How to implement backspace ?

Well what I am confused is that, how can the two languages be mixed :( ..in the same mode ..or am I being asked to detect the mode and then set the corresponding ascii value or kanji value ?

Thanks ozo for such prompt reply

H

"The most significant" means "the first" here. Thus, you should compare the byte before the last one to 128. If it is greater, you are dealing with a Kanji character and should delete two last bytes. Otherwise, you should delete only the last byte.

Now I am even more confused , should I have to detect whether it is a little or big endian and then accordingly read the MSB ?

Thanks
H

You're mixing up. It has nothing to do with endianess. It's all about the binary.

Bytes from 00000000 to 01111111 are ASCII.
Bytes from 10000000 to 11111111 AND the ones after those are Kanji.
binary(01111111)=decimal(127).

If byte value is 128 or more, it's the first byte in two-byte Kanji character sequence. That's the only thing you need to know.

You said "Now you are given an array of a characters "

create a new array one element smaller
load from the original array elemet 1 thru n-2
n being your index
load from the original array elemet thru the end

set old array = new array..

done

You said "Now you are given an array of a characters "
fixed a couple of typo's

create a new array one element smaller
load from the original array elemet 1 thru n-2
n being your index
load from the original array element n thru the end

set old array = new array..

done

If this is the actual interview question, it's ambiguous. It's always good to point that out during an interview. If it's an "array of characters", "characters" being defined just above as having variable lengths, then cdbeste has the answer. However, this is very unlikely. The question should read: "you are given a string of bytes, encoding characters (both ASCII and Kanji), and an index into the array." This is confirmed by the next sentence: the byte will be the start of variable-length encoding (on or two bytes). In that case, ozo was (of course) right from the start.

The second thing to point out is that the described encoding is *not* Unicode. It resembles a subset of UTF-8, but the standard allows up to four bytes per character, not two, and there are some special "noncharacters" to consider. For an interview, it would be good to read up on Unicode anyway.

http://en.wikipedia.org/wiki/Unicode
http://ja.wikipedia.org/wiki/Unicode (utf-8 encoded, btw)

(°v°)

Try looking at it like this. As ozo says, take the previous two bytes:

They can be any of:

0xxxxxxx 0xxxxxxx
0xxxxxxx 1xxxxxxx
1xxxxxxx 0xxxxxxx
1xxxxxxx 1xxxxxxx

Now you know that you are at the start of a 'character' so the two previous bytes must be one of:

<Ascii><Ascii>
<Kanji2><Ascii>
<Kanji1><Kanji2>

We know nothing about the structure of <Kanji2> but we do know that

<Kanji1> = 1xxxxxxx
<Ascii> = 0xxxxxxx

So

<Ascii><Ascii> = 0xxxxxxx 0xxxxxxx
<Kanji2><Ascii> = ?xxxxxxx 0xxxxxxx
<Kanji1><Kanji2> = 1xxxxxxx ?xxxxxxx

Or

0xxxxxxx 0xxxxxxx = <Ascii><Ascii> or <Kanji2><Ascii> -> Delete 1 byte!
0xxxxxxx 1xxxxxxx = impossible! -> Throw an exception!
1xxxxxxx 0xxxxxxx = <Kanji2><Ascii> or <Kanji1><Kanji2> -> Hmmm???????
1xxxxxxx 1xxxxxxx = <Kanji1><Kanji2> -> Delete 2 bytes!

So either we need at least 3 bytes and ozo is wrong or I've got it wrong.

Paul

Paul,
        I think you are right. I swear I had a hunch that it needed 3 bytes but since ozo
said that I did not give it a thought anymore...I think you are right.

You gave the correct solution :-)

Thanks,
-H

And, worse still, I think you have to go all the way back to the start of the buffer if you see the sequence 01010101... (going backwards).

Paul