replace it with the amount of times you've already seen it

First, English uses amount for the sort of quantities for which you would use real (floating point) numbers. For traditionally countable things, or units of continuously variable things, use the word number. (Examples: an amount of water, a number of bottles of water, an amount of rain, a number of centimeters of rain, a number of raindrops, etc.) As an aside, the same distinction applies to how much vs. how many, with many corresponding to a number.

Second, "the number of times you've already seen it" starts at zero. because "have already" implies the exclusion of this time. OK, you specifically force the first time to get a 1, but the second time it appears, I have already seen it just once, and yet you expect a 2. It's possibly confusing. From the examples, it becomes obvious, but then also it appears that the distinction between the first occurrence and every following occurrence is unnecessary. The line becomes consistent and agrees with the examples if you only change it to:

For each character in the string, replace it with the number of times that you have seen this same character [ now | as of now | thus far ] since the beginning of the string.

and/or something such as

From the beginning of the given string, change each character's Nth occurrence/appearance to that number N.

then the examples make clear it means 1-indexed and not 0-indexed.

Прохожу тесты, выводя время выполнения каждого из них, суммирую все полученные значения, получаю 9224. CodeWars даёт: Execution Timed Out (16000 ms), почему 9224 == 16 000 и почему так не считают зарплату? Решал на java

This kata is evil, cool but hard 😥

I found some testcases, where the Codewars environment shows different results than my program:

number of walls: 6
walls = (
[ (Point(6, 2),Point(7, 1))
, (Point(7, 4),Point(6, 3))
, (Point(1, 2),Point(3, 0))
, (Point(2, 1),Point(0, 3))
, (Point(9, 1),Point(8, 2))
, (Point(6, 4),Point(7, 5))
])
width = 10
height = 5
expected = False
Your answer: True

Here the 2 overlapping edges (Point(1, 2),Point(3, 0)) and (Point(2, 1),Point(0, 3)) form a partition with upper and left boundary walls.

Another testcase:

number of walls: 11
walls = (
[ ((22, 7),(19, 5))
, ((21, 6),(21, 4))
, ((24, 7),(24, 6))
, ((18, 8),(18, 8))
, ((17, 8),(17, 8))
, ((23, 8),(24, 9))
, ((17, 9),(17, 7))
, ((19, 8),(16, 8))
, ((16, 9),(14, 9))
, ((20, 9),(17, 7))
, ((18, 9),(17, 6))
])
width = 40
height = 16
expected = False
Your answer: True

There are 2 partitions in the center.

Thank you for this great kata. I learned a lot about how to handle graphs in python.

Haskell translation

remember to close your question when it is answered ;-)

Thank you, I was struggling with it and you cleared my mind with that insight!

see here

I think that may be the intended lesson of the kata (locality of reference for row-major matrices which makes more efficient use of the cache)

Hey, I have an issue and is I am not able to get the final test to work since I always get when I return the empty array, but when I add something to it there is no problem at all. I am really curious to know if I am getting something wrong and what could it be, can someone give me a little help?

C translation

nah this is more like a 7