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

Author: JawadSher

24 - Dynamic Programming Problems/07 - House Robber/02 - Top-Down | DP | Approach.cpp

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+class Solution {
+public:
+    // Function to solve the problem recursively with memoization
+    // `money`: vector of amounts of money in each house
+    // `n`: the current house index
+    // `dp`: memoization array to store the maximum money for each house index
+    int solve(vector<int>& money, int n, vector<int>& dp) {
+        // Base case: if the index `n` is less than 0, return 0 (no money left to rob)
+        if (n < 0) return 0;
+
+        // Base case: if we are at the first house (index 0), return the money at this house
+        if (n == 0) return money[0];
+
+        // If the result for this index `n` has already been calculated, return it from the `dp` array
+        if (dp[n] != -1) return dp[n];
+
+        // Option 1: Rob this house, so skip the next one (move to house `n-2`)
+        int include = solve(money, n - 2, dp) + money[n];
+
+        // Option 2: Skip this house, so move to the next house (house `n-1`)
+        int exclude = solve(money, n - 1, dp) + 0;
+
+        // Store the maximum of both choices in the memoization array `dp[n]`
+        dp[n] = max(include, exclude);
+
+        // Return the maximum money that can be robbed up to house `n`
+        return dp[n];
+    }
+
+    // Main function to calculate the maximum money that can be robbed
+    int rob(vector<int>& nums) {
+        int n = nums.size(); // Get the number of houses
+
+        // Initialize the dp array with -1 (indicating uncalculated values)
+        // dp[i] will store the maximum money that can be robbed from house 0 to house i
+        vector<int> dp(n + 1, -1);
+
+        // Call the helper function starting from the last house (index n-1)
+        int maxMoney = solve(nums, n - 1, dp);
+
+        // Return the maximum money that can be robbed
+        return maxMoney;
+    }
+};