Let $A$ be an array of integers $($positive or negative$).$ Assume that the array is of size $n.$ The subarray $A[i..j]$ is the part of the array that starts at index i and ends at index $j,$ where $0ijn-1.$ Let $s_{ij}$ equal the sum of the integers in $A[i.j].$

We wish to solve the following problem:

Important: Find the maximum value for $s_{ij}$ over all subarrays with a maximum length of $3$ in array $A,$ where $0ijn-1.$

The three algorithms given below solve this problem. NOTE: If all of the values in the array are negative, then the maximum value for $s_{ij}$ is $O$ by default.

Example: If the array contains the values $\{-1,12,-3,14,-4,3\},$ then the maximum sum over all subarrays is $23($for the subarray $\{12,-3,14\}).$ If the array contains the values $\{2,-3,5,-1,7\},$ then the maximum sum over all subarrays is $11($for the subarray $\{5,-1,7\}).$

You are to write a Java program that determines the amount of work each of these algorithms does to compute its answer for arrays of various sizes. Using this data, you are to determine the runtime complexity of each algorithm.

ALGORITHM $1$

Start with a maximum sum of $0.$ Compute the sum of each $1-$element subarray, then compute the sum of each $2$ element subarray, then compute the sum of each $3-$element subarray, etc. For each sum you compute, if it is larger than the maximum sum you've seen, then it becomes the maximum sum.

ALGORITHM $2$

Same as algorithm $1$ but now once you compute the sum of the subarray from $A[i]$ to $A[j],$ the sum of the subarray from $A[i]$ to $A[j+1]$ is just the previous sum you computed plus $A[j+1].$ Don't add up all of the previous values all over again.

ALGORITHM $3($Optional$)$

Same as algorithm $1$ but store the temporary steps in an array.