How to solve Arrfix ? Can someone explain in detail?
Knock Knock !! Can someone out there help me with this ??!!
Here is my solution 
Though the Problem was mention under DP , but I couldn’t find any overlapping substructure.
Example
Consider all Arrays as 0indexed.
A ={ 1, 8, 7, 6, 3, 4 }
B ={ 1, 8, 8, 8, 6, 5 }
F ={ 8, 8, 1, 5, 12}
Solution

Find the common elements of B and F including the repetitions i.e. the Common Elements are = { 8, 8, 1, 5}

Find the optimum position to paste those common elements from F to A such that the difference between A and B is least. For example Where should first 8 from F be pasted on A ? the optimum position is A[2] i.e. 7 is replaced by 8 . And where would second 8 from F be pasted on A ? The optimum position is A[3] i.e on 6.
But what if there is no optimum position ? like for 1 ? there is only one 1 in B and the corresponding A is 1 … that is , it already has the best match for it . But remember we have to use all the F elements , so to use the F[2] element i.e. 1 we can Replace A[0] with 1, though A[0] was already 1 but here the optimum position for F’s 1 is A[0] …because if we place anywhere else then it would just increase the difference 
Lastly you have to count the number of differences that exist between A and B even after doing the operation (mentioned in 2) for each of the common elements in F.

One More Check Step In the end you have to check whether the number of differences < the number of uncommon elements between B and F.
If so, the the least number of differences = the number of uncommon elements between B and F. What this means is that no matter what, you have to paste the F elements on A , so u have to paste the uncommon elements also , which increases the difference between A and B.
The Final Arrays are
A = { 1, 8, 8, 8, 12, 5 } modified
B = { 1, 8, 8, 8, 6, 5 } unchanged
Difference = 1
Complexity Analysis 
The main focus is to find the common elements between B and F.
Let n = B.size() and fnum = F.size()
Complexity to find the common elements after sorting B and F = O( n + fnum ). {I made a copy of B then I sort it}
But to sort them = O( max(nlogn, fnumlogfnum) )
Thus the complexity is O( max(nlogn, fnumlogfnum) )
which works fine by the constraints.
If you Like my solution , Give a THUMBS UP.
Thank you.