200 number of islands medium

200_number_of_islands_medium/README.md

@@ -0,0 +1,303 @@
+# 問題へのリンク
+[200. Number of Islands](https://leetcode.com/problems/number-of-islands/)
+
+# 言語
+Python
+
+## step1
+再帰関数を用いたDFS を用いる解法。
+```python
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid:
+            return 0
+        if not grid[0]:
+            return 0
+        num_rows = len(grid)
+        num_columns = len(grid[0])
+        num_islands = 0
+        seen_lands = set()
+
+        def is_land(row:int, column:int) -> bool:
+            return (0 <= row < num_rows
+                    and 0 <= column < num_columns
+                    and grid[row][column] == "1")
+
+        grid_diffs = ((1, 0), (-1, 0), (0, 1), (0, -1))
+        def traverse_connected_lands(row, column):
+            seen_lands.add((row, column))
+            for dr, dc in grid_diffs:
+                neighbor_row = row + dr
+                neighbor_column = column + dc
+                if not is_land(neighbor_row, neighbor_column):
+                    continue
+                if (neighbor_row, neighbor_column) in seen_lands:
+                    continue
+                traverse_connected_lands(neighbor_row, neighbor_column)
+            return
+
+        # find num_islands by multistart DFS
+        for row in range(num_rows):
+            for column in range(num_columns):
+                # skip water
+                if not is_land(row, column):
+                    continue
+                if (row, column) in seen_lands:
+                    continue
+                traverse_connected_lands(row, column)
+                num_islands += 1
+        return num_islands
+```
+
+再帰関数を用いたDFSは、スタックの分だけの空間を使う上、Pythonでは再起呼び出し回数の制限がデフォルトでは1000と小さいので、パフォーマンス上はBFSを用いる方が望ましい。が、再帰関数を用いたDFSの方がコードがシンプルになる場合が多い。
+- 時間計算量：`O(n*m)`
+- 空間計算量：`O(n*m)`
+
+## step2
+
+```python
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        num_rows = len(grid)
+        num_columns = len(grid[0])
+        num_islands = 0
+        visited_lands = set()
+
+        def is_land(row: int, column: int) -> bool:
+            return (
+                0 <= row < num_rows
+                and 0 <= column < num_columns
+                and grid[row][column] == "1"
+            )
+
+        # traverse connected lands by DFS
+        directions = ((1, 0), (-1, 0),  (0, 1), (0, -1))
+        def traverse_connected_lands(row, column):
+            # if inplace: grid[row][column] = "0"
+            visited_lands.add((row, column))
+            for dr, dc in directions:
+                neighbor_row = row + dr
+                neighbor_column = column + dc
+                if (neighbor_row, neighbor_column) in visited_lands:
+                    continue
+                if not is_land(neighbor_row, neighbor_column):
+                    continue
+                traverse_connected_lands(neighbor_row, neighbor_column)
+
+        for row in range(num_rows):
+            for column in range(num_columns):
+                if not is_land(row, column):
+                    continue
+                if (row, column) in visited_lands:
+                    continue
+                traverse_connected_lands(row, column)
+                num_islands += 1
+        return num_islands
+```
+
+- gridをinplaceに書き換えてOKならば、`seen_lands.add((row, column))` を`grid[row][column] = "0"` に変更することで、`seen_lands`の代わりに元のグリッドを利用でき、空間計算量を`O(1)`に削減できる。
+
+## step3
+
+```python
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        WATER = "0"
+        LAND = "1"
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        num_islands = 0
+        directions = ((1, 0), (-1, 0), (0, 1), (0, -1))
+        def traverse(row, col):
+            for dr, dc in directions:
+                neighbor_row = row + dr
+                neighbor_col = col + dc
+                if neighbor_row < 0 or neighbor_row >= num_rows:
+                    continue
+                if neighbor_col < 0 or neighbor_col >= num_cols:
+                    continue
+                if grid[neighbor_row][neighbor_col] == WATER:
+                    continue
+                grid[neighbor_row][neighbor_col] = WATER
+                traverse(neighbor_row, neighbor_col)
+
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if grid[row][col] == LAND:
+                    traverse(row, col)
+                    num_islands += 1
+        return num_islands
+```
+
+## step4 (FB)
+- DSUでは`directions` は `((0, 1), (1, 0))` の2方向のみでOK。なぜなら、左上から右下に向かって走査していくので、未訪問の陸地は右か下にしか存在し得ないから。
+- `if row < 0 or row >= num_rows:` のような境界チェックは 「数直線に沿った」条件式が望ましい。例えば `if not (0 <= row < num_rows):` のように書くようにする。
+- `return` は不要。
+
+`step4.py`
+```python
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        WATER = "0"
+        LAND = "1"
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        num_islands = 0
+        directions = ((1, 0), (0, 1), (0, -1), (-1, 0))
+        def traverse(row, col):
+            for dr, dc in directions:
+                neighbor_row = row + dr
+                neighbor_col = col + dc
+                if not (0 <= neighbor_row < num_rows):
+                    continue
+                if not (0 <= neighbor_col < num_cols):
+                    continue
+                if grid[neighbor_row][neighbor_col] == WATER:
+                    continue
+                grid[neighbor_row][neighbor_col] = WATER
+                traverse(neighbor_row, neighbor_col)
+
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if grid[row][col] == LAND:
+                    traverse(row, col)
+                    num_islands += 1
+        return num_islands
+```
+
+# 別解・模範解答
+## BFS を用いる解法
+
+`bfs.py`
+```python
+from collections import deque
+
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        num_rows = len(grid)
+        num_columns = len(grid[0])
+        num_islands = 0
+        num_islands = 0
+        visited_lands = set()
+
+        def is_land(row: int, column: int) -> bool:
+            return (
+                0 <= row < num_rows
+                and 0 <= column < num_columns
+                and grid[row][column] == "1"
+            )
+
+        directions = ((1, 0), (-1, 0),  (0, 1), (0, -1))
+        def traverse_connected_lands(initial_row, initial_column):
+            frontiers = deque([])
+            frontiers.appendleft((initial_row, initial_column))
+            visited_lands.add((initial_row, initial_column))
+            while frontiers:
+                row, column = frontiers.pop()
+                for dr, dc in directions:
+                    neighbor_row = row + dr
+                    neighbor_column = column + dc
+                    if not is_land(neighbor_row, neighbor_column):
+                        continue
+                    if (neighbor_row, neighbor_column) in visited_lands:
+                        continue
+
+                    visited_lands.add((neighbor_row, neighbor_column))
+                    frontiers.appendleft((neighbor_row, neighbor_column))
+
+        for row in range(num_rows):
+            for column in range(num_columns):
+                if not is_land(row, column):
+                    continue
+                if (row, column) in visited_lands:
+                    continue
+                traverse_connected_lands(row, column)
+                num_islands += 1
+        return num_islands
+```
+
+## Disjoint Set Union (Union-Find) を用いる解法
+
+`dsu.py`
+```python
+
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+
+        def is_land(row: int, column: int) -> bool:
+            return (
+                0 <= row < num_rows
+                and 0 <= column < num_cols
+                and grid[row][column] == "1"
+            )
+
+        parents: dict[tuple[int, int], tuple[int, int]] = {}
+        ranks: dict[tuple[int, int], int] = {}
+
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if is_land(row, col):
+                    parents[(row, col)] = (row, col)
+                    ranks[(row, col)] = 0
+
+        def find(node):
+            root = node
+            while root != parents[root]:
+                root = parents[root]
+            while node != root:
+                parent = parents[node]
+                parents[node] = root
+                node = parent
+            return root
+
+        def unite(node1, node2):
+            root1 = find(node1)
+            root2 = find(node2)
+            if ranks[root1] > ranks[root2]:
+                parents[root2] = root1
+            elif ranks[root2] > ranks[root1]:
+                parents[root1] = root2
+            else:
+                parents[root2] = root1
+                ranks[root1] += 1
+
+
+        directions = ((1, 0), (-1, 0),  (0, 1), (0, -1))
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if not is_land(row, col):
+                    continue
+                for dr, dc in directions:
+                    neighbor = (row + dr, col + dc)
+                    if is_land(neighbor[0], neighbor[1]):
+                        unite((row, col), neighbor)
+
+        roots = {find(cell) for cell in parents}
+        return len(roots)
+```
+
+- 時間計算量：`O(n*m*α(k))` （αはアッカーマン関数の逆関数、kはunion/findの呼び出し回数）
+- 空間計算量：`O(n*m)`
+- `mydsu.py` は DSU をクラスとして実装した例。`num_disjoint_sets` を逐次管理することで、最後にルートの集合を数える計算を省略できる。
+
+# 想定されるフォローアップ質問
+- Q. もしこのグリッドが非常に巨大で、例えば数テラバイトあり、メモリに一度に収まらない場合はどうしますか？
+    - A. グリッドが2行分ずつメモリに収まると仮定します。各行でDSUと島の数を逐一計算し、前の行と現在の行で接続されている島をマージしていきます。これにより、メモリ使用量を大幅に削減しつつ、正確な島の数を計算できます。
+
+
+# 次に解く問題の予告
+- [Validate Binary Search Tree](https://leetcode.com/problems/validate-binary-search-tree/)
+- [Word Ladder](https://leetcode.com/problems/word-ladder/description/)
+- [Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/description/)

200_number_of_islands_medium/bfs.py

@@ -0,0 +1,55 @@
+#
+# @lc app=leetcode id=200 lang=python3
+#
+# [200] Number of Islands
+#
+
+# @lc code=start
+from collections import deque
+
+class Solution:
+    def numIslands(self, grid: list[list[str]]) -> int:
+        if not grid or not grid[0]:
+            return 0
+        num_rows = len(grid)
+        num_columns = len(grid[0])
+        num_islands = 0
+        num_islands = 0
+        visited_lands = set()
+
+        def is_land(row: int, column: int) -> bool:
+            return (
+                0 <= row < num_rows
+                and 0 <= column < num_columns
+                and grid[row][column] == "1"
+            )
+
+        directions = ((1, 0), (-1, 0),  (0, 1), (0, -1))
+        def traverse_connected_lands(initial_row, initial_column):
+            frontiers = deque([])
+            frontiers.appendleft((initial_row, initial_column))
+            visited_lands.add((initial_row, initial_column))
+            while frontiers:
+                row, column = frontiers.pop()
+                for dr, dc in directions:
+                    neighbor_row = row + dr
+                    neighbor_column = column + dc
+                    if not is_land(neighbor_row, neighbor_column):
+                        continue
+                    if (neighbor_row, neighbor_column) in visited_lands:
+                        continue
+
+                    visited_lands.add((neighbor_row, neighbor_column))
+                    frontiers.appendleft((neighbor_row, neighbor_column))
+
+        for row in range(num_rows):
+            for column in range(num_columns):
+                if not is_land(row, column):
+                    continue
+                if (row, column) in visited_lands:
+                    continue
+                traverse_connected_lands(row, column)
+                num_islands += 1
+        return num_islands
+
+# @lc code=end