alternatively you could mod by 1e9 + 7. https://www.geeksforgeeks.org/modulo-1097-1000000007/

this looks like some hard graph theory

This is the best kata about palindromes, well done.

great, points 1 and 2 are most important

1. I will probably revamp the test cases later to make this a more constructive kata and gradually test each edge case.
2. I'll revert back to a lower constraint (<= 200) seeing how its a pain to scroll through 1000 characters (the point was to avoid people from hard coding)
3. Ill leave it up to future testers to vote if n, m = 1 is enough to warrant new edge cases. The main objective of this kata is within the amount of x and y palindromes.

• Either add more edge cases in small fixed tests or make sure the first batch of random tests uses small posters with lots of edge cases.
• I see no purpose in having big constraints, other than to make debugging harder.
• It's a shame you left out a bunch of edge cases with n or m < 2

I have specified 2 <= n, m <= 1000 and 0 <= x, y <= n, m and also raised the constraints

Input range is not specified, and it is implicitly assumed that 2 <= n, m <= 100 and 0 <= x, y <= n, m. There is a whole class of edge cases at n = 1/m = 1 but currently it's not tested.

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updated description

I think you mean "if the entire row/column is a palindrome".

i will specify a few things

• A 2d array of characters
• contains the lowercase English alphabet a - z
• Every letter in a row (left to right) or column (top to down) will form a string
• That string will be counted as one palindrome only if it is the same when read forwards and backwards.