grid47 Exploring patterns and algorithms
Sep 30, 2024
6 min read
Solution to LeetCode 375: Guess Number Higher or Lower II Problem
In this game, you must guess a number between 1 and n. Each wrong guess costs you the amount of the guessed number. Your goal is to minimize the total cost while guaranteeing a win. If you run out of money, you lose the game.
Problem
Approach
Steps
Complexity
Input: The input consists of a single integer n, which represents the maximum number in the guessing range.
Example: Input: n = 8
Constraints:
• 1 <= n <= 200
Output: Return the minimum amount of money you need to guarantee a win, regardless of what number is picked.
Example: Output: 12
Constraints:
• The solution should minimize the total cost of guesses while guaranteeing a win.
Goal: The goal is to determine the minimum cost by choosing optimal guesses and minimizing the maximum cost in the worst case.
Steps:
• 1. Use dynamic programming to calculate the minimum cost for each possible range of numbers.
• 2. For each number, calculate the maximum cost of guessing that number and continue with the next number based on whether the actual number is higher or lower.
• 3. Memoize the results to avoid redundant calculations and improve efficiency.
Goal: The problem requires efficient computation due to the potential range of n up to 200.
Steps:
• 1 <= n <= 200
Assumptions:
• The number picked will always be between 1 and n.
• Input: Input: n = 8
Output: 12
• Explanation: The optimal strategy involves guessing numbers in such a way that minimizes the worst-case cost, ensuring that you can win without spending too much.
• Input: Input: n = 3
Output: 2
• Explanation: With only three numbers, guessing 2 minimizes the maximum cost, which is 2.
Approach: We use dynamic programming to find the minimum cost for each range of numbers from 1 to n. The key is to calculate the cost for each possible number as a guess and then find the optimal strategy.
Observations:
• The game is a classic example of decision-making under uncertainty, where we aim to minimize the worst-case outcome.
• Using dynamic programming, we can break the problem into smaller subproblems, ensuring that we compute the optimal cost efficiently.
Steps:
• 1. Initialize a 2D table for memoization, where dp[i][j] represents the minimum cost to guess in the range [i, j].
• 2. Iterate over all possible ranges and for each number within the range, calculate the cost of guessing that number.
• 3. Store the results in the dp table to avoid redundant calculations.
Empty Inputs:
• n is always between 1 and 200, so there are no cases with empty inputs.
Large Inputs:
• The algorithm needs to efficiently handle cases where n is large, up to 200.
Special Values:
• If n = 1, the result is 0 because no guessing is needed.
Constraints:
• The solution must work within the constraints of n <= 200.
classSolution {
vector<vector<int>> table;
public:int getMoneyAmount(int n) {
table.resize(n +1, vector<int>(n +1));
returndpf(table, 1, n);
}
intdpf(vector<vector<int>>&dp, int s, int e) {
if(s >= e) return0;
if(dp[s][e] >0) return dp[s][e];
int res = INT_MAX;
for(int x = s; x <= e; x++) {
int tmp = (x + max(dpf(dp, s, x-1), dpf(dp, x +1, e)));
res = min(res, tmp);
}
return dp[s][e] = res;
}
1 : Class Definition
classSolution {
This line begins the definition of the `Solution` class, which will contain the methods needed to solve the problem.
2 : Variable Declaration
vector<vector<int>> table;
This declares a 2D vector `table` which will be used to store the intermediate results of the dynamic programming computation.
3 : Access Specifier
public:
The `public` access specifier marks the methods and variables that follow as accessible from outside the class.
4 : Function Declaration
intgetMoneyAmount(int n) {
This is the function declaration for `getMoneyAmount`, which computes the minimum money required to guarantee a win in a guessing game, taking `n` (the maximum number) as input.
5 : Dynamic Programming Table Initialization
table.resize(n +1, vector<int>(n +1));
The dynamic programming table `table` is resized to store values for all ranges of numbers from 1 to `n`, with each element initialized to 0.
6 : Function Call
returndpf(table, 1, n);
The function `dpf` is called with the parameters `table`, `1`, and `n` to compute the minimum cost of the game for the range `[1, n]`.
7 : Function Declaration
intdpf(vector<vector<int>>&dp, int s, int e) {
This is the declaration of the `dpf` function, which takes the dynamic programming table and the range `[s, e]` to compute the minimum money needed to guess within that range.
8 : Base Case
if(s >= e) return0;
The base case: if the start value `s` is greater than or equal to the end value `e`, no more guesses are needed, so the cost is 0.
9 : Memoization Check
if(dp[s][e] >0) return dp[s][e];
If the result for the range `[s, e]` has already been computed and stored in `dp`, it is returned to avoid redundant calculations (memoization).
10 : Variable Initialization
int res = INT_MAX;
A variable `res` is initialized to the maximum integer value, representing the minimum cost to guess for the range `[s, e]`.
11 : Loop Iteration
for(int x = s; x <= e; x++) {
This loop iterates over each possible guess `x` within the range `[s, e]`.
12 : Recursive Function Call
int tmp = (x + max(dpf(dp, s, x-1), dpf(dp, x +1, e)));
For each guess `x`, the function `dpf` is recursively called to compute the maximum cost for the left and right subranges (`[s, x-1]` and `[x+1, e]`). The result is the sum of `x` and the maximum of the two subranges.
13 : Result Update
res = min(res, tmp);
The variable `res` is updated to store the minimum result across all possible guesses `x`.
14 : Memoization Update
return dp[s][e] = res;
The computed minimum result for the range `[s, e]` is stored in the dynamic programming table `dp` for future use.
Best Case: O(n^2)
Average Case: O(n^2)
Worst Case: O(n^3)
Description: In the worst case, we perform calculations for each pair of numbers and then evaluate each guess within the range, leading to a cubic time complexity.
Best Case: O(n^2)
Worst Case: O(n^2)
Description: We use a 2D table to store the results for each pair of numbers, resulting in a space complexity of O(n^2).
