimax is a decision-making strategy used in game theory where two players have opposing goals. One player aims to minimize their maximum possible loss, while the other maximizes their gain. In this problem, you minimize your guessing cost, while the game rules (opponent) force you into the worst-case scenario.

Q: Why does the naive recursive approach cause timeouts?

A: The naive approach redundantly recalculates the same subproblems (e.g., solving for the same range [start, end] multiple times), leading to exponential time complexity.

Q: What is the main difference between memoization and tabulation?

A: Memoization uses a top-down approach (recursion) and caches results, while tabulation uses a bottom-up approach (iteration) to precompute all subproblems. Memoization avoids unnecessary subproblems but has recursion overhead, whereas tabulation computes all subproblems sequentially with no recursion.

Q: Why does the opponent always force the worst-case scenario?

A: The problem requires finding the minimum cost to guarantee a win, regardless of the chosen number. To ensure victory, you must assume the opponent (or game rules) will always steer you toward the most expensive path.

How to Prepare for Interviews with the Guess Number Higher or Lower II Problem

This problem assesses your ability to design and implement dynamic programming (DP) solutions and apply recursive thinking.

Tips for solving and preparing:

1. Master Minimax and Recursive Thinking: Understand the "I minimize, opponent maximizes" structure and express it recursively.

2. Identify and Optimize Redundancy: Recognize overlapping subproblems and explain how memoization/tabulation eliminates redundant computation.

3. Compare Top-down vs. Bottom-up: Discuss the pros/cons of memoization (flexibility) and tabulation (space efficiency).

4. Analyze Complexity Improvements: Clearly explain how DP reduces time complexity from O(2^N) to O(N^3).

5. Communicate Clearly and Handle Edge Cases: Walk through your logic step-by-step and handle base cases like start > end.

Similar Problems to Practice

• Predict the Winner

• Stone Game

• Difference Between Two Adjacent Numbers

How to review after solving this problem:

1. Collect problems where you minimize loss/maximize gain against an opponent and compare their DP structures.

2. Summarize the core solution in one sentence: "The minimum guaranteed cost for range [i, j] is the minimum of k + max(cost of left, cost of right) over all possible first guesses k."

3. Compare how the Minimax principle and DP state definitions apply to similar problems (e.g., Predict the Winner).

Coding Interview Prep Guide

• What you must know for real interviews:

  • Ability to model the problem using Minimax game theory and DP states.

  • Skill to formulate recurrence relations and implement them via memoization/tabulation.

  • Process to evaluate and improve efficiency through time/space complexity analysis.

• What you should be able to explain:

  • "Why is the cost for guessing k calculated as k + max(solve(start, k-1), solve(k+1, end))?" The opponent forces the worst-case scenario, so you must account for the more expensive subrange.

  • "What are the time/space complexities of pure recursion, memoization, and tabulation?" Pure recursion: O(2^N), Memoization: O(N^2), Tabulation: O(N^3).

  • "Why is the DP table filled in order of interval length?" Smaller subproblems must be solved before larger ones in bottom-up DP.

• Core principles of Minimax and DP:

  • This problem tests your ability to anticipate opponent moves via Minimax and optimize with DP for overlapping subproblems.

  • Observing patterns in small examples (e.g., n=3, n=4), generalizing to a recurrence, and optimizing with DP demonstrates strong problem-solving skills.