Find Robots by Blocker Distances

Given a non-empty rectangular grid grid and a four-element array query, return the coordinates of every robot whose nearest-blocker distances match query.
Each grid cell is one of:
O: a robot.
E: an empty cell.
X: a blocker.
The query order is [left, top, bottom, right]. In one direction, the distance is the number of steps from the robot to the first X cell. If there is no blocker before the grid edge, use the number of steps to the first out-of-bounds position. The blocker or out-of-bounds step is included in the distance.
Return zero-based [row, column] pairs in row-major order.

Function
findRobots(grid: String[], query: int[]) → int[][]

Examples
Example 1
grid = ["OEEEX","EOXXX","EEEEE","XEOEE","XEXEX"]
query = [2,2,4,1]
return = [[1,1]]
The robot at [1,1] is two steps from the left boundary, two steps from the top boundary, four steps from the bottom boundary, and one step from the blocker on its right.
Example 2
grid = ["O"]
query = [1,1,1,1]
return = [[0,0]]
Every direction reaches the first out-of-bounds position in one step.
Example 3
grid = ["XOXOX","XXXXX","XOXOX"]
query = [1,1,1,1]
return = [[0,1],[0,3],[2,1],[2,3]]
Each robot is immediately surrounded by blockers or a grid boundary, so all four coordinates match and are returned in row-major order.

Constraints
1 <= grid.length <= 200.
1 <= grid[i].length <= 200.
Every row has the same length.
Every cell is exactly O, E, or X.
query.length == 4.
Every query distance is positive and at most max(grid.length, grid[0].length) + 1.

The trick is precomputing four distance tables instead of walking from every robot. Sweep each row left to right, tracking the column of the last X seen, starting at -1 to represent the out-of-bounds position. For a cell at column c, left distance is c minus that last blocker column. Sweep right to left for the right distance, with the sentinel at width. Do the same down and up the columns for top and bottom. Then loop in row-major order, and for every O cell compare the four values to the query. Row-major iteration gives you the required output order for free. The pitfall is the off-by-one: the blocker or boundary step counts, so a robot at column 0 has left distance 1. Example 2 confirms it. Also remember the query order is left, top, bottom, right, not the clockwise order. If the live OA rattles you, StealthCoder is the hedge that catches a botched sentinel.

Drill it cold or hedge it with StealthCoder. Either way, don't walk into the OA hoping you remember the trick.