From c96298f60ce36bec638b8840e536f1061f656722 Mon Sep 17 00:00:00 2001 From: KentaroJay Date: Sat, 13 Sep 2025 12:07:28 +0900 Subject: [PATCH] Solved the problem --- find-minimum-in-rotated-sorted-array/memo.md | 221 ++++++++++++++++++ find-minimum-in-rotated-sorted-array/step1.py | 25 ++ find-minimum-in-rotated-sorted-array/step2.py | 20 ++ find-minimum-in-rotated-sorted-array/step3.py | 18 ++ 4 files changed, 284 insertions(+) create mode 100644 find-minimum-in-rotated-sorted-array/memo.md create mode 100644 find-minimum-in-rotated-sorted-array/step1.py create mode 100644 find-minimum-in-rotated-sorted-array/step2.py create mode 100644 find-minimum-in-rotated-sorted-array/step3.py diff --git a/find-minimum-in-rotated-sorted-array/memo.md b/find-minimum-in-rotated-sorted-array/memo.md new file mode 100644 index 0000000..d6a1633 --- /dev/null +++ b/find-minimum-in-rotated-sorted-array/memo.md @@ -0,0 +1,221 @@ +## ステップ 1 + +問題文でアルゴリズムの時間計算量を `O(log n)` にするよう求められているため、 +解法は二分探索的なアプローチに関係しているはずです。 + +* 実験 + +[4,5,6,7,0,1,2] + ^l ^mid ^r + +もし `mid` > `r` なら、`l` = `mid` + +[4,5,6,7,0,1,2] + ^l^m ^r + +もし `mid` < `r` なら、`r` = `mid` + +[4,5,6,7,0,1,2] + ^l^r + ^mid + +終了条件：`l` > `r` を満たす隣接する `l` と `r` を見つける。 + +問題文では配列内のすべての要素が一意であると述べられていますが、 +要素が同一の場合についても考えました。 + +## ステップ 2 + +### 参考資料 + +* [https://github.com/thonda28/leetcode/pull/7/files](https://github.com/thonda28/leetcode/pull/7/files) + * `left, right = -1, len(nums) - 1` の背後にある考え方がわかりません。 + * left より前はすべて配列の大きい部分で、 + right より後はすべて配列の小さい部分であるという考え方は理解できます。 + * しかし、配列が最小値から始まるときに起こる問題を + `left = -1` を割り当てることでなぜ修正できるのか理解できません。 + （または、`left = 0` に設定したときに問題が発生する場合に、 + `left = -1` に設定するというアイデアをどのように思いつくかがわからないというべきか） + * インデックス0が大きい部分でないときに `left = 0` に設定するのが + 間違っていることには気づけると思いますが、`left = -1` に設定すれば + それが修正されることには自分では気づけないと思いました。 + + ```python + class Solution: + def findMin(self, nums: List[int]) -> int: + left = 0 # `nums` が最小値から始まる場合にパスしない + right = len(nums) - 1 + while right - left > 1: + mid = (left + right) // 2 + if nums[mid] > nums[right]: + left = mid + else: + right = mid + return nums[right] + ``` + +* [https://github.com/seal-azarashi/leetcode/pull/39/files](https://github.com/seal-azarashi/leetcode/pull/39/files) + * `middle` の値を除外できるので、 + `left = middle + 1` とできることに気づきました + +全体的に、`left` と `right` の間の範囲が閉区間か開区間かを +考慮していなかったことがわかりました。 + +また、与えられた配列が空の場合について考えることができませんでした。 +そのため、そのためのロジックも追加しました。 + +その他可能な代替解法： + +```python +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[mid] <= nums[-1]: + right = mid + else: + left = mid + 1 + return nums[left] +``` + +```python +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[right] < nums[mid]: + left = mid + 1 + if nums[left] < nums[right]: + return nums[left] + elif nums[mid] < nums[right]: + right = mid + else: # nums[mid] == nums[right] + # 与えられた配列に重複がある + raise ValueError + return nums[left] +``` + +## ステップ3 + +問題を2分50秒で解きました。 + +--- + +## step 1 + +Since the problem description says to write an algorithm so that its +time complexity is `O(log n)`, the solution must have something to +do with binary search-ish approach. + +* Experiment + +[4,5,6,7,0,1,2] + ^l ^mid ^r + +if `mid` > `r`, `l` = `mid` + +[4,5,6,7,0,1,2] + ^l^m ^r + +if `mid` < `r`, `r` = `mid` + +[4,5,6,7,0,1,2] + ^l^r + ^mid + +Termination condition: find adjacent `l` and `r` that meets `l` > `r`. + +I also thought about, even though the problem description states +that all items in the array are unique, a case where items are identical. + +## step 2 + +### Referrence + +* [https://github.com/thonda28/leetcode/pull/7/files](https://github.com/thonda28/leetcode/pull/7/files) + * I don't know the idea behind `left, right = -1, len(nums) - 1`. + * I get the idea that + anything before left is the larger part of the array and + anything after right is the smaller part of the array. + * However, I don't understand why assigning `left = -1` fix the + problem that happens when the array starts with the smallest value. + (Or I should say how to come up with the idea to set `left = -1` + when the problem happens if you have set `left = 0`) + * I think I can notice that setting `left = 0` when index 0 is not + the larger part is wrong, but I don't think I can notice setting + `left = -1` would fix it. + + ```python + class Solution: + def findMin(self, nums: List[int]) -> int: + left = 0 # does not pass when `nums` starts with the smallest value + right = len(nums) - 1 + while right - left > 1: + mid = (left + right) // 2 + if nums[mid] > nums[right]: + left = mid + else: + right = mid + return nums[right] + ``` + +* [https://github.com/seal-azarashi/leetcode/pull/39/files](https://github.com/seal-azarashi/leetcode/pull/39/files) + * I realized I could do `left = middle + 1` + since I can exclude the value at `middle` + +Overall, I understand that I did not concern about if the range +between `left` and `right` is closed space or open. + +Also, I failed to think about when the given array is empty. So I added +a logic for that as well. + +Possible alternative solutions are: + +```python +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[mid] <= nums[-1]: + right = mid + else: + left = mid + 1 + return nums[left] +``` + +```python +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[right] < nums[mid]: + left = mid + 1 + if nums[left] < nums[right]: + return nums[left] + elif nums[mid] < nums[right]: + right = mid + else: # nums[mid] == nums[right] + # There is a duplication in the given array + raise ValueError + return nums[left] +``` + +## step3 + +Solved the problem in 2 mins 50 secs. diff --git a/find-minimum-in-rotated-sorted-array/step1.py b/find-minimum-in-rotated-sorted-array/step1.py new file mode 100644 index 0000000..f3ce47a --- /dev/null +++ b/find-minimum-in-rotated-sorted-array/step1.py @@ -0,0 +1,25 @@ +from typing import List + + +class Solution: + def findMin(self, nums: List[int]) -> int: + left, right = 0, len(nums) - 1 + while right - left >= 1: + if right - left == 1 and nums[right] < nums[left]: + left = right + continue + mid = (left + right) // 2 + if nums[right] < nums[mid]: + left = mid + elif nums[mid] < nums[right]: + right = mid + else: # nums[mid] == nums[right] + # There is a duplication in the given array + raise ValueError + return nums[left] + + +if __name__ == "__main__": + solution = Solution() + assert solution.findMin([0,1,2,4,5,6,7]) == 0 + diff --git a/find-minimum-in-rotated-sorted-array/step2.py b/find-minimum-in-rotated-sorted-array/step2.py new file mode 100644 index 0000000..72bbbb0 --- /dev/null +++ b/find-minimum-in-rotated-sorted-array/step2.py @@ -0,0 +1,20 @@ +from typing import List + + +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[right] < nums[mid]: + left = mid + 1 + elif nums[mid] < nums[right]: + right = mid + else: # nums[mid] == nums[right] + # There is a duplication in the given array + raise ValueError + return nums[left] + diff --git a/find-minimum-in-rotated-sorted-array/step3.py b/find-minimum-in-rotated-sorted-array/step3.py new file mode 100644 index 0000000..fbe7f9c --- /dev/null +++ b/find-minimum-in-rotated-sorted-array/step3.py @@ -0,0 +1,18 @@ +from typing import List + + +class Solution: + def findMin(self, nums: List[int]) -> int: + if not nums: + raise ValueError + + left, right = 0, len(nums) - 1 + while left < right: + mid = (left + right) // 2 + if nums[mid] < nums[right]: + right = mid + elif nums[right] < nums[mid]: + left = mid + 1 + else: # nums have identical items + raise ValueError + return nums[left]