skypenguins · skypenguins · Jul 26, 2026 · nodchip · Jul 30, 2026 · skypenguins
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+# 695. Max Area of Island
+- 問題: https://leetcode.com/problems/max-area-of-island/
+- 言語: Python
+
+## Step1
+### 方針
+- `200. Number of Islands` の類題と考えて、Union-Findで島の数を数えるときに島の面積を求める処理を入れてみる
+- unionするときに島の面積をどう求めるか考えていたら、15分経過してしまったので正答を見る
+
+### 正答
+- 以下の正答を読んで `200. Number of Islands` の時のDFSの方法で解いていれば、素直に面積を求められたかもしれないと思った。
+- Union-Findは実装が複雑になりがち
+
+#### 方針1: Union-Find
+- それぞれの根に対応するsize配列を持ち、unionのたびに合算する
+- waterのセルはunionされないため、対応する `size` の値は使われないまま残る
+
+##### コード
+```py
+class Solution:
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        WATER = 0
+        LAND = 1
+
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        num_cells = num_rows * num_cols
+        parent = list(range(num_cells))
+        rank = [0] * num_cells
+        size = [1] * num_cells  # size[i]: セルiが属す連結成分の要素数(初期値1)
+
+        def flatten_index(x, y):
+            return x * num_cols + y
+
+        def find(x):
+            while parent[x] != x:
+                parent[x] = parent[parent[x]]
+                x = parent[x]
+            return parent[x]
+
+        def union(x, y):
+            root_x = find(x)
+            root_y = find(y)
+
+            if root_x == root_y:
+                return
+
+            if rank[root_x] < rank[root_y]:
+                root_x, root_y = root_y, root_x
+
+            parent[root_y] = root_x
+            size[root_x] += size[root_y]  # マージ先root_xにサイズを足し込む
+
+            if rank[root_x] == rank[root_y]:
+                rank[root_x] += 1
+
+        for row_index in range(num_rows):
+            for col_index in range(num_cols):
+                if grid[row_index][col_index] == WATER:
+                    continue
+                if row_index + 1 < num_rows and grid[row_index + 1][col_index] == LAND:
+                    union(
+                        flatten_index(row_index, col_index),
+                        flatten_index(row_index + 1, col_index),
+                    )
+                if col_index + 1 < num_cols and grid[row_index][col_index + 1] == LAND:
+                    union(
+                        flatten_index(row_index, col_index),
+                        flatten_index(row_index, col_index + 1),
+                    )
+
+        max_area = 0
+        for row_index in range(num_rows):
+            for col_index in range(num_cols):
+                if grid[row_index][col_index] == LAND:
+                    root = find(flatten_index(row_index, col_index))
+                    max_area = max(max_area, size[root])
+
+        return max_area
+```
+- 時間計算量: $O(mn)$
+- 空間計算量: $O(mn)$
+
+#### 方針2: DFS
+- 各LAND未訪問セルを起点に「繋がっている陸セルを全部辿って数える」を全セルに対して行い、最大値を取る
+- 訪問済みセルは二度と数えないように `visited` で管理
+
+##### iterative DFS
+```py
+class Solution:
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        WATER = 0
+        LAND = 1
+
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        visited = [[False] * num_cols for _ in range(num_rows)]
+
+        def area_of_island(start_row, start_col):
+            to_visit_cells = [(start_row, start_col)]
+            visited[start_row][start_col] = True
+            area = 0
+
+            while to_visit_cells:
+                row, col = to_visit_cells.pop()
+                area += 1
+
+                for d_row, d_col in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+                    next_row, next_col = row + d_row, col + d_col
+                    if (
+                        0 <= next_row < num_rows
+                        and 0 <= next_col < num_cols
+                        and not visited[next_row][next_col]
+                        and grid[next_row][next_col] == LAND
+                    ):
+                        visited[next_row][next_col] = True
+                        to_visit_cells.append((next_row, next_col))
+
+            return area
+
+        max_area = 0
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if grid[row][col] == LAND and not visited[row][col]:
+                    max_area = max(max_area, area_of_island(row, col))
+
+        return max_area
+```
+- 時間計算量: $O(mn)$
+- 空間計算量: $O(mn)$
+
+##### 再帰DFS
+```py
+class Solution:
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        WATER = 0
+        LAND = 1
+
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        visited = [[False] * num_cols for _ in range(num_rows)]
+
+        def area_of_island(row, col):
+            if (
+                row < 0
+                or row >= num_rows
+                or col < 0
+                or col >= num_cols
+                or visited[row][col]
+                or grid[row][col] == WATER
+            ):
+                return 0
+
+            visited[row][col] = True
+            area = 1
+
+            for d_row, d_col in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+                area += area_of_island(row + d_row, col + d_col)
+
+            return area
+
+        max_area = 0
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if grid[row][col] == LAND and not visited[row][col]:
+                    max_area = max(max_area, area_of_island(row, col))
+
+        return max_area
+```
+- 時間計算量: $O(mn)$
+- 空間計算量: $O(mn)$
+- 再帰DFSは、繋がった陸地が細長く伸びている場合（例：1000×1000のグリッドが蛇行した1本の細い陸地でほぼ埋まっている場合）、再帰の深さが $m·n$ に達するため再帰上限のエラーになる可能性がある
+
+#### 方針3: BFS
+- 方針はDFSとほぼ同じ
+
+##### iterative BFS
+```py
+class Solution:
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        WATER = 0
+        LAND = 1
+
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        visited = [[False] * num_cols for _ in range(num_rows)]
+
+        def area_of_island(start_row, start_col):
+            to_visit_cells = deque([(start_row, start_col)])
+            visited[start_row][start_col] = True
+            area = 0
+
+            while to_visit_cells:
+                row, col = to_visit_cells.popleft()
+                area += 1
+
+                for d_row, d_col in ((1, 0), (-1, 0), (0, 1), (0, -1)):
+                    next_row, next_col = row + d_row, col + d_col
+                    if (
+                        0 <= next_row < num_rows
+                        and 0 <= next_col < num_cols
+                        and not visited[next_row][next_col]
+                        and grid[next_row][next_col] == LAND
+                    ):
+                        visited[next_row][next_col] = True
+                        to_visit_cells.append((next_row, next_col))
+
+            return area
+
+        max_area = 0
+        for row in range(num_rows):
+            for col in range(num_cols):
+                if grid[row][col] == LAND and not visited[row][col]:
+                    max_area = max(max_area, area_of_island(row, col))
+
+        return max_area
+```
+- 時間計算量: $O(mn)$
+- 空間計算量: $O(mn)$
+
+
+## Step2
+- 典型コメント集: https://docs.google.com/document/d/11HV35ADPo9QxJOpJQ24FcZvtvioli770WWdZZDaLOfg/edit?tab=t.0#heading=h.f28i04p206ak
+
+- https://github.com/YukiMichishita/LeetCode/pull/6
+  - Python
+  - やはり `search_land(x + 1, y)` 、 `search_land(x - 1, y)` 、 `search_land(x, y + 1)` 、 `search_land(x, y - 1)` のように方向ごとに再帰する方が分かりやすいか？
+  - `nonlocal` の議論: https://github.com/YukiMichishita/LeetCode/pull/6#discussion_r1555974201
+
+- https://github.com/colorbox/leetcode/pull/32
+  - C++
+  - この方もスタックに、方向を格納した配列のiterativeではなくハードコードで方向ごとに積んでいる
+    - 配列の方がシンプルとの見解もある: https://github.com/colorbox/leetcode/pull/32#discussion_r1898537718
+  - スタックに追加前に範囲チェックする考え方: https://github.com/colorbox/leetcode/pull/32#discussion_r1898178545
+
+- https://github.com/t0hsumi/leetcode/pull/19
+  - Python
+  - 同じようなチェックを関数化しているが実装が複雑になりそう
+
+- https://github.com/ryoooooory/LeetCode/pull/21
+  - Java
+  - `addToQueue` を4回呼び出していれば、それは4方向に探索すると伝わりやすいなと思った
+   - cf. https://github.com/ryoooooory/LeetCode/pull/21#discussion_r1966729356
+  - Javaの `record` は便利そう
+
+- https://github.com/Fuminiton/LeetCode/pull/18
+  - Python
+  - 方向を書き下すか、配列で持つかは趣味の範囲っぽそう: https://github.com/Fuminiton/LeetCode/pull/18#discussion_r1986038739
+
+## Step3
+### 方針: iterative DFS
+```py
+class Solution:
+    def maxAreaOfIsland(self, grid: List[List[int]]) -> int:
+        WATER = 0
+        LAND = 1
+
+        num_rows = len(grid)
+        num_cols = len(grid[0])
+        visited = [[False] * num_cols for _ in range(num_rows)]
+
+        def get_area_of_island(start_row_index, start_col_index):
+            to_visit_cells = [(start_row_index, start_col_index)]
+            visited[start_row_index][start_col_index] = True
+            area = 0
+
+            while len(to_visit_cells) != 0:
+                row_index, col_index = to_visit_cells.pop()
+                area += 1
+
+                for direction_row, direction_col in [(1, 0), (-1, 0), (0, 1), (0, -1)]:
+                    next_row_index = row_index + direction_row
+                    next_col_index = col_index + direction_col
+                    if (
+                        0 <= next_row_index < num_rows
+                        and 0 <= next_col_index < num_cols
+                        and not visited[next_row_index][next_col_index]
+                        and grid[next_row_index][next_col_index] == LAND
+                    ):
+                        visited[next_row_index][next_col_index] = True
+                        to_visit_cells.append((next_row_index, next_col_index))
+
+            return area
+
+        max_area = 0
+        for row_index in range(num_rows):
+            for col_index in range(num_cols):
+                if (
+                    grid[row_index][col_index] == LAND
+                    and not visited[row_index][col_index]
+                ):
+                    max_area = max(max_area, get_area_of_island(row_index, col_index))
+
+        return max_area
+```
+- 所要時間:
+  - 1回目: 8:11
+  - 2回目: 7:19
+  - 3回目: 8:16