Question: I'm working on an iOS app, that should work fine with both iphones and ipads.
As I know we can build the app as universal or convert iphone storyboard into ipad storyboard.What I want to know is, what is the best way from these and, when we launch app to app store, is it a problem to have two storyboards one for iphone and another for ipad.

Answer: 
Your approach is too complicated. You don't need to examine every array element. The first value is `4`, so `7` is *at least* `7-4` elements away, and you can skip those.

```
#include <stdio.h>
#include <stdlib.h>

int main (void)
{
    int array[] = {4,5,6,5,4,3,2,3,4,5,6,7,8};
    int len = sizeof array / sizeof array[0];
    int i = 0;
    int steps = 0;
    while (i < len && array[i] != 7) {
        i += abs(7 - array[i]);
        steps++;
    }

    printf("Steps %d, index %d\n", steps, i);
    return 0;
}

```

Program output:

```
Steps 4, index 11

```

Edit: improved after comments from @Martin Zabel.
A variation of the conventional linear search could be a good way to go. Let us pick an element say `array[i] = 2`. Now, `array[i + 1]` will either be 1 or 3 (odd), `array[i + 2]` will be (positive integers only) 2 or 4 (even number). 

On continuing like this, a pattern is observable - `array[i + 2*n]` will hold even numbers and so all these indices can be ignored.

Also, we can see that

```
array[i + 3] = 1 or 3 or 5
array[i + 5] = 1 or 3 or 5 or 7

```

so, index `i + 5` should be checked next and a while loop can be used to determine the next index to check, depending on the value found at index `i + 5`.

While, this has complexity `O(n)` (linear time in terms of asymptotic complexity), it is better than a normal linear search in practical terms as all the indices are not visited.

Obviously, all this will be reversed if `array[i]` (our starting point) was odd.