63. Unique Paths II

63_unique-paths-ii.md

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+# 63. Unique Paths II
+
+## 1st
+
+### ①
+
+pathsという変数名は、num_pathsにするとnum_rows, num_colsと似てるけど別物になるのが嫌だったので変えた。変だと思うんだが良い変数名が出てこない。num_paths_per_colとか...？
+
+1行目とそれ以外を区別して処理してしまったが、その必要はない。
+
+obstacleGridの0, 1は定数にした方が良かった。
+
+所要時間: 13:16
+
+gridのサイズ: m x n
+- 時間計算量: O(mn)
+- 空間計算量: O(n)
+
+```py
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        # precondition: len(obstacleGrid) > 0, len(obstacleGrid[0]) > 0, len(obstacleGrid[i]) == len(obstacleGrid[j]) forall i != j
+        if obstacleGrid[0][0] == 1:
+            return 0
+        num_rows = len(obstacleGrid)
+        num_cols = len(obstacleGrid[0])
+        paths = [0] * num_cols
+        paths[0] = 1
+        for col in range(1, num_cols):
+            if obstacleGrid[0][col] == 0:
+                paths[col] += paths[col - 1]
+        for row in range(1, num_rows):
+            if obstacleGrid[row][0] == 1:
+                paths[0] = 0
+            for col in range(1, num_cols):
+                if obstacleGrid[row][col] == 1:
+                    paths[col] = 0
+                else:
+                    paths[col] += paths[col - 1]
+        return paths[-1]
+```
+
+### ②
+
+メモ化再帰。2次元で長方形であることを保証しようとすると、assertでやるなら `assert len(grid) > 0`, `assert len(grid[0]) > 0` とループで `assert len(grid[i]) == len(grid[0])` みたいなことをしないといけないはずだが、やりすぎな気がしてやっていない。しかし前提にして書いていいのかはいつも迷ってしまう。
+
+所要時間: 2:58
+
+gridのサイズ: m x n
+- 時間計算量: O(mn)
+- 空間計算量: O(mn)
+
+```py
+class Solution:
+    OBSTACLE = 1
+
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        num_rows = len(obstacleGrid)
+        num_cols = len(obstacleGrid[0])
+
+        @cache
+        def count_paths_from(row: int, col: int) -> int:
+            if obstacleGrid[row][col] == Solution.OBSTACLE:
+                return 0
+            if row == num_rows - 1 and col == num_cols - 1:
+                return 1
+            count = 0
+            if row < num_rows - 1:
+                count += count_paths_from(row + 1, col)
+            if col < num_cols - 1:
+                count += count_paths_from(row, col + 1)
+            return count
+
+        return count_paths_from(0, 0)
+```
+
+## 2nd
+
+### 参考
+
+- https://discord.com/channels/1084280443945353267/1233603535862628432/1257363629272072282
+- https://discord.com/channels/1084280443945353267/1192736784354918470/1252607758604308532
+  ①で困っていたpathsという名前。こちらではunique_paths_tableにしている。 自分のに合わせるとnum_paths_rowになるか。pathの数が入っている列、なら何が入っているか分かるかも。
+- https://discord.com/channels/1084280443945353267/1227073733844406343/1244288660309020712
+
+これが一番素直かもしれない。テーブルを作って上と左からpathの数を足していく。
+
+```py
+class Solution:
+    OBSTACLE = 1
+
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        assert len(obstacleGrid) > 0 and len(obstacleGrid[0]) > 0
+        num_rows = len(obstacleGrid)
+        num_cols = len(obstacleGrid[0])
+        num_paths_table = [[0] * num_cols for _ in range(num_rows)]
+        num_paths_table[0][0] = 1
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if obstacleGrid[row][col] == Solution.OBSTACLE:
+                    num_paths_table[row][col] = 0
+                    continue
+                if row == 0 and col == 0:
+                    continue
+                num_paths = 0
+                if row > 0:
+                    num_paths += num_paths_table[row - 1][col]
+                if col > 0:
+                    num_paths += num_paths_table[row][col - 1]
+                num_paths_table[row][col] = num_paths
+        return num_paths_table[-1][-1]
+```
+
+- https://discord.com/channels/1084280443945353267/1225849404037009609/1236740338937368737
+
+
+## 3rd
+
+```py
+class Solution:
+    OBSTACLE = 1
+
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        num_rows = len(obstacleGrid)
+        num_cols = len(obstacleGrid[0])
+        num_paths_row = [0] * num_cols
+        num_paths_row[0] = 1
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if obstacleGrid[row][col] == Solution.OBSTACLE:
+                    num_paths_row[col] = 0
+                    continue
+                if col == 0:
+                    continue
+                num_paths_row[col] += num_paths_row[col - 1]
+        return num_paths_row[-1]
+```