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This approach involves using dynamic programming with recursion to solve the problem. The main idea is to break the problem into subproblems: determining the cost of guessing any number within a given range. For each number in the current range, calculate the worst-case cost of guessing that number, accounting for both possibilities where the picked number is less or greater than the guessed number. Store solutions to subproblems to avoid recalculating them.
Time Complexity: O(n^3) due to the three nested loops through possible ranges and guesses.
Space Complexity: O(n^2) for storing results of subproblems.
1class Solution:
2 def getMoneyAmount(self, n: int) -> int:
3 dp = [[0] * (n+1) for _ in range(n+1)]
4 for length in range(2, n+1):
5 for start in range(1, n-length+2):
6 min_cost = float('inf')
7 for pivot in range(start, start+length-1):
8 cost = pivot + max(dp[start][pivot-1] if pivot > start else 0,
9 dp[pivot+1][start+length-1] if pivot < start+length-1 else 0)
10 min_cost = min(min_cost, cost)
11 dp[start][start+length-1] = min_cost
12 return dp[1][n]
13
14if __name__ == "__main__":
15 sol = Solution()
16 print("Minimum money needed: ", sol.getMoneyAmount(10))The Python solution closely follows the dynamic programming paradigm. The function getMoneyAmount initializes a 2D array to hold the minimum expenses for each subproblem. It iterates over all possible intervals and pivot choices, updating the dp table with minimized worst-case scenario costs. The function ultimately returns the answer at dp[1][n].
A more advanced recursive approach with memoization achieves similar results, storing results of previously solved subproblems in a cache for fast retrieval. This is especially useful for large inputs, reducing computational overhead by avoiding repeated calculations of the same conditions and ranges.
Time Complexity: O(n^2), reduced with memoization.
Space Complexity: O(n^2), for memoizing results.
1 private int[,] dp;
private int MinCost(int start, int end) {
if (start >= end) return 0;
if (dp[start, end] != 0) return dp[start, end];
int res = int.MaxValue;
for (int piv = start; piv <= end; piv++) {
int cost = piv + Math.Max(MinCost(start, piv - 1), MinCost(piv + 1, end));
res = Math.Min(res, cost);
}
dp[start, end] = res;
return res;
}
public int GetMoneyAmount(int n) {
dp = new int[n+1, n+1];
return MinCost(1, n);
}
public static void Main(string[] args) {
Solution sol = new Solution();
System.Console.WriteLine("Minimum amount to guarantee a win: " + sol.GetMoneyAmount(10));
}
}This C# implementation utilizes recursive memoization in its code structure. It efficiently calculates required costs for all possible number ranges by storing intermediate results and retrieving them as needed, enhancing the overall speed and reducing redundant processing.