WAP to Delete a Character before the Given Index in Character Array (You Have Given Array of Ascii & kanji Characters)

First some definitions for this problem: a) An ASCII character is one byte long and the most significant bit in the byte is always '0'. b) A Kanji character is two bytes long. The only characteristic of a Kanji character is that in its first byte the most significant bit is '1'.

Now you are given an array of a characters (both ASCII and Kanji) and, an index into the array. The index points to the start of some character. Now you need to write a function to do a backspace (i.e. delete the character before the given index).

We have to write an algorithm & a quality working code for this.


Explanation

1. Look at MSB of the previous character. If 1, then delete the previous two bytes. This is because an MSB of 1 can only be part of a Kanjii character.
2. If the MSB is 0 then all you know is that the previous byte ENDs a character. The previous byte could be an Ascii character, the single byte is the start and the end of the character OR the previous byte is the second byte of a Kanjii character. Look the following figures which show the characters to the left the given character, denoted by ‘|’.

xxx10|xxxxx. You cannot decide on the previous character based on the scan so far because there are two possibilities after this
1. xx010|xxxxx - The previous character is a Kanjii character. After first 0 from left we are left with ‘10’ which is Kanjii.
2. xx110|xxxxx - We have to scan even further. Leads to two choices
x1110|xxxxx - We have to scan even further.
x0110|xxxxx - after leftmost 0, we have ‘110’ which break into ‘11’ and ‘0’. So the previous character is Ascii.

Continuing on this logic you will notice that you have to first find the end byte of a previous character and the only way to do that is to find the previous ‘0’ OR start of the file. After you have found the previous end byte of a character, you count the number of ‘1’ in between the 0 before the cursor and the previous end byte.
1. If number of ‘1’s is even then the previous character is Ascii so the backspace should delete one byte.
2. If the number of ‘1’s is odd then the previous character is Kanjii and two bytes should be deleted.


Data Structure Used: Array

Algorithm

Working Code(In progress)


Time Complexity
Space Complexity
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