Create 04 - Space Optimized | DP | Approach.cpp

Author: JawadSher

24 - Dynamic Programming Problems/07 - House Robber/04 - Space Optimized | DP | Approach.cpp

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+class Solution {
+public:
+    // Function to solve the problem using dynamic programming (optimized space approach)
+    // `money`: vector containing the amounts of money in each house
+    // `n`: the number of houses
+    int solve(vector<int>& money, int n) {
+        // Base cases:
+        // prev2 represents the maximum money robbed from houses up to index 0
+        // prev1 represents the maximum money robbed from houses up to index 1
+        int prev2 = money[0];  // Only one house, so rob it
+        int prev1 = max(money[0], money[1]);  // Take the maximum of robbing the first or second house
+
+        // Iterate through the remaining houses starting from index 2
+        for (int i = 2; i < n; i++) {
+            // Option 1: Rob the current house (house i), and add money from house i to the result of robbing house i-2 (prev2)
+            int include = prev2 + money[i];
+
+            // Option 2: Skip the current house (house i), and keep the maximum from the previous house (prev1)
+            int exclude = prev1 + 0;
+
+            // Calculate the maximum money we can rob up to house i by choosing either to include or exclude house i
+            int current = max(include, exclude);
+
+            // Update prev2 and prev1 for the next iteration
+            prev2 = prev1;  // Move prev1 to prev2 (we are shifting one step to the right)
+            prev1 = current;  // Update prev1 to be the maximum of robbing up to house i
+        }
+
+        // After iterating through all houses, prev1 will hold the maximum money that can be robbed
+        return prev1;
+    }
+
+    // Main function to start solving the problem
+    int rob(vector<int>& nums) {
+        int n = nums.size(); // Get the number of houses
+
+        // Handle edge cases where there are no houses or only one house
+        if (n == 0) return 0; // No houses to rob
+        if (n == 1) return nums[0]; // Only one house to rob, return its value
+
+        // Call the helper function to calculate the maximum money that can be robbed
+        return solve(nums, n);
+    }
+};