Run Time for Linear Probing on Hash table

wish help you to fix your issue You don't have to modify the hash function for quadratic probing. The simplest form of quadratic probing is really just adding consequent squares to the calculated position instead of linear 1, 2, 3.
There's a good resource here. The following is taken from there. This is the simplest form of quadratic probing when the simple polynomial c(i) = i^2 is used:

wish of those help When we talk about Asymptotic complexities we generally take into account very large n. Now for collision handling in a Hash Table some of the methods are chained hashing & linear probing. In both the cases two things may happen (that will help in answering your question): 1. You may require resizing of the hash table due to it getting full 2. Collisions may happen.
In the worst case it will depend on how you have implemented your hash table, say in linear probing you dont find the number,you keep on moving and the number you were looking for was at the end. Here comes the O(n) worst case running time. Coming to chained hashing technique, when a collision happens, to handle them say we have stored the keys in a balanced binary tree so the worst case running time would be O(log n).

To fix the issue you can do As long as you do it consistently, it does not matter. The effect of adding one (or any other constant, for that matter) to hash code has no effect on the composition of the table, except that the bucket numbering would be "shifted off" by a constant "offset". Since bucket numbering is a private matter of your has table, nobody should care.
In essence, a linear probing hash function is

H(x, i) = (H(x) + i) % N

hope this fix your issue Notice how the question doesn't specify the order in which the keys are inserted, so your answer is only correct assuming that the keys are actually inserted in the order A-B-C-D-E-F-G, but since the question doesn't explicitly state the order, you need to dig deeper.
What you do know, however, is that one of those keys will be inserted first and it will go to its designated slot as shown in the Key-to-Hash diagram, since the hash table is initially empty. This immediately discards option choice 2 because none of the keys are in their designated array entry, leaving you with choice 1 and 3.