Find next Aloha Number - objective-c

Question

I found this problem in one of coding practice sites...Problem goes as below.
A number is said to be an aloha number if it is comprises entirely of 4s and 7s. Below are few examples of Aloha Numbers
4477
47744477
Your program will get an integer number of a non Aloha Number and your output must return the next Aloha number (in the form of string) for that input.
Example 1
Input : 1234
Output: 4444
Example 2
Input : 4472
Output: 4474
Input number represent the integer between 1 and 2*(10^9)
Input number is not Aloha Number
I tried to solve this problem as below..
- (NSString *) nextAlohaNumber:(NSUInteger) number{
// method convertNumberIntoDigit return a array of digits present in input number
NSMutableArray *numberDigits = [self convertNumberIntoDigit:number];
NSUInteger tempNum;
for(int index = 0; index < numberDigits.count; index++){
tempNum = [[numberDigits objectAtIndex:index] intValue];
if(4 != tempNum && 7 != tempNum){
if(4 > tempNum){
[numberDigits replaceObjectAtIndex:index withObject:[NSNumber numberWithInt:4]];
}else if (7 > tempNum){
[numberDigits replaceObjectAtIndex:index withObject:[NSNumber numberWithInt:7]];
}
else if(7 < tempNum){
[numberDigits replaceObjectAtIndex:index withObject:[NSNumber numberWithInt:4]];
int previousIndex = index-1;
int previousNum = [[numberDigits objectAtIndex:previousIndex] integerValue];
if(previousIndex ){
if(4 == previousNum){
[numberDigits replaceObjectAtIndex:previousIndex withObject:[NSNumber numberWithInt:7]];
}
else if(7 == previousNum){
[numberDigits replaceObjectAtIndex:index withObject:[NSNumber numberWithInt:4]];
}
}
}
}
}
return [self numberArrayToStr:numberDigits]; // convert strings with digit in Array
}
But this method doesn't pass all the test cases, for example it fails on input 4478, 790 etc.
Can any one please help me to solve this problem for all inputs with better time and space complexity.
PS: I'm ok with C,C++ code as well.
Thanks in advance.

Answer 1

Despite my comment above, I wondered about the "time and space complexity". It turned out to be trivial; this only takes 3 long integer variables and a single loop over each input digit (which can be incremented by this algorithm to use 1 more digit; and this will not cause an integer overflow).
Starting from the lowest digit, if it's lower than 4 replace it with 4. If it's lower than 7, replace it with 7. For 8 and 9, replace it with 4 and increment the next digit. Repeat until all digits are processed.
The following code returns the number itself for a proper "aloha" number, the next higher if not.
It requires long variables (using more than 32 bits), because the aloha number for the largest allowed input, 2*(10^9), is 4444444444 -- outside the range of a 32-bit integer.
#include <stdio.h>
#include <stdlib.h>
int main (int argc, char **argv)
{
long not_aloha, aloha, digit;
if (argc != 2)
{
printf ("please give an aloha number\n");
return -1;
}
not_aloha = atoi (argv[1]);
printf ("not aloha: %ld\n", not_aloha);
aloha = 0;
digit = 1;
while (not_aloha)
{
if ( (not_aloha % 10) <= 4)
aloha += 4*digit;
else if ( (not_aloha % 10 ) <= 7)
aloha += 7*digit;
else
{
aloha += 4*digit;
not_aloha += 10;
}
digit *= 10;
not_aloha /= 10;
}
printf ("aloha: %ld\n", aloha);
return 0;
}