Well, you also had the solution in your last blog. Therefore, this blog makes no sense.

You can review your input string as a decimal number. So, answer will be your string as number + input number. Base is 3, not 10.

The answer to the problem as follow (this solution works even for |x| <= 10^6 and n <= 10^18):

if |x| < 1001 then add (char('a' - 1)) to the beginning of x until we get |x| == 1001 .

char ('a' - 1) means the character that is before the letter 'a' .

Now we will use simple math:

Suppose the index of the last letter of x is j ( j = 1000 ) ;

Moving x[j] forward by one will cost (1) .

Moving x[j-1] forward by one will cost (3) .

so moving any letter x[i] forward one step will cost 3^(j-i).

So the problem now is easy:

for(i=1000; i>=0; i--) {
	while (n > 0 && x[i] != 'c') {
		if ( 1000-i <= 20 && n >= power(3, 1000-i)) {
			x[i]++; 
			n-= power(3, 1000-1); 
		} else {
			//break the two loops we can't go further
		}
	}
	if (x[i] == 'c') x[i-1]++;
}  

i++;
while (i <= 1000) {
	x[i] = 'a';
	while (n > 0) { 
		if (1000-i <= 20 && n >= power(3, 1000-i))  {
			x[i]++;
			n-=power(3, 1000-1);
		} else {
			break;
		}
	}
	i++;
}
//print x without leading ('a'-1) characters

The reason behind (1000-i) <= 20 because (10^9 <= 3^20)

Simple modification on the code above you can solve the problem for (|x| <= 10^6)

Please notify me if you see something wrong or I've missed something .