Maximum score possible after performing given operations on an Array

Given an array A of size N, the task is to find the maximum score possible of this array. The score of an array is calculated by performing the following operations on the array N times: 

1. If the operation is odd-numbered, the score is incremented by the sum of all elements of the current array.
    
2. If the operation is even-numbered, the score is decremented by the sum of all elements of the current array. 
    
3. After every operation, either remove the first or the last element of the remaining array. 
    

Examples: 

Input: A = {1, 2, 3, 4, 2, 6} 
Output: 13 
Explanation: 
The optimal operations are: 
1. Operation 1 is odd. 
-> So add the sum of the array to the score: Score = 0 + 18 = 18 
-> remove 6 from last, 
-> new array A = [1, 2, 3, 4, 2] 
2. Operation 2 is even. 
-> So subtract the sum of the array from the score: Score = 18 – 12 = 6 
-> remove 2 from last, 
-> new array A = [1, 2, 3, 4] 
3. Operation 1 is odd. 
-> So add the sum of the array to the score: Score = 6 + 10 = 16 
-> remove 4 from last, 
-> new array A = [1, 2, 3] 
4. Operation 4 is even. 
-> So subtract the sum of the array from the score: Score = 16 – 6 = 10 
-> remove 1 from start, 
-> new array A = [2, 3] 
5. Operation 5 is odd. 
-> So add the sum of the array to the score: Score = 10 + 5 = 15 
-> remove 3 from last, 
-> new array A = [2] 
6. Operation 6 is even. 
-> So subtract the sum of the array from the score: Score = 15 – 2 = 13 
-> remove 2 from first, 
-> new array A = [] 
The array is empty so no further operations are possible.

Input: A = [5, 2, 2, 8, 1, 16, 7, 9, 12, 4] 
Output: 50 

Naive approach 
 

1. In each operation, we have to remove either the leftmost or the rightmost element. A simple way would be to consider all possible ways to remove elements and for each branch compute the score and find the maximum score out of all. This can simply be done using recursion. 
    
2. The information we need to keep in each step would be 
   • The remaining array [l, r], where l represents the leftmost index and r the rightmost,
   • The operation number, and
   • The current score.
3. In order to calculate the sum of any array from [l, r] in each recursive step optimally, we will keep a prefix sum array. 
   Using prefix sum array, new sum from [l, r] can be calculated in O(1) as: 
    

Sum(l, r) = prefix_sum[r] – prefix_sum[l-1] 
 

Below is the implementation of the above approach: 
 

13

Time complexity: O(2^N)
Auxiliary space: O(N)

Efficient approach 
 

• In the previous approach it can be observed that we are calculating same subproblems many times, i.e. it follows the property of Overlapping Subproblems. So we can use Dynamic programming to solve the problem 
   
• In the recursive solution stated above, we only need to add memoization using a dp table. The states will be: 
   

DP table states = dp[l][r][num]
where l and r represent the endpoints of the current array and num represents the operation number. 
 

Below is the implementation of the Memoization approach of the recursive code: 
 

13

Time complexity: O(N^3)

Auxiliary space: O(N^3)

Efficient approach : DP tabulation (iterative)

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memorization(top-down) because memorization method needs extra stack space of recursion calls. In this approach we use 3D DP store computation of subproblems and find the final result using iteration and not with the help of recursion.

Steps that were to follow the above approach:

• Initialize a 3D DP array dp of size n x n x n.
• Calculate the prefix sum of the input array a in an array prefix_sum.
• Fill the dp table diagonally, considering all possible lengths of subarrays.
• For each subarray of length len, consider all possible starting indices i, ending indices j, and even-numbered operations num.
• Calculate the current sum of the subarray based on prefix_sum and the even/odd nature of the operation.
• If the subarray has only one element (i.e., i == j), store the current sum in dp[i][j][num].
• Otherwise, calculate the maximum score by exploring all possible paths, and store the maximum value in dp[i][j][num] to avoid further repetitive recursive calls.
• Return dp[0][n – 1][1] as the maximum score.

Below is the implementation of the above approach:

Output:

13

Time complexity: O(N^3)
Auxiliary space: O(N^3)

Efficient Approach: Space Optimization

In previous approach the current value dp[i][j][num] is only depend upon the current and previous row values of DP. So to optimize the space complexity we use a single 2D matrix instead of 3d array to store the computations.

Follow the step below to implement the above idea:

• Initialize a 2D array dp of size n x n to store maximum scores.
• Calculate the prefix sum array prefix_sum by iterating through the input array a[] and adding the previous prefix sum.
• Iterate over different lengths of subarrays (len) and starting indices of subarrays (i).
• Iterate over the number of elements to consider in the subarray (num).
• Calculate the current sum of the subarray and handle even/odd number conditions.
• Determine the maximum score by choosing between including or excluding the current element.
• Return the maximum score for the entire array considering 1 element (dp[0][1]).

Below is the implementation of the above approach:

13

Time complexity: O(N^3)
Auxiliary space: O(N^2)