63. Unique Paths II

medium/63/answer.md

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+# Step1
+
+かかった時間：9min
+
+計算量：
+obstacleGrid.length = M
+obstacleGrid[i].length = Nとして、
+
+時間計算量：O(M*N)
+
+空間計算量：O(M*N)
+
+DP
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        num_ways = [[0] * n for _ in range(m)]
+
+        if obstacleGrid[0][0] == 1\
+            or obstacleGrid[-1][-1] == 1: 
+            return 0
+
+        num_ways[0][0] = 1
+        for row in range(m):
+            for col in range(n):
+                if row == 0 and col == 0:
+                    continue
+                if obstacleGrid[row][col] == 1:
+                    continue
+                if row == 0:
+                    num_ways[row][col] = num_ways[row][col - 1]
+                    continue
+                if col == 0:
+                    num_ways[row][col] = num_ways[row - 1][col]
+                    continue
+                num_ways[row][col] = (
+                    num_ways[row][col - 1] + num_ways[row - 1][col]
+                )
+
+        return num_ways[-1][-1]
+```
+思考ログ：
+- 前回のUniquePathsIのやり方を継承しつつ、obstacleGridを参照しながら行けない場所を考慮できるようにする
+- ばーっと書いてしまったが、for文の中の処理を切り出して見通しをよくしたい（初手これを書いたので一応載せている）
+
+上を少しまとめた
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        def _count_unique_paths(row: int, col: int) -> int:
+            # gridのセル(row, col)までのパス数を数える
+            if row == 0:
+                return num_ways[row][col - 1]
+            if col == 0:
+                return num_ways[row - 1][col]
+
+            return num_ways[row][col - 1] + num_ways[row - 1][col]
+
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        num_ways = [[0] * n for _ in range(m)]
+
+        if obstacleGrid[0][0] == 1\
+            or obstacleGrid[-1][-1] == 1: 
+            return 0
+
+        num_ways[0][0] = 1
+        for row in range(m):
+            for col in range(n):
+                if row == 0 and col == 0:
+                    continue
+                if obstacleGrid[row][col] == 1:
+                    continue
+                num_ways[row][col] = _count_unique_paths(row, col)
+
+        return num_ways[-1][-1]
+```
+思考ログ：
+
+再帰
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        @cache
+        def _count_unique_paths(row: int, col: int) -> int:
+            assert row >= 0 or col >= 0, 'row and col must be positive integer values.'
+
+            if obstacleGrid[row][col] == 1:
+                return 0
+            if row == 0 and col == 0:
+                return 1
+            if row == 0:
+                return _count_unique_paths(row, col - 1)
+            if col == 0:
+                return _count_unique_paths(row - 1, col)
+
+            return _count_unique_paths(row - 1, col)\
+                + _count_unique_paths(row, col - 1)
+
+        return _count_unique_paths(
+            len(obstacleGrid) - 1,
+            len(obstacleGrid[0]) - 1
+```
+思考ログ：
+
+1-DP
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        # 11:59
+        def _check_if_valid_cell(row: int, col: int, is_transposed: bool) -> bool:
+            if is_transposed:
+                return obstacleGrid[col][row] == 0
+            else:
+                return obstacleGrid[row][col] == 0
+
+        if obstacleGrid[0][0] == 1\
+            or obstacleGrid[-1][-1] == 1: 
+            return 0
+
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        is_transposed = False
+        if m < n:
+            n, m = m, n
+            is_transposed = True
+
+        num_ways = [0] * n
+        num_ways[0] = 1
+        for row in range(m):
+            for col in range(n):
+                if _check_if_valid_cell(row, col, is_transposed):
+                    if col == 0:
+                        continue
+                    num_ways[col] += num_ways[col - 1]
+                else:
+                    num_ways[col] = 0
+
+        return num_ways[-1]
+```
+思考ログ：
+- メモリ節約ver
+- 今回は、n, mを入れ替えた時に参照するobstacleGridにも影響が及ぶ
+  - これも転置しちゃえば良いのだが、破壊的な変更をしたくないし、転置した情報を別に用意したらこれはこれで本末転倒なことに
+  - 転置フラグで管理するようにしたが、どうだろうか
+- ```num_ways```の0番目の要素の扱いで少しハマった
+  - ```num_ways```がどんな要素を持つ配列なのか、それは意図したものなのか、確認がまだ甘い
+  - colのループの前で```col = 0```だけ処理してしまってもいいが、うーん
+
+# Step2
+
+講師役目線でのセルフツッコミポイント：
+- マジックナンバー（obstacle）
+
+参考にした過去ログなど：
+- https://github.com/Ryotaro25/leetcode_first60/pull/40
+- https://github.com/seal-azarashi/leetcode/pull/32
+- https://github.com/Yoshiki-Iwasa/Arai60/pull/49
+  - BFS
+- https://github.com/nittoco/leetcode/pull/27
+- https://github.com/fhiyo/leetcode/pull/35
+- https://github.com/goto-untrapped/Arai60/pull/33
+  - 以下の条件のまとめ方が頭になかった  
+    - https://github.com/goto-untrapped/Arai60/pull/33/files#r1665011732
+- https://github.com/YukiMichishita/LeetCode/pull/15
+- https://github.com/sakupan102/arai60-practice/pull/35
+  - シンプルな1DP
+- https://github.com/SuperHotDogCat/coding-interview/pull/17
+- https://github.com/shining-ai/leetcode/pull/34
+- https://github.com/hayashi-ay/leetcode/pull/44
+  - 値を受け取る、渡す  
+
+DP
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE \
+            or obstacleGrid[-1][-1] == OBSTACLE:
+            return 0
+
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        num_ways = [[0] * n for _ in range(m)]
+        num_ways[0][0] = 1
+        for row in range(m):
+            for col in range(n):
+                if row == 0 and col == 0:
+                    continue
+                if obstacleGrid[row][col] == OBSTACLE:
+                    continue
+                if row > 0:
+                    num_ways[row][col] += num_ways[row - 1][col]
+                if col > 0:
+                    num_ways[row][col] += num_ways[row][col - 1]
+
+        return num_ways[-1][-1]
+```
+思考ログ：
+- 少し処理の順序を変えた
+- マジックナンバーの解消
+- DP受け取りの分岐を変更
+
+DP、事前に初期化
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE \
+            or obstacleGrid[-1][-1] == OBSTACLE:
+            return 0
+
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        def initialize_num_ways() -> list[list[int]]:
+            num_ways = [[0] * n for _ in range(m)]
+            for row in range(m):
+                if obstacleGrid[row][0] == OBSTACLE:
+                    break
+                num_ways[row][0] = 1
+            for col in range(n):
+                if obstacleGrid[0][col] == OBSTACLE:
+                    break
+                num_ways[0][col] = 1
+
+            return num_ways
+
+        num_ways = initialize_num_ways()
+        for row in range(1, m):
+            for col in range(1, n):
+                if obstacleGrid[row][col] == OBSTACLE:
+                    continue
+                num_ways[row][col] = num_ways[row - 1][col] + num_ways[row][col - 1]
+
+        return num_ways[-1][-1]
+```
+思考ログ：
+- 後半がシンプルに書けて良い
+
+# Step3
+
+かかった時間：6min
+
+```python
+class Solution:
+    def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:
+        OBSTACLE = 1
+        if obstacleGrid[0][0] == OBSTACLE \
+            or obstacleGrid[-1][-1] == OBSTACLE:
+            return 0
+
+        m = len(obstacleGrid)
+        n = len(obstacleGrid[0])
+        def initialize_num_ways():
+            num_ways = [[0] * n for _ in range(m)]
+            for row in range(m):
+                if obstacleGrid[row][0] == OBSTACLE:
+                    break
+                num_ways[row][0] = 1
+            for col in range(n):
+                if obstacleGrid[0][col] == OBSTACLE:
+                    break
+                num_ways[0][col] = 1
+
+            return num_ways
+
+        num_ways = initialize_num_ways()
+        for row in range(1, m):
+            for col in range(1, n):
+                if obstacleGrid[row][col] == OBSTACLE:
+                    continue
+                num_ways[row][col] = num_ways[row - 1][col] + num_ways[row][col - 1]
+
+        return num_ways[-1][-1]
+```
+思考ログ：
+- 個人的には後半の処理がすっきりするのが良かった
+- 1DPは```min(n, m)```の最適化までやりたくなるが、コードが込み入ってくるので2DPでも良い気がする
+
+# Step4
+
+```python
+```
+思考ログ：
+