Minimum Cleaning Robot Runs
Reported by candidates from HSBC's online assessment. Pattern, common pitfall, and the honest play if you blank under the timer.
The mistake that sinks a first attempt on this HSBC problem is counting every room instead of only the dirty ones. It was reported in September 2026, and it's a grid connected-components question dressed up as a robot story. You get a plan of '.', '*' and '#' cells, and you return how many rooms contain at least one dirty field. Example 3 returns 0 for a reason. If you blank on the traversal during the live OA, StealthCoder can sit invisibly on your screen as a safety net, but the logic here is short enough to own tonight.
The problem
A building plan is represented by a rectangular array of strings. Each field contains one of three symbols: . is clean floor. * is dirty floor. # is a wall. A room is a maximal group of non-wall fields connected in the four cardinal directions. In one run, a cleaning robot starts in one room and cleans every dirty field in that room. It cannot cross a wall or leave the plan. Return the minimum number of robot runs needed to clean every dirty field. Function solution(plan: String[]) → int Examples Example 1 plan = ["..#..",".*#*.","..#.."] return = 2 The vertical wall separates the floor into two rooms. Each room contains a dirty field, so the robot needs one run in each room. Example 2 plan = ["....",".##.",".*#.","...."] return = 1 All non-wall fields are connected around the wall block. The single dirty field is therefore cleaned in one run. Example 3 plan = ["###","#.#","###"] return = 0 The only floor field is already clean, so no robot run is required. Constraints 1 <= plan.length and every row has the same positive length. plan.length * plan[0].length <= 200000. Every character is., *, or #.
Reported by candidates. Source: FastPrep
Pattern and pitfall
Treat each non-wall cell as a node and connect the four neighbors. Scan the grid once. When you hit an unvisited non-wall cell, flood fill its whole room with BFS or an iterative DFS, and track a flag for whether you saw a '*'. If the flag is set, add one to the answer. The pitfall is counting all components, which fails Example 3 and any grid with clean-only rooms. The second pitfall is recursion depth. With up to 200000 cells, a recursive DFS can overflow the stack, so use an explicit stack or queue. Mark visited in a boolean array or mutate a copy. Time is O(rows*cols). If the traversal slips your mind mid-assessment, StealthCoder is the hedge that hands you the working version in real time.
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Minimum Cleaning Robot Runs FAQ
What's the trick in the HSBC Minimum Cleaning Robot Runs problem?+
Count connected components of non-wall cells, but only those containing at least one '*'. Clean-only rooms cost nothing. Flood fill each unvisited room, track whether a dirty cell appeared, and increment the answer only when it did.
How hard is this really?+
Easy to medium. It's the classic number-of-islands pattern with one extra flag. If you've written a grid BFS before, it takes about ten minutes. The only real risks are counting clean rooms and blowing the stack with recursion.
Should I use BFS or DFS?+
Either works. Use BFS with a queue or DFS with an explicit stack. With up to 200000 cells, a deep recursive DFS can crash on a snake-shaped room, so avoid plain recursion unless you're sure about the stack limit.
What edge cases should I test?+
A grid with no dirty cells returns 0, like Example 3. Test a single row, a single column, an all-wall grid, and a room wrapped around a wall block like Example 2, where one run covers everything because the floor is connected.
How do I prepare for this in 48 hours?+
Write the grid flood fill from scratch twice. Use a direction array, a visited grid, and a queue. Then add the dirty flag. Run the three examples by hand. That covers this problem and most of its variants.