Here is how I would approach the problem.

Step 1: What are the possible divisors of A[i] given that they are of range 1 to 10^5. Only 10^5 possibilities, a number we can work with!

Step 2: For every element can we find all its divisors? Yes. This is a O(rootN) algo per element.

Step 3: For every divisor can we record its first occurence and last occurence in array? Yes.

From Step 2 and 3 we get 2 maps:

- Map 1 -> Divisors and Frequency
- Map 2 -> Divisors and pair(first occurence, last occurence).

Determine the largest gap between last occurence and first occurence. Lets call it gapMax. Similarly get the largest divisor. lets called gcdMAX. Lets get a third variable called Max which is max(gapMax, gcdMax). Final answer cannot be larger than Max.

Step 3. Go through Map 1 -> For each divisor greater than Max, get the last occurence - first occurence. Add it to GCD, if this value is greater than Max, this is the new Max.