Added House Robber II (Dynamic Programming, C++)

dynamic_programming/HouseRobberII.cpp

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+/*
+=========================================================
+🏠 Problem: House Robber II
+=========================================================
+🔗 LeetCode Link:
+https://leetcode.com/problems/house-robber-ii/
+
+📘 Problem Statement:
+You are a professional robber planning to rob houses along a street.
+Each house has a certain amount of money. All houses are arranged in a circle,
+which means the first and last houses are adjacent.
+You cannot rob two adjacent houses.
+
+Return the maximum amount of money you can rob tonight without alerting the police.
+
+---------------------------------------------------------
+🧠 Approach: Dynamic Programming (Tabulation)
+---------------------------------------------------------
+- Because the first and last houses are adjacent, we cannot include both.
+- So, we compute two scenarios:
+  1️⃣ Rob houses from index [0 ... n-2] (skip last)
+  2️⃣ Rob houses from index [1 ... n-1] (skip first)
+- Take the maximum of both cases.
+
+---------------------------------------------------------
+⏱️ Time Complexity:  O(n)
+💾 Space Complexity: O(1)
+---------------------------------------------------------
+📌 Points to Remember:
+- For circular arrangements, always consider two linear cases.
+- DP base cases are mandatory (at least one).
+- Transition: dp[i] = max(dp[i-1], dp[i-2] + nums[i])
+---------------------------------------------------------
+*/
+
+#include <bits/stdc++.h>
+using namespace std;
+
+class Solution {
+public:
+    int robLinear(vector<int>& nums, int start, int end){
+        int n = end - start + 1;  // we don’t always use the full array (so we can't write nums.size())
+
+        // Edge cases:
+        if(n == 0) return 0;
+        if(n == 1) return nums[start];
+
+        // Initialize the first two states
+        int prev2 = nums[start];  // dp[i-2]
+        int prev1 = max(nums[start], nums[start + 1]);  // dp[i-1]
+
+        // Iterate over the rest of the houses
+        for(int i = start + 2; i <= end; i++){
+            int curr = max(prev2 + nums[i], prev1);
+            prev2 = prev1;  // move the window forward
+            prev1 = curr;
+        }
+
+        return prev1;
+    }
+
+    int rob(vector<int>& nums){
+        int n = nums.size();
+
+        // Edge cases
+        if(n == 1) return nums[0];
+        if(n == 2) return max(nums[0], nums[1]);
+
+        // Case 1: Rob houses from index 0 to n-2 (exclude last)
+        int skipLast = robLinear(nums, 0, n - 2);
+
+        // Case 2: Rob houses from index 1 to n-1 (exclude first)
+        int skipFirst = robLinear(nums, 1, n - 1);
+
+        // Take maximum of both cases
+        return max(skipLast, skipFirst);
+    }
+};
+
+
+/*
+=========================================================
+✅ Example Test (Uncomment to run locally)
+=========================================================
+int main() {
+    Solution sol;
+    vector<int> nums = {2, 3, 2};
+    cout << "Maximum money robbed: " << sol.rob(nums) << endl;
+    return 0;
+}
+=========================================================
+
+Time complexity = O(n) and Space complexity = O(1) using Dynamic Programming (Tabulation approach)
+
+First & Last houses are adjacent so we can either skip first house or last
+Solve for both separately and then return 'maximum' of it
+
+Thumb rule of DP:
+    - We always need 'at least one' base case (bcz DP is recurrence based)
+    - Always >= 1 base cases
+
+If absolute indices --> i<=end
+If relative indices --> i<n (i = 2 ... n-1)
+=========================================================
+*/