Efficient way to search an element

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 11but 9 first "occurs" at index -1

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 @Raphael Miedl and @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 5array[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.