Maximum Subarray sum of A[] by adding any element of B[] at any end

Given two arrays A[] and B[] having lengths N and M respectively, Where A[] and B[] can contain both positive and negative elements, the task is to find the maximum sub-array sum obtained by applying the below operations till the B[] has no element left in it:

• Choose either the leftmost or rightmost element from B[] and add it to any end of A[].
• Delete the element chosen from B[].

Examples:

Input: N = 4, A[] = {72, 23, -56, 80}, M = 2, B[] = {50, 10} 
Output: 179
Explanation: First Operation: Get element 50 at first index from B[] and prepend it in A[] to make A[] = {50, 72, 23 -56, 80} and B[] = {10}
Second Operation: Get element 10 at last index from B[] and append it in A[] to make A[] = {50, 72, 23 -56, 80, 10}. 
Now, B[] has no elements. The maximum value of sum that can be obtained by perform given operation at most K times is: 179 by subarray (50, 72, 23, -56, 80, 10).

Input: N = 2, A[] = {50, -45}, M = 2, B[] = {90},  
Output: 140

Approach: The problem can be solved based on the following idea:

Problem is based on Kadane's Algorithm and Greedy approach. Find maximum subarray sum in A[] using Kadane's Algorithm let's say it as Z. Now there can be three cases for the subarray with maximum sum:

• The subarray has one end at the front: In this case adding the positive elements of B[] at the front of A[] gives the maximum possible subarray sum after the operations.
• The subarray of A[] with sum Z has one end at the end of A[]: Here appending positive elements of B[] to the end of A[] will give maximum sum.
• The subarray with maximum sum lies somewhere in between: In this case the choices are to add the positive elements of B[] to the beginning (say the maximum subarray sum from start becomes P) or to the end (say the maximum subarray sum ending at end is Q). The maximum sum will then be max among Z, P and Q.

Follow the steps given below to solve the problem:

• Create a variable max_sum to store the maximum sub-array sum of X[]. 
• Add positive elements from B[] to the starting of A[] and find the max subarray sum starting from the beginning of A[].
• Add positive elements from B[] to the end of A[] and find the max subarray sum finishing at the end of A[].
• The maximum of the above three is the required answer. 

Below is the implementation of the above approach.

179

Time Complexity: O(max(N, M))
Auxiliary Space: O(1)

Another Approach: 

• Initialize two pointers, left and right, to the leftmost and rightmost indices of array A[].
  Initialize a variable currsum to 0 and maxsum to INT_MIN value.
  Iterate through the elements of array B[] and then A[].
• Append all positive elements of B[] to front of A[] and negative to back of A[].
• Append all positive elements of B[] to back of A[] and negative to front of A[]
• Get maxSubArray in both cases and the maximum of those two will be the answer.
• Return maxSum.

179

Time Complexity: O(N)

Auxiliary Space: O(1)