Island Count and Maximum Area

Given a rectangular binary matrix grid, find all islands. An island is a maximal group of 1 cells connected horizontally or vertically.
Return [islandCount, maximumArea]. The area of an island is its number of cells. Return [0,0] when the grid contains no land.

Function
islandStats(grid: int[][]) → int[]

Examples
Example 1
grid = [[1,1,0,0],[0,1,0,1],[1,0,0,1]]
return = [3,3]
There are three islands; the island in the upper-left has area 3 and is the largest.
Example 2
grid = [[0,0],[0,0]]
return = [0,0]
No land cell exists.

Constraints
1 <= grid.length, grid[i].length <= 300.
Every row has the same length.
Every cell is 0 or 1.

The trick is one pass over the grid. Scan every cell. When you hit a 1, that's a new island: increment the count, then flood fill from it with BFS or DFS, counting cells as you go and marking each visited cell as 0 so you never revisit it. After the fill, update maxArea with the size you counted. Each cell gets visited a constant number of times, so it's O(rows*cols) time. The common pitfall is recursion depth. The grid can reach 300 by 300, which is 90,000 cells, and a recursive DFS on one giant island can blow the stack in some languages. Use an iterative BFS with a queue or an explicit stack. Other mistakes: counting diagonals, forgetting to mark cells on push instead of on pop, and returning [0,0] wrongly. If you freeze on the live OA, StealthCoder is the hedge that surfaces the working flood fill.

Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.