From 5ee7aa0cf96bf8eb26b31653f41cff9d12b7bb88 Mon Sep 17 00:00:00 2001
From: Rytxxx <k.ryuto.5@icloud.com>
Date: Sun, 30 Nov 2025 16:59:01 -0800
Subject: [PATCH] 153-find-minimum-in-rotated-sorted-array

---
 .../memo.md                                   | 56 +++++++++++++++++++
 .../step1.py                                  | 11 ++++
 .../step2.py                                  | 15 +++++
 .../step3.py                                  | 14 +++++
 4 files changed, 96 insertions(+)
 create mode 100644 153-find-minimum-in-rotated-sorted-array/memo.md
 create mode 100644 153-find-minimum-in-rotated-sorted-array/step1.py
 create mode 100644 153-find-minimum-in-rotated-sorted-array/step2.py
 create mode 100644 153-find-minimum-in-rotated-sorted-array/step3.py

diff --git a/153-find-minimum-in-rotated-sorted-array/memo.md b/153-find-minimum-in-rotated-sorted-array/memo.md
new file mode 100644
index 0000000..3cd965b
--- /dev/null
+++ b/153-find-minimum-in-rotated-sorted-array/memo.md
@@ -0,0 +1,56 @@
+## 問題文
+Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:
+
+[4,5,6,7,0,1,2] if it was rotated 4 times.
+[0,1,2,4,5,6,7] if it was rotated 7 times.
+Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].
+
+Given the sorted rotated array nums of unique elements, return the minimum element of this array.
+
+You must write an algorithm that runs in O(log n) time.
+
+ 
+
+Example 1:
+
+Input: nums = [3,4,5,1,2]
+Output: 1
+Explanation: The original array was [1,2,3,4,5] rotated 3 times.
+Example 2:
+
+Input: nums = [4,5,6,7,0,1,2]
+Output: 0
+Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.
+Example 3:
+
+Input: nums = [11,13,15,17]
+Output: 11
+Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 
+
+## step1で考えたこと
+1回rotateすると、一番後ろのnumberが先頭に来て、それ以降が一つずれる
+1ずつIndexをずらすには？
+Binary Searchを使った解法が思いつかない
+
+return the minimum element of this array.
+-> この部分に関してはsortしたListの先頭をただreturnすれば良いのでは？
+
+## step1でのコード
+```py
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        sorted_nums = sorted(nums)
+        min = sorted_nums[0]
+        rotate_time = 0
+
+        while nums != sorted_nums:
+            sorted_nums = 
+            rotate_time += 1
+        
+        return min
+
+```
+
+## step2での気づき
+やはり、大小を比較するときに、=を含めるか否かが判断できていない。
+returnの位置を間違えていた。
\ No newline at end of file
diff --git a/153-find-minimum-in-rotated-sorted-array/step1.py b/153-find-minimum-in-rotated-sorted-array/step1.py
new file mode 100644
index 0000000..1318552
--- /dev/null
+++ b/153-find-minimum-in-rotated-sorted-array/step1.py
@@ -0,0 +1,11 @@
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        sorted_nums = sorted(nums)
+        min = sorted_nums[0]
+        rotate_time = 0
+
+        while nums != sorted_nums:
+            sorted_nums = 
+            rotate_time += 1
+        
+        return min
diff --git a/153-find-minimum-in-rotated-sorted-array/step2.py b/153-find-minimum-in-rotated-sorted-array/step2.py
new file mode 100644
index 0000000..f877568
--- /dev/null
+++ b/153-find-minimum-in-rotated-sorted-array/step2.py
@@ -0,0 +1,15 @@
+## 解答を見た後に自力で書いたコード
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        left = 0
+        right = len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if nums[mid] < nums[right]:
+                right = mid
+            else:
+                left = mid + 1
+
+            return nums[left]
\ No newline at end of file
diff --git a/153-find-minimum-in-rotated-sorted-array/step3.py b/153-find-minimum-in-rotated-sorted-array/step3.py
new file mode 100644
index 0000000..1e51b7f
--- /dev/null
+++ b/153-find-minimum-in-rotated-sorted-array/step3.py
@@ -0,0 +1,14 @@
+class Solution:
+    def findMin(self, nums: List[int]) -> int:
+        left = 0
+        right = len(nums) - 1
+
+        while left < right:
+            mid = (left + right) // 2
+
+            if nums[mid] < nums[right]:
+                right = mid
+            else:
+                left = mid + 1
+
+        return nums[left]
\ No newline at end of file