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: 
You can do a linear search with steps that are often greater than 1. The crucial observation is that if e.g. `array[i] == 4` and 7 hasn't yet appeared then the next candidate for 7 is at index `i+3`. Use a while loop which repeatedly goes directly to the next viable candidate.

Here is an implementation, slightly generalized. It finds the first occurrence of `k` in the array (subject to the +=1 restriction) or `-1` if it doesn't occur:

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

int first_occurence(int k, int array[], int n);

int main(void){
    int a[] = {4,3,2,3,2,3,4,5,4,5,6,7,8,7,8};
    printf("7 first occurs at index %d\n",first_occurence(7,a,15));
    printf("but 9 first \"occurs\" at index %d\n",first_occurence(9,a,15));
    return 0;
}

int first_occurence(int k, int array[], int n){
    int i = 0;
    while(i < n){
        if(array[i] == k) return i;
        i += abs(k-array[i]);
    }
    return -1;
}

```

output:

```
7 first occurs at index 11
but 9 first "occurs" at index -1

```
The approach presented by John Coleman is what the interviewer was hoping for, in all probability.  

If you are willing to go quite a bit more complicated, you can increase expected skip length:  

Call the target value *k*. Start with the first element's value *v* at position *p* and call the difference k-v *dv* with absolute value *av*. To speed negative searches, have a peek at the last element as the other value *u* at position *o*: if dv×du is negative, k is present (if any occurrence of k is acceptable, you may narrow down the index range here the way binary search does). If av+au is greater than the length of the array, k is absent. (If dv×du is zero, v or u equals k.)  

Omitting index validity: Probe the ("next") position where the sequence might return to v with k in the middle: `o = p + 2*av`.  

If dv×du is negative, find k (recursively?) from p+av to o-au;  

if it is zero, u equals k at o.  

If du equals dv and the value in the middle isn't k, or au exceeds av,  

or you *fail* to find k from p+av to o-au,  

let `p=o; dv=du; av=au;` and keep probing.  

(For a full flash-back to '60ies texts, view with Courier. My "1st 2nd thought" was to use `o = p + 2*av - 1`, which precludes *du equals dv*.)