Running-time complexity of two nested loops: quadratic or linear?

Question

int Solution::diffPossible(vector<int> &A, int B) {
    for (int i = 0; i < A.size(); i++) {
       for (int j = i+1; j < A.size(); j++)
          if ((A[j]-A[i]) == B)
             return 1;      
    }
    return 0;
}

This is the solution to a simple question where we are supposed to write a code with time complexity less than or equal to O(n). I think the time complexity of this code is O(n^2) but still it got accepted. So, I am in doubt please tell me the right answer.

Answer 1

Let's analyze the worst-case scenario, ie when the condition of the if -statement in the inner loop, (A[j]-A[i]) == B , is never fulfilled, and therefore the statement return 1 is never executed.

If we denote A.size() as n , the comparison in the inner loop is performed n-1 times for the first iteration of the outer loop, then n-2 times for the second iteration, and so on...

So, the number of the comparisons performed in the inner loop for this worst-case scenario is (by calculating the sum of the resulting arithmetic progression below):

n-1 + n-2 + ... + 1 = (n-1)n/2 = (n^2 - n)/2
^                 ^
|_________________|
     n-1 terms

Therefore, the running-time complexity is quadratic, ie, O(n^2), and not O(n).