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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.
1function getMoneyAmount(n) {
2 const dp = Array.from({length: n+1}, () => Array(n+1).fill(0));
3
4 for (let len = 2; len <= n; len++) {
5 for (let start = 1; start <= n - len + 1; start++) {
6 const end = start + len - 1;
7 let minCost = Infinity;
8 for (let piv = start; piv <= end; piv++) {
9 const cost = piv + Math.max(
10 piv - 1 >= start ? dp[start][piv - 1] : 0,
11 piv + 1 <= end ? dp[piv + 1][end] : 0
12 );
13 minCost = Math.min(minCost, cost);
14 }
15 dp[start][end] = minCost;
16 }
17 }
18
19 return dp[1][n];
20}
21
22console.log(`Minimum money needed: ${getMoneyAmount(10)}`);This JavaScript solution uses a 2D array dp to determine the minimum cost of guaranteeing a win. It calculates costs for increasing intervals, evaluating all possible guesses for each, and storing best-worst case costs. The console statement at the end prints the minimum amount required, concluding this dynamically programmed approach.
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
In this Python approach, memoization is achieved using an instance class variable dp initialized to zeros. The recursive function ensures subproblems are evaluated only once per unique input values, avoiding overlapping calculations and thus improving efficiency.