CareerCup

I think doing % and / is not the correct approach to this, the algorithm basically work, but if we consider a number with millions of digits, the datatype should be some sort of linked list containing a single digit peer bucket, in that way the "big" number can grow to million digits, so in my opinion the correct way should be visit from back to beginning the linked list an put each digit into a string buffer then return it.

Can you provide the method's signature to perform this task? In what form is the integer provided?

how the integer presented to us? Could be its own datatype with huge memory used to store integers.
int lastdigit= number mod 10;
number = number/10;
use switch case to determine character value of lastdigit.
Store the character in a stack. Do this until the number is 0.
Empty the stack into a string.

string convertInt(int number)
{
if (number == 0)
return "0";
string temp="";
string returnvalue="";
while (number>0)
{
temp+=number%10;
number/=10;
}
for (int i=0;i<temp.length();i++)
returnvalue+=temp[temp.length()-i-1];
return returnvalue;
}

Could you give an example.So that we can understand in a good manner

Well.. I also have a doubt on this question.. if it is what U explained above.. all you are doing is pointing to the memory location where the integer is and framing it to string or char *.

Probably the acsii string means an equivalent character string .. is it so ?

Do we need to consider extended AScii codes too which goes till 255 or normal ascii which
spans till 127.For any of the case convert them into first that base number i.e for non-extended vesion, make 1 million digits to a number equivalent to base 128,and then group them from left.

in decimal or hex format string. It is easy for hex.

/* Basic prototype implementation in C.*/

#include<stdio.h>
#include <string.h>

void main() {
int number = 12345678;
int data = number;
int index = 0;
char ch[16];
char * chp = ch;
memset(chp, 0, sizeof(ch));

while(number != 0) {
*chp++ = number%10 + 48;
number = number/10;
index++;
}
printf("Ascii ver of number %d is \n", data);
for(index=15; index>=0 ; --index)
printf("%d ", ch[index]);
printf("\n");
return;
}

Read carefully: how would you convert this integer to an ASCII string
It does not say it has to be DEC string. You can just convert it to a HEX or OCT ASCII string.

since the integer has millions of digits, we can divide it by 1000 or more, and base on the remainder of 1000, we can use this remainder as an index of our pre-stored string-array to get the corresponding string, it sacrifices the memory for time.

How will you represent 'this very very long integer'? is it going to be a bit string of size needed to represent a number?

32 bit represent a maximum of 10 digit number, so for a number of with 1 million digit what representation would be best?

One way could be to go for BCD representation, each byte representing a decimal from 0 - 9 or group of 4 bits representing a decimal from 0-9 (10,11,12,13,14,15 ignored).

In either case we need to know size of a byte stream representing a number. Now if we know number of bytes and way of representation we can compute size of a string needed beforehand. In first case, it will be same but for second it will be twice.

The way of conversion is now trivial. Just run through a byte array, interpreting entire byte or half-byte depending on a representation of long long type and then fill character array with it as usual ('0'+number)?

let me know if you think this won't work:
suppose this data is saved in RAM in a big endian system, then part of this big number can be like "beef1234", what it is saved in RAM will be like this "be " take 1st byte, "ef" take 2nd byte,12 take 3rd byte, 34 take 4th byte, then

so we make a union:
union byteto4bit
{
unsigned char data;
struct 4bits{
top4bit:4;
btm4bit:4;
} Bits;
}
}

then we can access like this
while(i<sizeof(BigInteger))
{
union byteto4bit g.data=*(pBigInterger+i);
itoa(g.Bits.top4bit);
itoa(g.Bits.btm4bit);
}

void convert(long long number)
{
while(number)
{
long long tmp = nombre & 255;
cout << tmp << endl;
nombre = nombre>>8;
}
}

I think they are testing how good you know C.

char str[];
long long value =123456789101112;
sprintf(str,"%ld",value);

This solution should be sufficient.

I think this question is asking given a integer number, say 101102103, output "ABC", since A is 101 in the ascii table...

As long as the solution goes, assuming the input integer comes in the reverse order (this is the simple version) and each digit is in a node of a linkedlist.
public static String findString(node head){
int count = 0;

and

String result = "";

and

String buffer = "";

and

while(head != null){

and

and

char thisDigit = head.value;

and

and

count ++;

and

and

buffer = thisDigit + buffer;

and

and

if(count == 3){

and

and

and

result = (char)(Integer.parseInt(buffer)) + result;

and

and

and

count = 0;

and

and

and

buffer = "";

and

and

}

and

and

head = head.next;

and

}
}