From 3833af71393cc4fd464f1408d154422676eaa8ce Mon Sep 17 00:00:00 2001
From: skypenguins <poetry.painter-7w@icloud.com>
Date: Mon, 13 Jul 2026 21:51:54 +0900
Subject: [PATCH] 209. Minimum Size Subarray Sum

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+# 209. Minimum Size Subarray Sum
+- 問題: https://leetcode.com/problems/minimum-size-subarray-sum/
+- 言語: Python
+
+## Step1
+### 方針
+- 見積: $1 <= nums.length <= 10^{5}$ 、計算量: $O(n)$ 、 Pythonの実行時間: $10^{7}$ ステップ/秒 の時、最大 $10^{-2}$ 秒
+- 先頭から順に数値を足していって、合計が `target` 以上になったらその時の足した数値（要素）の個数の最小値を求める
+- `target = 11, nums = [1,2,3,4,5]` のようなテストケースでWA、正しい方法は尺取り法だったと思い出すも15分経過したため正答を見る
+
+### WA
+```py
+class Solution:
+    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
+        if len(nums) <= 1:
+            return len(nums)
+        
+        if sum(nums) < target:
+            return 0
+        
+        min_length = float("inf")
+        current_length = 0
+        current_sum = 0
+        for n in nums:
+            current_sum += n
+            current_length += 1
+
+            if current_sum >= target:
+                min_length = min(min_length, current_length)
+                current_sum = 0
+                current_length = 0
+
+        return min_length
+```
+
+### 正答
+- 尺取り法（Sliding Window）
+  - 右端を伸ばし、条件を満たす間、左端を縮める
+```py
+class Solution:
+    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
+        if sum(nums) < target:
+            return 0
+        
+        i = 0
+        j = 0
+        current_sum = 0
+        min_length = float("inf")
+        while j < len(nums):
+            current_sum += nums[j]
+
+            while current_sum >= target:
+                current_sum -= nums[i]
+                min_length = min(min_length, j - i + 1)
+                i += 1
+            
+            j += 1
+
+        return min_length
+```
+- `nums` の要素はすべて正の整数なので、次の単調性が成立
+  - ウィンドウの右端 `j` を伸ばす → 和は増える
+  - ウィンドウの左端 `i` を縮める → 和は減る
+- 時間計算量: $O(n)$
+  - `i`、 `j` がそれぞれ高々 $n$ 回しか動かないため
+- 空間計算量: $O(1)$
+
+## Step2
+- 典型コメント集: https://docs.google.com/document/d/11HV35ADPo9QxJOpJQ24FcZvtvioli770WWdZZDaLOfg/edit?tab=t.0#heading=h.p6d6fndbrthh
+
+- https://github.com/SuperHotDogCat/coding-interview/pull/31
+  - Python
+  - `i`, `j`の命名は `left`, `right` の方が読みやすい
+  - 論理的には問題ないが、以下の順序が自然か
+    - ```py
+        while current_sum >= target:
+            min_length = min(min_length, j - i + 1)
+            current_sum -= nums[i]
+            i += 1
+        ```
+
+- https://github.com/olsen-blue/Arai60/pull/50
+  - Python
+  - 累積和の配列を作れば、要素の値は単調増加であるため二分探索（bisect_left）でも解ける
+  - 二分探索（bisect_left）での方法
+    ```py
+    class Solution:
+        def minSubArrayLen(self, target: int, nums: List[int]) -> int:
+            prefix_sums = [0] * (len(nums) + 1)
+            for i in range(1, len(nums) + 1):
+                prefix_sums[i] = prefix_sums[i-1] + nums[i-1]
+    
+            min_length = sys.maxsize
+            for from_index in range(len(prefix_sums)):
+                target_sum = prefix_sums[from_index] + target
+                target_index = bisect.bisect_left(prefix_sums, target_sum)
+                if target_index == len(prefix_sums):
+                    break
+                min_length = min(min_length, target_index - from_index)
+    
+            if min_length == sys.maxsize:
+                return 0
+            return min_length
+    ```
+
+## Step3
+### 読みやすく書き直したコード
+```py
+class Solution:
+    def minSubArrayLen(self, target: int, nums: List[int]) -> int:
+        if sum(nums) < target:
+            return 0
+        
+        left = 0
+        right = 0
+        current_sum = 0
+        min_length = float("inf")
+        while right < len(nums):
+            current_sum += nums[right]
+
+            while current_sum >= target:
+                min_length = min(min_length, right - left + 1)
+                current_sum -= nums[left]
+                left += 1
+            
+            right += 1
+
+        return min_length
+```
+- 所要時間: 
+  - 1回目: 2:24
+  - 2回目: 2:59
+  - 3回目: 2:07