Maximum Damage Patch
Reported by candidates from Gecko Robotics's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The Gecko Robotics OA reported in September 2023 is titled "Maximum Damage Patch," but strip the theme and it's a max island area problem on a binary grid. The trap isn't the algorithm. It's the input that makes a naive solution fall over: an all-water grid, a single-row grid, or a 300 by 300 grid of all ones. That last one is where recursion gets ugly. If you've seen flood fill before, you'll finish fast. If you blank on the traversal, StealthCoder is the safety net running invisibly during the live OA. Here's the script.
The problem
Given a rectangular binary matrix grid, return the maximum area of an island. An island is a maximal group of cells containing 1 connected horizontally or vertically. Its area is its number of cells. Return 0 when the matrix contains no land. Function maxIslandArea(grid: int[][]) → int Examples Example 1 grid = [[0,0,1,0],[1,1,1,0],[0,1,0,1]] return = 5 The center island contains five orthogonally connected land cells. The bottom-right cell is a separate island of area 1. Example 2 grid = [[0,0],[0,0]] return = 0 There are no land cells. Constraints 1 <= grid.length, grid[i].length <= 300. Every row has the same length. grid[i][j] is either 0 or 1.
Reported by candidates. Source: FastPrep
Pattern and pitfall
The trick is flood fill with a running count. Scan every cell. When you hit a 1, start a DFS or BFS, count the connected cells, mark them visited as you go, and keep the max. Mark visited by flipping the cell to 0 so you skip a separate visited array. Only four directions count, up, down, left, right. Diagonals don't connect, and example 1 proves it with the bottom-right cell. The pitfall is depth. A 300 by 300 grid of ones means up to 90,000 nested recursive calls, which can overflow the stack in some languages. Use an explicit stack or a BFS queue if you're unsure. Initialize your max to 0 so the no-land case returns correctly. Total work is O(rows * cols). If you freeze mid-assessment, StealthCoder can hand you the iterative version while you keep typing like nothing happened.
Memorize the pattern. If you can't, run StealthCoder. The proctor sees the IDE. They don't see what's behind it.
You can drill Maximum Damage Patch cold, or you can hedge it. StealthCoder runs invisibly during screen share and surfaces a working solution in under 2 seconds. The proctor sees the IDE. They don't see what's behind it. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge.
Get StealthCoderRelated leaked OAs
This OA pattern shows up on LeetCode as max area of island. If you have time before the OA, drill that.
You've seen the question.
Make sure you actually pass Gecko Robotics's OA.
Gecko Robotics reuses patterns across OAs. Made by an engineer who treats the OA as theater. If yours is tonight, you don't have time to grind. You have time to hedge. Works on HackerRank, CodeSignal, CoderPad, and Karat.
Maximum Damage Patch FAQ
How hard is the Gecko Robotics Maximum Damage Patch problem really?+
It's medium at most. It's a standard connected-components count on a grid. The theme name sounds exotic, but the actual task is returning the largest group of orthogonally connected 1s. If you've done any grid flood fill, expect to finish quickly.
What's the trick to solving it?+
Run DFS or BFS from each unvisited land cell, count the cells in that component, and track the maximum. Flip visited cells to 0 so you never count them twice. Check four directions only. The whole solution is about 15 lines.
What edge cases should I test before submitting?+
Test an all-zero grid, which must return 0. Test a single cell grid with a 1. Test a single row or column. Test a full grid of ones at 300 by 300 for stack depth. Also confirm diagonal-only neighbors are counted as separate islands, like in example 1.
Should I use recursion or an explicit stack?+
Prefer an explicit stack or BFS queue. With up to 300 by 300 cells, a worst-case all-land grid produces about 90,000 levels of recursion, which risks overflow in Python and some other languages. Iterative traversal avoids the risk and costs almost nothing extra to write.
How do I prepare for this in 48 hours?+
Write the flood fill from scratch two or three times until the direction loop and bounds check are automatic. Then run it against both examples and the edge cases above. Learn the pattern rather than memorizing code, because the grid and theme change from one assessment to the next.