I agree with that this solution is not right.

Yeah, my understanding of string.count is that any arguments it's given are basically reduced to a unique set of letters that are in all arguments.

So whether the search_text is 'c' or 'cc' it will return the same number.

This works for the third test because all 'c's are part of the substring 'cc', but as shown by lzw120's example, if you search 'bb' but have isolated 'b's as in 'baa_bb_cc_bdd_bb_eb' this solution will give you all instances of 'b' and then divide by 2, the length of 'bb'.

Would suggest adding a test to account for this, like:

Test.expect(solution('ccddeeccddeecc', 'cd') == 2 #This solution would return 5, the total number of c's and d's (10) divided by the length of 'cd' (2).