CareerCup

Base 26 representation from 1 to 26.

void print(int n)
{
	if(n>0)
	{
		n--;
		print(n/26);
		printf("%c",(n%26)+65);
	}
}

nerd's solution is almost correct, just need to take care of some offset

template<class T>
string Translate(T input) {
  static const char kLookUp[26] = {'A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'};
  string res = "";
  while (input > 0) {
    T idx = (input - 1) % 26;
    res = kLookUp[idx] + res;
    input = (input - 1) / 26;
  }
  return res;
}

Btw, the team I interviewed with uses C/C++ predominantly hence I mentioned in the beginning to write a code in C.

Convert the decimal Base Number to 26 Base Number whose digits are A-Z

Following is the recursive method to change a number into base 26 and then converting
the resultant number to string.

Recursive because: I wanted to print the last remainder's corresponding char.
This because recall that in binary conversion of decimal number, the last remainder obtained was the most significant.
There can be other versions along with iterative equvalents of it.

void convertToAlpha(long long int num)
{
	long long int q = num;
	long long int r = 0;
	if (num < 1)
		return;
	r = q % 26;
	q = q / 26;
	if (r == 0)
	{
		q = q - 1;
		r = 26;
	}
	if (q == 0)
		printf("%c", (char) (r + 'A' - 1));
	else
	{
		convertToAlpha(q);
		printf("%c", (char) (r + 'A' - 1));
	}
	return;

}

This comment has been deleted.

we need to do %26 but need to take care that 0 has no representation over here....

i.e. when we do % 8 out values are from 0 to 7

but here 0 has no representation ... so every time remainder is 0 we take it as 26 and print 'Z' and subtract 1 from Quotient ....

#include<stdio.h>
#include<math.h>
 
int main()
{
   long long int x = 704;
   char *ptr = (char *)malloc(sizeof(char)); 
 
   // we are skipping any of the decimal points 
   //from input... i.e. error handling is not there
   //right now

   int i=0;
   while(x > 0)
   {
        int r = x%26;
        x = x/26;

        if(r==0)
        {
                *(ptr+i)='Z';
                 i++;
                x = x-1;
        }
        else
        {
                *(ptr+i)=(char)r+'A'-1;
                i++;
        }
 
   }
   *(ptr+i)='\0';
for(i=i-1;i>=0;i--)
   printf("%c",*(ptr+i));
return 0;
}

/* Disclaimer: This Solution will work upto 2-characters i.e. from A,B,C ------ upto: ZZ */

#include <stdio.h>

#define INVALID_VAL -1

int main()
{
int numb = INVALID_VAL;
char array_alphabets[26] = {'A','B','C','D','E','F','G','H','I','J','K',
'L','M','N','O','P','Q','R','S','T','U','V',
'W','X','Y','Z'};

int Q = INVALID_VAL;
int R = INVALID_VAL;

int row_index = INVALID_VAL;
int col_index = INVALID_VAL;

printf ("\nEnter number: ");
scanf ("%d", &numb);

/* Calculate the position of this number in the Hypothetical 2-D Array
* Hypothetical 2-D Array:
* Row-0: A | B | C | D | .......................... | Z |
* Row-1: AA | AB | AC | AD | .......................... | AZ |
* Row-2: BA | BB | BC | BD | .......................... | BZ |
*
*/

/* Algorithm to determine the Index of Number in Hypothetical 2-D Array:
*
* Step-1: Divide Number by 26, collect the Quotient (Q) & Remainder (R)
Example: Number = 55

55/26 => Q = 2, R = 3

Q => Row No. in Hypothetical Array
R => Column No. in Hypothetical Array

* Step-2: Allocate an Actual 1-D Array of 26 characters

| A | B | C | D | ......................... | Z |

Row-Index of Hypothetical Array = Q-1; [Map to 1-D Array]
Column-Index of Hypothetical Array = R-1; [Map to 1-D Array]
*/


if (numb>26)
{
Q = numb/26;
R = numb%26;

row_index = Q-1;

if (0==R)
{
col_index = 0;
}
else
{
col_index = R-1;
}

printf ("\nAlphabet Equivalent to the Number %d is: %c%c\n", numb,array_alphabets[row_index],array_alphabets[col_index]);
}
else if (numb<=26)
{
printf ("\nAlphabet Equivalent to the Number %d is: %c\n", numb,array_alphabets[numb-1]);
}

return 0;
}

This is the same question as posted here: careercup.com/question?id=14097726
See my implementation at the link above.

#include "stdafx.h"
#include <string.h>

void reverseOutput(char *output)
{
int last = strlen(output);
int first=0;
while(first < last )
{
char str = *(output+first);
*(output+first) = *(output+last-1);
*(output+last-1)=str;
last--;first++;
}
}


int _tmain(int argc, _TCHAR* argv[])
{
long int input,store;
printf("Enter the number: ");
scanf("%ld",&input);

store = input;
char output[100];
int j=0;
memset(output,0,100);
while(store != 0)
{
char c = (char)(((store - 1)%26)+97);
output[j++] = c;
store = (store-1)/26;
}
reverseOutput(output);
printf("%s\n",output);
return 0;
}

import java.io.*;
class A
{
   int i=0,j=-1,x,m=1;
    
  void intToAplha(int x)
     {

        int n=0,k=0,counter=0;

     
       while(true)
         {

            n=i*26;

            if(x>n)
              {
                i++;
                j++;
                 
              }
             else
                 break;

          }


      while(true)
       {
           k=702*m ;
            if(x>k)
              {
                 counter++;
                  j=j%26;
                 m++;
              }
            else
              { 
                break;
              }
         }

        System.out.print("The Equivalent String: ");
 
        if(counter>0)
          {
              for(int t=0;t<counter;t++)
                 System.out.print("A");
              if(j==0)
                System.out.print("Z");
           } 
     

        n=x%26;

        if(j>0)
         {
              System.out.print((char)(64+j));
              
              if(n==0)
               System.out.print("Z");
              else
                System.out.print((char)(64+n));
          }
        else
          {
             if(n==0)
               System.out.print("Z");
             else
                 System.out.print((char)(64+n));
          }

       
 
    
  

    
 }

void input()
{
    try{
         BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
     
         System.out.print("Enter Int Val : ");
          String s=br.readLine();

          x=Integer.parseInt(s);
          
         
          br.close();  
       }
     catch(Exception e){System.out.println(e);}

       
 
       intToAplha(x);
 }
      

      
    

public static void main(String args[])
 {
    A a=new A();
      
      a.input();
      
 }
}

by firoz

void conv(unsigned long int x)
{
unsigned long int y=x%26;
unsigned long int z=x/26;
if(z>1)
{
if(y==0)
ola(z-1);
else
ola(z);
}
else
if(z==1)
if(y!=0)
ola(z);
if(y==0)
y=26;
printf("%c",(y+64));
}

i tried test cases those obvious ones and result is satisfactory. this is simplest and complete code for the given problem. let me know if it voilate some test cases.

void encrypt(int n)
{
	int r,q=n;
	char *res="";
	while(q>0){
		r=q%26;
		if(r==0){
			q=q-1;
			r=26;
		}
		if(r>0 && r<=26){
			res=appendAtLeft(res,r);
			if(q>0){
				q=q/26;	
			}
		}
	}
	printf("%s\n",res);
}

find the length assign A = 1, B =2,.... Z=26
AAA = 703 = A*(26 square) + A*(26 power 1) + A*(26 power 0) = 1*(26 power 2) + 1*(26) + 1
ZZZ = 26 *( 26 power 2) + 26 *( 26 power 1) + 26 *(26 power 0)

public static string ConvertToAlphabet(int number)
        {
            int temp = number;
            char[] alphabets = new char[] { 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z' };
            string Alphabet = string.Empty;
            int Index = 0;

            if (number <= 0)
            {

                Console.WriteLine("input number {0} is invalid", temp);

                return string.Empty;
            }
            
            if (number < 27)
            {

                Console.WriteLine("input number {0} , output alphabet {1}", temp, alphabets[number - 1]);

                return alphabets[number - 1].ToString();
            }
            while (number > 0)
            {
                Index = number % 26;
                number = number / 26; ;

                if (Index == 0)
                {
                    Index = 26;
                    number = number - 1;
                }
                Alphabet = alphabets[Index - 1] + Alphabet;

            }
            
            Console.WriteLine("input number {0} , output alphabet {1}", temp, Alphabet);
            return Alphabet;
        }

input
Microsoft.ConvertToAlphabet(-11);
Microsoft.ConvertToAlphabet(1);
Microsoft.ConvertToAlphabet(26);
Microsoft.ConvertToAlphabet(52);
Microsoft.ConvertToAlphabet(28);
Microsoft.ConvertToAlphabet(702);
Microsoft.ConvertToAlphabet(703);
output

input number -11 is invalid
input number 1 , output alphabet A
input number 26 , output alphabet Z
input number 52 , output alphabet AZ
input number 28 , output alphabet AB
input number 702 , output alphabet ZZ
input number 703 , output alphabet AAA

public static String convertToInt(int num)
{
String alpha ="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char character = 0;String res ="";
while( num > 0)
{
int id = ( num-1)%26;
res = alpha.charAt(id)+res;
num = (num -1)/26;}

return res;
}

int x,y;
y= n-1;
while(n>0)
{
n = n-1;
if(x = n/26)
printf("%c",((x-1)%26)+65);
else
printf("%c",(y%26)+65);
y = n%26;
n = n/26;
}

Try this one:
void convertAphabet(int intx, int base)
{
int inttx = intx;
if(inttx/base > 26)
{
if(inttx%(26*base) == 0)
{

convertAphabet(inttx-26*base, base*26);
putchar('Z');
}else
{
convertAphabet(inttx, base*26);
putchar('A' + (inttx/base)%26 -1);
}
}else
{
putchar('A' + (inttx/base -1));
}
}

public class ConvertToAlphabet {
	static char[] array;

	static {
		array = new char[27];
		for (int i = 1; i <= 26; i++) {
			array[i] = (char) (i + 96);
		}
	}

	public static String convert(int n) {
		if (n <= 0) {
			return "";
		}
		int remainder = n % 26;
		if (remainder == 0) {
			return "z" + convert((n - 1) / 26);
		} else {
			return array[remainder] + convert(n / 26);
		}
	}

	public static void main(String[] args) {
		System.out.println(convert(702));
	}
}

char* ConvertToAlphabet(int x)
{
        char* str = NULL;
	int i = 0;
	int j;

        str = (char* )malloc(N*sizeof(char));
	memset(str, N, 0);

	do {
	    x--;
	    str[i++] = x%26 + 'A';
	} while ( x /= 26 );

	str[i] = '\0';

	printf("str: %s \n", str);

	Reverse(str);

	return str;
}

a similar code here

codes-to-problem.blogspot.in/2012/07/wap-to-return-numbered-index-if-input.html

easily you can call the function like this:

printf("%s", number2alpha(x));

char *number2alpha(long long int x)
{
	char *s;
	s=(char *)malloc(sizeof(char) * (x/26)+2 );
	int i=0; 
	while(x>0)
	{
		x--;
		int k=x%26;
		x/=26;
		s[i++]=k+'A';
	}
	s[i]='\0';

	return strrev(s);
}