There are n servers. Each server i is currently processing a(i) amount of requests. There is another array b in which b(i) represents the number of incoming requests that are scheduled to server i. Reschedule the incoming requests in such a way that each server i holds an equal amount of requests after rescheduling. An incoming request to server i can be rescheduled only to server i-1, i, i+1. If there is no such rescheduling possible then output -1 else print number of requests hold by each server after rescheduling.
Examples: 
 

Input : a = {6, 14, 21, 1}
        b = {15, 7, 10, 10}
Output : 21
b(0) scheduled to a(0) --> a(0) = 21
b(1) scheduled to a(1) --> a(1) = 21
b(2) scheduled to a(3) --> a(3) = 11
b(3) scheduled to a(3) --> a(3) = 21
a(2) remains unchanged --> a(2) = 21

Input : a = {1, 2, 3}
        b = {1, 100, 3}
Output : -1
No rescheduling will result in equal requests.

Approach: Observe that each element of array b is always added to any one element of array a exactly once. Thus the sum of all elements of array b + sum of all elements of old array a = sum of all elements of new array a. Let this sum be S. Also all the elements of new array a are equal. Let each new element is x. If array a has n elements, this gives 
 

 x * n = S
  => x = S/n     ....(1)

Thus all the equal elements of new array a is given by eqn(1). Now to make each a(i) equals to x we need to add x-a(i) to each element. We will iterate over entire array a and check whether a(i) can be made equal to x. There are multiple possibilities: 
1. a(i) > x: In this case a(i) can never be made equal to x. So output -1. 
2. a(i) + b(i) + b(i+1) = x. Simply add b(i) + b(i+1) to a(i) and update b(i), b(i+1) to zero. 
3. a(i) + b(i) = x. Add b(i) to a(i) and update b(i) to zero. 
4. a(i) + b(i+1) = x. Add b(i+1) to a(i) and update b(i+1) to zero.
After array a is completely traversed, check whether all elements of array b are zero or not. If yes then print a(0) otherwise print -1.
Why b(i) is updated to zero after addition? 
Consider a test case in which b(i) is neither added to a(i-1) nor a(i). In that case, we are bounded to add b(i) to a(i+1). Thus while iterating over the array a when we begin performing computations on element a(i), first we add element b(i-1) to a(i) to take into consideration above possibility. Now if b(i-1) is already added to a(i-1) or a(i-2) then, in that case, it cannot be added to a(i). So to avoid this double addition of b(i) it is updated to zero. 
The stepwise algorithm is: 
 

1. Compute sum S and find x = S / n
2. Iterate over array a
3. for each element a(i) do:
   a(i) += b(i-1)
   b(i-1) = 0;
   if a(i) > x:
      break
   else:
     check for other three possibilities
     and update a(i) and b(i).
4. Check whether all elements of b(i) are
   zero or not.

Implementation: 
 

Output: 
 

21

Time Complexity: O(n) 
Auxiliary Space : O(1) If we are not allowed to modify original arrays, then O(n)
 

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