The entire string shouldn't necessarily be lowered either; chars could be lowered as you go. It's pointless lowering the whole of aABCDEFGHIJKLMNOPQRSTUVWXYZ if you're going to detect a non-isogram on str[1]

"but not so fast as char comparison" - explain better what you mean by this

That is true, I actually looked for an Idea (somehow I tried the most upvoted answer but for some reason, my counter didn't increased/decreased, idk why honestly lol)

But anyways, after trying, I looked for some answer and found the same problem but more complex (with other kind of braces) and it was cool, seeing the same problem but with a stack.

That enables code to exist to save from further iterations.

Indeed, but this solution would be easily extensible to one validating nesting of different types of brackets.

Allocating a stack increases space complexity more than necessary. A single int variable can be used for a more space efficient solution.

From the description "There will always be only one integer that appears an odd number of times."

Your sequence does not fulfil this promise. Integers 5 and 11 both appear an odd number of times (3 and 1 respectively).

Nice :)