Count Straight-Line Word Occurrences in a Grid

Given a rectangular grid of characters and a nonempty word, count every occurrence that lies entirely in one row or one column.
Search left to right, right to left, top to bottom, and bottom to top. Matches may overlap and characters are compared exactly. When the word has length one, count each matching cell once rather than once per direction.

Function
countStraightWords(grid: String[], word: String) → int

Examples
Example 1
grid = ["ABCA","BCAB","CABC"]
word = "ABC"
return = 4
There are two left-to-right row matches and two top-to-bottom column matches.
Example 2
grid = ["AAAA"]
word = "AAA"
return = 4
Two overlapping matches appear in each horizontal direction.
Example 3
grid = ["AX","YA"]
word = "A"
return = 2
Each matching cell is counted once for a one-character word.

Constraints
1 <= grid.length, grid[i].length <= 500.
All rows have equal length.
1 <= word.length <= 500.
The grid and word contain visible ASCII characters.

Skip BFS. Nothing branches, so there's no graph to traverse. Treat each row and each column as a string. For every one of them, count overlapping matches of the word going forward. Then reverse the word and count again, which covers right to left and bottom to top. Check Example 2: AAAA with AAA gives 2 forward and 2 backward, so 4. The pitfall is double counting. If the word is a palindrome, the forward and reverse scans find the same positions, but the statement still counts each direction, as Example 2 shows. The special case is length one, where you count matching cells once. With a 500 by 500 grid and a 500-length word, naive comparison costs about 500*500*500*4 character checks, which is fine. Use KMP only if you want to be safe. Slide the start index by one so overlaps count. If you freeze under the timer, StealthCoder is the hedge that hands you the loop structure live.

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