From 50579e53ce2a27fac07b071b1e594bb0255ee86a Mon Sep 17 00:00:00 2001
From: Hiro Funatsuka <hiro.funatsuka@gmail.com>
Date: Thu, 7 May 2026 22:21:11 +0900
Subject: [PATCH] Solve house robber ii

---
 213_house_robber_ii/problem.md |  1 +
 213_house_robber_ii/step1.md   | 29 +++++++++++++++++++++++++++++
 213_house_robber_ii/step2.md   | 26 ++++++++++++++++++++++++++
 213_house_robber_ii/step3.md   | 28 ++++++++++++++++++++++++++++
 4 files changed, 84 insertions(+)
 create mode 100644 213_house_robber_ii/problem.md
 create mode 100644 213_house_robber_ii/step1.md
 create mode 100644 213_house_robber_ii/step2.md
 create mode 100644 213_house_robber_ii/step3.md

diff --git a/213_house_robber_ii/problem.md b/213_house_robber_ii/problem.md
new file mode 100644
index 0000000..babd4b9
--- /dev/null
+++ b/213_house_robber_ii/problem.md
@@ -0,0 +1 @@
+## 問題: [213 House Robber II](https://leetcode.com/problems/house-robber-ii/description/)
diff --git a/213_house_robber_ii/step1.md b/213_house_robber_ii/step1.md
new file mode 100644
index 0000000..0c87640
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+++ b/213_house_robber_ii/step1.md
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+# Step 1
+
+- $i$番目まで訪れた時の最大利益$maxAmount[i]$は$max(maxAmount[i - 1], maxAmount[i - 2] + nums[i])$
+- 円形のため、$0\le i \le nums.size -2$の範囲の最大利益と$1\le i \le nums.size -1$の範囲の最大利益のうち大きい方を返す
+
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) <= 2:
+            return max(nums)
+        first_max_amount_prev_2 = nums[0]
+        first_max_amount_prev_1 = max(nums[0], nums[1])
+        for i in range(2, len(nums) - 1):
+            current_max_amount = max(first_max_amount_prev_1, first_max_amount_prev_2 + nums[i])
+            first_max_amount_prev_2 = first_max_amount_prev_1
+            first_max_amount_prev_1 = current_max_amount
+
+        second_max_amount_prev_2 = nums[1]
+        second_max_amount_prev_1 = max(nums[1], nums[2])
+        for i in range(3, len(nums)):
+            current_max_amount = max(second_max_amount_prev_1, second_max_amount_prev_2 + nums[i])
+            second_max_amount_prev_2 = second_max_amount_prev_1
+            second_max_amount_prev_1 = current_max_amount
+        return max(first_max_amount_prev_1, second_max_amount_prev_1)
+```
+
+時間計算量: $O(n)$
+
+空間計算量: $O(1)$
diff --git a/213_house_robber_ii/step2.md b/213_house_robber_ii/step2.md
new file mode 100644
index 0000000..8ac66a7
--- /dev/null
+++ b/213_house_robber_ii/step2.md
@@ -0,0 +1,26 @@
+# Step 2
+
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) <= 2:
+            return max(nums)
+
+        def rob_linear(start, end):
+            profit_two_back = nums[start]
+            profit_one_back = max(nums[start], nums[start + 1])
+
+            for i in range(start + 2, end):
+                current_profit = max(profit_one_back, profit_two_back + nums[i])
+                profit_two_back = profit_one_back
+                profit_one_back = current_profit
+
+            return profit_one_back
+
+        return max(rob_linear(0, len(nums) - 1), rob_linear(1, len(nums)))
+
+```
+
+時間計算量: $O(n)$
+
+空間計算量: $O(1)$
diff --git a/213_house_robber_ii/step3.md b/213_house_robber_ii/step3.md
new file mode 100644
index 0000000..5ff7786
--- /dev/null
+++ b/213_house_robber_ii/step3.md
@@ -0,0 +1,28 @@
+# Step 3
+
+```python
+class Solution:
+    def rob(self, nums: List[int]) -> int:
+        if len(nums) <= 2:
+            return max(nums)
+
+        def rob_linear(start, end):
+            profit_two_back = nums[start]
+            profit_one_back = max(nums[start], nums[start + 1])
+
+            for i in range(start + 2, end):
+                current_profit = max(profit_one_back, profit_two_back + nums[i])
+                profit_two_back = profit_one_back
+                profit_one_back = current_profit
+
+            return profit_one_back
+
+        return max(rob_linear(0, len(nums) - 1), rob_linear(1, len(nums)))
+
+```
+
+1回目: 2分　57秒
+
+2回目: 2分　18秒
+
+3回目: 2分　14秒