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

this looks like some hard graph theory

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.

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

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updated description

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.

The check for palindromes function reveals part of the solution. Im not sure how to import it into the example test cases without being accessable

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nice solution :)

how are you so fast 0-0. anyway, fixed

floor division. ex. 800 / 400 will return 2.0, which is a float number

sorry. what?