🚀Day 19. Stickler Thief II 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

You are given an array arr[] representing houses arranged in a circular fashion. Each house contains a certain value, and a thief aims to maximize the stolen amount while ensuring that no two adjacent houses are robbed.

Since the houses form a circle, the first and last house are also considered adjacent.

🔍 Example Walkthrough:

Example 1:

Input:

arr[] = [2, 3, 2]

Output:

3

Explanation:

The houses at index 0 and 2 are adjacent, so they cannot be robbed together. The maximum amount that can be stolen is 3 (robbing the house at index 1).

Example 2:

Input:

arr[] = [1, 2, 3, 1]

Output:

4

Explanation:

The optimal strategy is to rob houses at index 0 and 2, which gives 1 + 3 = 4.

Example 3:

Input:

arr[] = [2, 2, 3, 1, 2]

Output:

5

Explanation:

Possible choices:

1. Rob house at index 0 and 2 → 2 + 3 = 5

2. Rob house at index 2 and 4 → 3 + 2 = 5 The maximum amount that can be stolen is 5.

Constraints:

• $(2 \leq \text{arr.size()} \leq 10^5)$

• $(0 \leq \text{arr}[i] \leq 10^4)$

🎯 My Approach:

Optimized Dynamic Programming

1. Since the houses are arranged in a circle, the first and last houses cannot be robbed together.

2. The problem reduces to two linear House Robber problems:

   • Robbing from house 0 to n-2 (excluding the last house).

   • Robbing from house 1 to n-1 (excluding the first house).

3. We solve the House Robber problem for both cases using Dynamic Programming with space optimization.

4. The final answer is the maximum of both cases.

Algorithm Steps:

1. Define a helper function robRange(l, r):

   • Use two variables (prev1, prev2) to track maximum money stolen up to the previous two houses.

   • Iterate through the houses from l to r, updating the maximum amount stolen.

   • Return prev1 (max stolen amount).

2. Compute the result by taking the maximum of:

   • robRange(0, n-2) (excluding last house).

   • robRange(1, n-1) (excluding first house).

3. Edge case: If there's only one house, return its value.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(N), since we iterate through the houses twice.

• Expected Auxiliary Space Complexity: O(1), since we use only a few variables (constant space).

📝 Solution Code

Code (C++)

class Solution {
public:
    int maxValue(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return nums[0];

        auto robRange = [&](int l, int r) {
            int prev2 = 0, prev1 = 0;
            for (int i = l; i <= r; i++) {
                int curr = max(prev1, prev2 + nums[i]);
                prev2 = prev1;
                prev1 = curr;
            }
            return prev1;
        };

        return max(robRange(0, n - 2), robRange(1, n - 1));
    }
};

🛠 Alternative Solutions

2️⃣ Dynamic Programming with Array (O(N) Time, O(N) Space)

Approach:

Instead of using two variables, maintain a DP array where dp[i] stores the maximum amount that can be stolen up to house i.

Code (C++)

class Solution {
public:
    int maxValue(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return nums[0];

        auto robRange = [&](int l, int r) {
            vector<int> dp(r - l + 2, 0);
            dp[1] = nums[l];
            for (int i = l + 1; i <= r; i++) {
                dp[i - l + 1] = max(dp[i - l], dp[i - l - 1] + nums[i]);
            }
            return dp[r - l + 1];
        };

        return max(robRange(0, n - 2), robRange(1, n - 1));
    }
};

3️⃣ Memoization (Top-Down DP)

Approach:

Instead of using iterative DP, we use recursion with memoization:

1. Define a recursive function robRange(l, r, dp).

2. Use memoization (dp[i]) to avoid recomputation.

3. Solve for both cases:

   • Excluding the last house (robRange(0, n-2)).

   • Excluding the first house (robRange(1, n-1)).

4. Return the maximum of both cases.

Code (C++)

class Solution {
public:
    int robRange(vector<int>& nums, int i, int r, vector<int>& dp) {
        if (i > r) return 0;
        if (dp[i] != -1) return dp[i];
        return dp[i] = max(robRange(nums, i + 1, r, dp), nums[i] + robRange(nums, i + 2, r, dp));
    }

    int maxValue(vector<int>& nums) {
        int n = nums.size();
        if (n == 1) return nums[0];

        vector<int> dp1(n, -1), dp2(n, -1);
        return max(robRange(nums, 0, n - 2, dp1), robRange(nums, 1, n - 1, dp2));
    }
};

Comparison of Approaches

Approach	⏱️ Time Complexity	🗂️ Space Complexity	✅ Pros	⚠️ Cons
Optimized DP (Two Variables)	🟡 O(N)	🟢 O(1)	Space efficient, faster	Slightly harder to understand
DP with Array	🟡 O(N)	🟡 O(N)	Easier to implement	Extra space for DP array
Memoization (Top-Down DP)	🟡 O(N)	🔴 O(N)	Good for recursion lovers	Higher memory consumption

💡 Best Choice?

• ✅ Optimized DP (O(1) Space) is the best solution due to minimal space usage.

• ✅ DP with Array is useful for educational purposes, but not efficient for large inputs.

• ✅ Memoization? Useful for recursion-based approaches.

Code (Java)

class Solution {
    public int maxValue(int[] nums) {
        int n = nums.length;
        if (n == 1) return nums[0];

        return Math.max(rob(nums, 0, n - 2), rob(nums, 1, n - 1));
    }

    private int rob(int[] nums, int l, int r) {
        int prev2 = 0, prev1 = 0;
        for (int i = l; i <= r; i++) {
            int curr = Math.max(prev1, prev2 + nums[i]);
            prev2 = prev1;
            prev1 = curr;
        }
        return prev1;
    }
}

Code (Python)

class Solution:
    def maxValue(self, nums):
        if len(nums) == 1:
            return nums[0]

        def rob(l, r):
            prev2 = prev1 = 0
            for i in range(l, r + 1):
                prev2, prev1 = prev1, max(prev1, prev2 + nums[i])
            return prev1

        return max(rob(0, len(nums) - 2), rob(1, len(nums) - 1))

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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