From 51cb35133c416fd8a98a5a6bafd35e7a4f758525 Mon Sep 17 00:00:00 2001
From: fuminiton <fumi_8823@icloud.com>
Date: Mon, 21 Apr 2025 23:52:39 +0900
Subject: [PATCH] new file:   problem34/memo.md

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 create mode 100644 problem34/memo.md

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+## 取り組み方
+- step1: 5分以内に空で書いてAcceptedされるまで解く + テストケースと関連する知識を連想してみる
+- step2: 他の方の記録を読んで連想すべき知識や実装を把握した上で、前提を置いた状態で最適な手法を選択し実装する
+- step3: 10分以内に1回もエラーを出さずに3回連続で解く
+
+## step1
+2次元配列のsum_pathsを基本的に、`sum_paths[r][c] = sum_paths[r-1][c] + sum_paths[r][c-1]`として考える。
+今の座標に障害物がない時は上と左の和を組み合わせの数として、障害物があるときは0とすれば良いだけ。
+
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE:
+            return 0
+
+        width = len(obstacleGrid[0])
+        height = len(obstacleGrid)
+        sum_paths = [[0] * width for _ in range(height)]
+        sum_paths[0][0] = 1
+        for h in range(1, height):
+            if obstacleGrid[h][0] == OBSTACLE:
+                continue
+            sum_paths[h][0] = sum_paths[h - 1][0]
+        for w in range(1, width):
+            if obstacleGrid[0][w] == OBSTACLE:
+                continue
+            sum_paths[0][w] = sum_paths[0][w - 1]
+        
+        for h in range(1, height):
+            for w in range(1, width):
+                if obstacleGrid[h][w] == OBSTACLE:
+                    continue
+                sum_paths[h][w] = sum_paths[h - 1][w] + sum_paths[h][w - 1]
+        return sum_paths[-1][-1]
+```
+
+## step2
+### 読んだコード
+- https://github.com/hayashi-ay/leetcode/pull/44/files
+- https://github.com/SuperHotDogCat/coding-interview/pull/17/files
+- https://github.com/sakupan102/arai60-practice/pull/35/files
+
+### 感想
+- h>0, w>0 に加算を分けるとかなりスッキリ変形できて良い
+- 地味なところだが、 意図がない限りは height -> width の順で操作することが統一されてる方が適切だと思った
+
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE:
+            return 0
+
+        height, width = len(obstacleGrid), len(obstacleGrid[0])
+        sum_paths = [[0] * width for _ in range(height)]
+        sum_paths[0][0] = 1
+        for h in range(height):
+            for w in range(width):
+                if obstacleGrid[h][w] == OBSTACLE:
+                    continue
+                if h > 0:
+                    sum_paths[h][w] += sum_paths[h - 1][w]
+                if w > 0:
+                    sum_paths[h][w] += sum_paths[h][w - 1]
+        return sum_paths[-1][-1]
+```
+
+## step3
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE:
+            return 0
+        num_rows = len(obstacleGrid)
+        num_cols = len(obstacleGrid[0])
+        sum_paths = [[0] * num_cols for _ in range(num_rows)]
+        sum_paths[0][0] = 1
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if obstacleGrid[row][col] == OBSTACLE:
+                    continue
+                if row > 0:
+                    sum_paths[row][col] += sum_paths[row - 1][col]
+                if col > 0:
+                    sum_paths[row][col] += sum_paths[row][col - 1]
+        return sum_paths[num_rows - 1][num_cols - 1]
+```