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 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 were 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.

• Example:

  Example 1:
  
  Input: nums = [2,3,2]
  Output: 3
  Explanation: You cannot rob house 1 (money = 2) and then rob house 3 (money = 2), because they are adjacent houses.
  Example 2:
  
  Input: nums = [1,2,3,1]
  Output: 4
  Explanation: Rob house 1 (money = 1) and then rob house 3 (money = 3).
  Total amount you can rob = 1 + 3 = 4.
  Example 3:
  
  Input: nums = [1,2,3]
  Output: 3

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Solution: Space Optimized

int solve(vector<int> &nums) {
    int n = nums.size();
    if(n == 0) return 0;

    long long prev = nums[0], prev2 = 0;

    for(int i = 1; i < n; i++) {
        long long pick = nums[i] + prev2;
        long long nonPick = prev;
        long long curr_i = max(pick, nonPick);
        prev2 = prev;
        prev = curr_i;
    }

    return prev;
}

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

    vector<int> arr1(nums.begin() + 1, nums.end());     // exclude first
    vector<int> arr2(nums.begin(), nums.end() - 1);     // exclude last

    return max(solve(arr1), solve(arr2));
}

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Solution: Memoization (Top-Down DP)

int houseRobMemo(int i, vector<int>& nums, vector<int>& dp) {
    if(i == 0) return nums[i];
    if(i < 0) return 0;
    if(dp[i] != -1) return dp[i];

    int pick = nums[i] + houseRobMemo(i - 2, nums, dp);
    int nonPick = houseRobMemo(i - 1, nums, dp);

    return dp[i] = max(pick, nonPick);
}

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

    vector<int> arr1(nums.begin() + 1, nums.end());
    vector<int> arr2(nums.begin(), nums.end() - 1);

    vector<int> dp1(arr1.size(), -1);
    vector<int> dp2(arr2.size(), -1);

    int ans1 = houseRobMemo(arr1.size() - 1, arr1, dp1);
    int ans2 = houseRobMemo(arr2.size() - 1, arr2, dp2);

    return max(ans1, ans2);
}

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Solution: Tabulation (Bottom-Up DP)

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

    vector<int> dp(n);
    dp[0] = nums[0];
    dp[1] = max(nums[0], nums[1]);

    for(int i = 2; i < n; i++) {
        dp[i] = max(dp[i - 1], nums[i] + dp[i - 2]);
    }

    return dp[n - 1];
}

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

    vector<int> arr1(nums.begin() + 1, nums.end());
    vector<int> arr2(nums.begin(), nums.end() - 1);

    return max(houseRobTab(arr1), houseRobTab(arr2));
}

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✅ How It Works

• You cannot rob both first and last house due to the circular arrangement.
• So divide the array into two linear parts:
  • Exclude first house → [1...n-1]
  • Exclude last house → [0...n-2]
• Apply House Robber I logic (non-adjacent max sum) on both.
• Return the maximum of the two results.

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🧠 Key Points

• If you rob house 0, you cannot rob n-1 (and vice versa).
• Final answer = max(rob(1 to n-1), rob(0 to n-2))
• Use space optimization to reduce from O(n) → O(1)
• Handle n == 0 and n == 1 explicitly.

⏱️ Time & Space Complexity

Approach	Time	Space
Memoization	O(n)	O(n)
Tabulation	O(n)	O(n)
Space Optimized ✅	O(n)	O(1)

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⚠️ Edge Cases

• n == 0 → return 0
• n == 1 → return nums[0]
• All elements are 0
• Very large values → use long long to avoid overflow

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💡 Other Approaches

Approach	Time	Space
Recursion (TLE) ❌	2ⁿ	O(1)
Memoization ✅	O(n)	O(n)
Tabulation ✅	O(n)	O(n)
Space Optimization ✅	O(n)	O(1)

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🔁 Related Problems

• House Robber I
• Frog Jump
• Maximum Non-Adjacent Subset Sum
• Paint House / Cost Minimization DP