Finding whether a number exists as sum of two array elements in a sorted array

Given the array{-5,-2,7,8,9,10,12}

Find whether sum 8 is possible

As element no.2 (-2) and element no.7(10) are present the sum of the two is 8

Solution:  As the array is sorted, we can leverage this property and come up with an approach which is similar to binary searching

Start with two markers at start and end of the array

Sum the two numbers at start and end marker

if sum is greater than given number, decrease the end marker 

if sum is less than given number, advance the start marker

Break if desired sum is found or start becomes greater than or equal to end

Complexity: O(n)

#include <stdio.h>
int main()
{
    int a[] = {-5,-2,7,8,9,10,12};
    
    int sum =8;
    
    int found_flag = 0;
    int size = sizeof a / sizeof a[0];
    printf("\nSize of array %d", size);
    int start = 0;
    int end = size-1;
    
    while(start<=end) {
        if((a[start]+a[end])<sum) {
            start++;
        }
        else if ((a[start]+a[end]>sum)) {
            end--;
        }
        else {
            found_flag++;
            break;
        }
    }
    
    if(found_flag) {
     printf("\nElement at location(%d):%d and location(%d):%d has sum:%d", start+1,a[start],end+1,a[end],sum);
    }
    else {
        printf("\nNot found !!");
    }
    return 0;
}

Output:

Size of array 7                                                                                                               

Element at location(2):-2 and location(6):10 has sum:8