Merge pull request #171 from Abe0770/master

Added Leetcode Solution for Q-213

Cpp/0213-house-robber-II/problem.md

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+<h1><a href = "https://leetcode.com/problems/house-robber-ii/description/">213. House Robber II</h1>
+
+## Problem Statement
+
+You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses have a security system connected, and it will automatically contact the police if two adjacent houses are broken into on the same night.
+
+Given an integer array `nums` representing the amount of money of each house, return the maximum amount of money you can rob tonight without alerting the police.
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<pre>
+<strong>Input: </strong> nums = [2,3,2]
+<strong>Output:</strong> 3
+<strong>Explanation:</strong> You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), 
+because they are adjacent houses.
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<pre>
+<strong>Input:</strong> nums = [1,2,3,1]
+<strong>Output:</strong> 4
+<strong>Explanation:</strong> Rob house 1 (money = 1) and then rob house 3 (money = 3). 
+Total amount you can rob = 1 + 3 = 4.
+</pre>
+
+<p><strong class="example">Example 3:</strong></p>
+<pre>
+<strong>Input:</strong> nums = [1,2,3]
+<strong>Output:</strong> 3
+</pre>
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+- `1 <= nums.length <= 100`
+- `0 <= nums[i] <= 1000`

Cpp/0213-house-robber-II/solution.cpp

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+class Solution {
+public:
+    int rob(vector<int>& nums) {
+        int n = nums.size();
+
+        // Handle edge cases
+        if (n == 0) return 0; // No houses
+        if (n == 1) return nums[0]; // Only one house
+
+        // Helper function to calculate max loot for a linear array
+        auto robLinear = [](const vector<int>& nums, int start, int end) {
+            int prev1 = 0, prev2 = 0; // Initialize previous two max values
+            for (int i = start; i < end; ++i) {
+                int pick = nums[i] + prev1; // Value if we pick current house
+                int notPick = prev2; // Value if we do not pick current house
+                int curr = max(pick, notPick); // Max of both options
+                prev1 = prev2; // Update previous for the next iteration
+                prev2 = curr;
+            }
+            return prev2; // Return the maximum value obtained
+        };
+
+        // Rob houses excluding the last house and then excluding the first house
+        int maxLoot1 = robLinear(nums, 0, n - 1); // From house 0 to house n-2
+        int maxLoot2 = robLinear(nums, 1, n); // From house 1 to house n-1
+
+        // Return the maximum of both scenarios
+        return max(maxLoot1, maxLoot2);
+    }
+};