Create 02 - Top-Down | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/29 - Guess Number Higher or Lower II/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Recursive function to calculate the minimum money required to guarantee a win
+    // for the range [start, end]. We use dynamic programming to memoize the results.
+    int solve(int start, int end, vector<vector<int>>& dp) {
+        // Base case: If start is greater than or equal to end, no money is needed.
+        if (start >= end) return 0;
+
+        // Check if the result is already computed (memoized). If so, return it.
+        if (dp[start][end] != -1) return dp[start][end];
+
+        // Initialize maxi with a very large value to store the minimum amount of money.
+        int maxi = INT_MAX;
+
+        // Try every number i in the range [start, end] as a potential guess.
+        for (int i = start; i <= end; i++) {
+            // Calculate the cost of guessing i:
+            // - i is the cost of guessing i.
+            // - The two ranges are [start, i-1] and [i+1, end].
+            // We want to minimize the worst-case scenario, so we take the maximum
+            // of the two subranges and add the cost of guessing i.
+            maxi = min(maxi, i + max(solve(start, i - 1, dp), solve(i + 1, end, dp)));
+        }
+
+        // Memoize the result for this range [start, end] and return the calculated value.
+        return dp[start][end] = maxi;
+    }
+
+    // Function to initiate the process for the full range [1, n]
+    int getMoneyAmount(int n) {
+        // Create a 2D DP table to store the minimum money required for each range.
+        // Initialize all values to -1 to signify that they are not computed yet.
+        vector<vector<int>> dp(n + 1, vector<int>(n + 1, -1));
+
+        // Start solving for the range [1, n]
+        return solve(1, n, dp);
+    }
+};