Create max-area-of-island.py

Python/max-area-of-island.py

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+# Time:  O(m * n)
+# Space: O(1)
+
+# Given a non-empty 2D array grid of 0's and 1's,
+# an island is a group of 1's (representing land) connected 4-directionally (horizontal or vertical.)
+# You may assume all four edges of the grid are surrounded by water.
+#
+# Find the maximum area of an island in the given 2D array. (If there is no island, the maximum area is 0.)
+#
+# Example 1:
+# [[0,0,1,0,0,0,0,1,0,0,0,0,0],
+#  [0,0,0,0,0,0,0,1,1,1,0,0,0],
+#  [0,1,1,0,1,0,0,0,0,0,0,0,0],
+#  [0,1,0,0,1,1,0,0,1,0,1,0,0],
+#  [0,1,0,0,1,1,0,0,1,1,1,0,0],
+#  [0,0,0,0,0,0,0,0,0,0,1,0,0],
+#  [0,0,0,0,0,0,0,1,1,1,0,0,0],
+#  [0,0,0,0,0,0,0,1,1,0,0,0,0]]
+#
+# Given the above grid, return 6. Note the answer is not 11,
+# because the island must be connected 4-directionally.
+#
+# Example 2:
+# [[0,0,0,0,0,0,0,0]]
+#
+# Given the above grid, return 0.
+#
+# Note: The length of each dimension in the given grid does not exceed 50.
+
+class Solution(object):
+    def maxAreaOfIsland(self, grid):
+        """
+        :type grid: List[List[int]]
+        :rtype: int
+        """
+        directions = [[-1,  0], [ 1,  0], [ 0,  1], [ 0, -1]]
+
+        def dfs(i, j, grid, area):
+            if not (0 <= i < len(grid) and \
+                    0 <= j < len(grid[0]) and \
+                    grid[i][j] > 0):
+                return False
+            grid[i][j] *= -1
+            area[0] += 1
+            for d in directions:
+                dfs(i+d[0], j+d[1], grid, area)
+            return True
+
+        result = 0
+        for i in xrange(len(grid)):
+            for j in xrange(len(grid[0])):
+                area = [0]
+                if dfs(i, j, grid, area):
+                    result = max(result, area[0])
+        return result