CareerCup

void numchars(int nth)
{
        int n = 1;


        while(nth > pow(26,n))
        {
                n++;
        }


        cout<<"num of chars="<<n<<endl;
}

int main()
{
        numchars(26);
        numchars(676);
        numchars(800);
        return 0;
}

void Find(int num)
{
	char a[26]="ABCDEFGHIJKLMNOPQRSTUVWXYZ", result[1000]={'\0'};
	int i=0, n=num;
	if(n <= 0)
	{
		printf("invalid number");
		return;
	}
	while(n)
	{
		result[i++]=a[n%26 - 1];
		n/=26;
	}
	printf("Number of characters in %d th member: %d\n",num, i-1);
	printf("%d th member: ");
	for(i--;i>=0;i--)
	{
		printf("%c",result[i]);
	}
}

Approach 1:

Divide the number 'n' with 26^1, if the remainder is <26 then return 1. if it is greater, then divide (n-26^1) with 26^2. if it is <26^2 return 2. else divide( n-26^1 - 26^2) with 26^3..... and so on.
So if the sequence has m characters, the last iteration computes (n-26^1-26^2...26^(m-1)) divided by 26^m.

Approach 2:

This is the easiest. find ceil((ln n)/(ln 26)).This is nothing but finding log n to the base 26. ceil of that gives the number of characters.

Please have a look at below code and share your suggestion,,,

int num = 706;
char str[10] = {0};
int i=0;
while( num )
{
str[i++] =(char) ((num % 26) + 64);
num /= 26;
}
strrev(str);

In the above solution handle the case for input 26 .something like

int num = 1408;
char str[10] = {0};
int i=0;

while( num )
{
if ( num == 26) {
str[i++] = 'z';
break;
}
else{
str[i++] =(char) ((num % 26) + 64);
}

num /= 26;
}
strrev(str);

// Only handles finding out the number of chars in the nth term. n is the input

int d = 26, cnt = 0;
while(n > 0) {
n = n - d;
d = d * 26;
cnt++;
}
printf("\n%d", cnt);

void Find(int num)
{
char ch[27]="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
char *pResult=NULL;
int count = 0;
int reminder = num % 26;
if(reminder)
count = num/26 + 1 ;
else
count = num/26 ;

pResult = (char*)malloc(count + 1); // count characters + NULL character

memset(pResult, 'A', count - 1);// count-1 numbers of 'A' will be there
pResult[count-1] = ch[reminder];
pResult[count] = '\0';
printf("Number of characters in nth member: %d & member is : %s", count, pResult);
}

The series is of form n + n^2 + n^3 + n^4 + ....

where n is 26

now apply the logic!

int ReturnChar(int n)
/*
Find number of characters in nth member in the following series.
A,B,C,D....,AA,AB,......,ZY,ZZ,AAA,AAB,............
e.g for n<=26 chars are 1 for n=27 it is 2

Additional Question: Find nth member

here seq is getting generated 26,26^2,26^3 so on

*/
#include<math.h>
float y=26;
int ReturnChar(int n)
{
if(n==0)
return 0;
if(n<=26)
return 1;

return(1 + ReturnChar(ceil(n/y)));
}
int main()
{
int n;
scanf("%d",&n)
printf("Number of Chars=%d",ReturnChar(n));
return 0;
}

// Using the simple i + I^2 + i^3+..

int numChars (int n) {

// nth element
int c=1; // number of chars
int i=26;

while (n > i) {
c++;
i = i + i ^c;
}

return c;

}

public static String findNth(int index){
String result="";
while(index>0){
int currentBit=index%26;
result=getWhat(currentBit)+result;
index=index/26;
}
return result;
}
public static String getWhat(int currentBit){
switch(currentBit){
case 1:
return "A";
case 2:
return "B";
case 3:
return "C";
case 4:
return "D";
case 5:
return "E";
case 6:
return "F";
case 7:
return "G";
case 8:
return "H";
case 9:
return "I";
case 10:
return "J";
case 11:
return "K";
case 12:
return "L";
case 13:
return "M";
case 14:
return "N";
case 15:
return "O";
case 16:
return "P";
case 17:
return "Q";
case 18:
return "R";
case 19:
return "S";
case 20:
return "T";
case 21:
return "U";
case 22:
return "V";
case 23:
return "W";
case 24:
return "X";
case 25:
return "Y";
default:
return "Z";
}
}

I like this question, very similar to binary calculation

final static char[] chars = { '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' };


static String findNthNumber(int n) {
StringBuilder builder = new StringBuilder();
while (n > 0) {
int index = n % 26;
if (index == 0) {
index = chars.length;
n = n - 1;
}
builder.append(chars[index - 1]);
n = n / 26;
}
return builder.reverse().toString();
}

public static void main(String[] args) {
int max = 26 * 26 * 26 + 26 * 26 + 26;
for (int i = 1; i <= max; i++) {
System.out.print(findNthNumber(i) + " ");
if (i % 26 == 0) {
System.out.println();
}
}
}

final static char[] chars = { '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' };

   public static void main(String[] args) {
      int max = 26 * 26 * 26 + 26 * 26 + 26;
      for (int i = 1; i <= max; i++) {
         System.out.print(findNthNumber(i) + " ");
         if (i % 26 == 0) {
            System.out.println();
         }
         if (i == 26 * 26 + 26) {
            System.out.println();
         }
      }
   }

   static String findNthNumber(int n) {
      StringBuilder builder = new StringBuilder();
      while (n > 0) {
         int index = n % 26;
         if (index == 0) {
            index = chars.length;
            n = n - 1;
         }
         builder.append(chars[index - 1]);
         n = n / 26;
      }
      return builder.reverse().toString();
   }

public static int numChar(int n) {
            int count = 0, sub = 1;
            while(n > 0) {
                count++;
                sub *= 26;
                n -= sub;
            }
            return count;
        }

public static void solution(int n) {
            StringBuilder result = new StringBuilder();
            int count = 0;
            while(n > 0){
                count++;
                n--;
                result.append((char) ('A' + n%26));
                n /= 26;
            }
            System.out.println("number of char: " + count);
            System.out.println(n+"th member: " + result.reverse().toString()); 
        }