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: 

C++

// C++ program to find the maximum
// score after given operations

#include <bits/stdc++.h>
using namespace std;

// Function to calculate
// maximum score recursively
int maxScore(
    int l, int r,
    int prefix_sum[],
    int num)
{

    // Base case
    if (l > r)
        return 0;

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
          - (l - 1 >= 0
                 ? prefix_sum[l - 1]
                 : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, by removing
    // leftmost and rightmost element
    // and selecting the maximum value
    return current_sum
           + max(
                 maxScore(
                     l + 1, r,
                     prefix_sum,
                     num + 1),
                 maxScore(
                     l, r - 1,
                     prefix_sum,
                     num + 1));
}

// Function to find the max score
int findMaxScore(int a[], int n)
{
    // Prefix sum array
    int prefix_sum[n] = { 0 };

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    return maxScore(0, n - 1,
                    prefix_sum, 1);
}

// Driver code
int main()
{
    int n = 6;
    int A[n] = { 1, 2, 3, 4, 2, 6 };

    cout << findMaxScore(A, n);
    return 0;
}

Java

// Java program to find the maximum
// score after given operations
import java.util.*;

class GFG{

// Function to calculate
// maximum score recursively
static int maxScore(
    int l, int r,
    int prefix_sum[],
    int num)
{

    // Base case
    if (l > r)
        return 0;

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
        - (l - 1 >= 0
                ? prefix_sum[l - 1]
                : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, by removing
    // leftmost and rightmost element
    // and selecting the maximum value
    return current_sum
        + Math.max(maxScore(l + 1, r,
                            prefix_sum,
                            num + 1),
                    maxScore(l, r - 1,
                            prefix_sum,
                            num + 1));
}

// Function to find the max score
static int findMaxScore(int a[], int n)
{
    // Prefix sum array
    int prefix_sum[] = new int[n];

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    return maxScore(0, n - 1,
                    prefix_sum, 1);
}

// Driver code
public static void main(String[] args)
{
    int n = 6;
    int A[] = { 1, 2, 3, 4, 2, 6 };

    System.out.print(findMaxScore(A, n));
}
}

// This code is contributed by sapnasingh4991

Python3

# Python3 program to find the maximum
# score after given operations

# Function to calculate maximum
# score recursively
def maxScore(l, r, prefix_sum, num):

    # Base case
    if (l > r):
        return 0;

    # Sum of array in range (l, r)
    if((l - 1) >= 0):
        current_sum = (prefix_sum[r] -
                       prefix_sum[l - 1])
    else:
        current_sum = prefix_sum[r] - 0

    # If the operation is even-numbered
    # the score is decremented
    if (num % 2 == 0):
        current_sum *= -1;

    # Exploring all paths, by removing
    # leftmost and rightmost element
    # and selecting the maximum value
    return current_sum + max(maxScore(l + 1, r,
                                      prefix_sum,
                                      num + 1),
                             maxScore(l, r - 1,
                                      prefix_sum,
                                      num + 1));

# Function to find the max score
def findMaxScore(a, n):

    # Prefix sum array
    prefix_sum = [0] * n

    prefix_sum[0] = a[0]

    # Calculating prefix_sum
    for i in range(1, n):
        prefix_sum[i] = prefix_sum[i - 1] + a[i];

    return maxScore(0, n - 1, prefix_sum, 1);

# Driver code
n = 6;
A = [ 1, 2, 3, 4, 2, 6 ]
ans = findMaxScore(A, n)

print(ans)

# This code is contributed by SoumikMondal

C#

// C# program to find the maximum
// score after given operations
using System;

class GFG{

// Function to calculate
// maximum score recursively
static int maxScore(
    int l, int r,
    int []prefix_sum,
    int num)
{

    // Base case
    if (l > r)
        return 0;

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
        - (l - 1 >= 0
                ? prefix_sum[l - 1]
                : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, by removing
    // leftmost and rightmost element
    // and selecting the maximum value
    return current_sum
        + Math.Max(maxScore(l + 1, r,
                            prefix_sum,
                            num + 1),
                    maxScore(l, r - 1,
                            prefix_sum,
                            num + 1));
}

// Function to find the max score
static int findMaxScore(int []a, int n)
{
    // Prefix sum array
    int []prefix_sum = new int[n];

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    return maxScore(0, n - 1,
                    prefix_sum, 1);
}

// Driver code
public static void Main(String[] args)
{
    int n = 6;
    int []A = { 1, 2, 3, 4, 2, 6 };

    Console.Write(findMaxScore(A, n));
}
}

// This code is contributed by 29AjayKumar

Javascript

<script>

// JavaScript program to find the maximum
// score after given operations

// Function to calculate
// maximum score recursively
function maxScore(l, r, prefix_sum, num)
{

    // Base case
    if (l > r)
        return 0;

    // Sum of array in range (l, r)
    let current_sum
        = prefix_sum[r]
        - (l - 1 >= 0
                ? prefix_sum[l - 1]
                : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, by removing
    // leftmost and rightmost element
    // and selecting the maximum value
    return current_sum
        + Math.max(
                maxScore(
                    l + 1, r,
                    prefix_sum,
                    num + 1),
                maxScore(
                    l, r - 1,
                    prefix_sum,
                    num + 1));
}

// Function to find the max score
function findMaxScore(a, n)
{
    // Prefix sum array
    let prefix_sum = new Uint8Array(n);

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (let i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    return maxScore(0, n - 1,
                    prefix_sum, 1);
}

// Driver code

    let n = 6;
    let A = [ 1, 2, 3, 4, 2, 6 ];

    document.write(findMaxScore(A, n));


// This code is contributed by Surbhi Tyagi.

</script>

Output

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: 

C++

// C++ program to find the maximum
// Score after given operations

#include <bits/stdc++.h>
using namespace std;

// Memoizing by the use of a table
int dp[100][100][100];

// Function to calculate maximum score
int MaximumScoreDP(int l, int r,
                   int prefix_sum[],
                   int num)
{
    // Bse case
    if (l > r)
        return 0;

    // If the same state has
    // already been computed
    if (dp[l][r][num] != -1)
        return dp[l][r][num];

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
          - (l - 1 >= 0
                 ? prefix_sum[l - 1]
                 : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, and storing
    // maximum value in DP table to avoid
    // further repetitive recursive calls
    dp[l][r][num] = current_sum
                    + max(
                          MaximumScoreDP(
                              l + 1, r,
                              prefix_sum,
                              num + 1),
                          MaximumScoreDP(
                              l, r - 1,
                              prefix_sum,
                              num + 1));

    return dp[l][r][num];
}

// Function to find the max score
int findMaxScore(int a[], int n)
{
    // Prefix sum array
    int prefix_sum[n] = { 0 };

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    // Initialising the DP table,
    // -1 represents the subproblem
    // hasn't been solved yet
    memset(dp, -1, sizeof(dp));

    return MaximumScoreDP(
        0, n - 1,
        prefix_sum, 1);
}

// Driver code
int main()
{
    int n = 6;
    int A[n] = { 1, 2, 3, 4, 2, 6 };

    cout << findMaxScore(A, n);
    return 0;
}

Java

// Java program to find the maximum
// Score after given operations


class GFG{

// Memoizing by the use of a table
static int [][][]dp = new int[100][100][100];

// Function to calculate maximum score
static int MaximumScoreDP(int l, int r,
                   int prefix_sum[],
                   int num)
{
    // Bse case
    if (l > r)
        return 0;

    // If the same state has
    // already been computed
    if (dp[l][r][num] != -1)
        return dp[l][r][num];

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
          - (l - 1 >= 0
                 ? prefix_sum[l - 1]
                 : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, and storing
    // maximum value in DP table to avoid
    // further repetitive recursive calls
    dp[l][r][num] = current_sum
                    + Math.max(
                          MaximumScoreDP(
                              l + 1, r,
                              prefix_sum,
                              num + 1),
                          MaximumScoreDP(
                              l, r - 1,
                              prefix_sum,
                              num + 1));

    return dp[l][r][num];
}

// Function to find the max score
static int findMaxScore(int a[], int n)
{
    // Prefix sum array
    int []prefix_sum = new int[n];

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    // Initialising the DP table,
    // -1 represents the subproblem
    // hasn't been solved yet
    for(int i = 0;i<100;i++){
       for(int j = 0;j<100;j++){
           for(int l=0;l<100;l++)
           dp[i][j][l]=-1;
       }
   }

    return MaximumScoreDP(
        0, n - 1,
        prefix_sum, 1);
}

// Driver code
public static void main(String[] args)
{
    int n = 6;
    int A[] = { 1, 2, 3, 4, 2, 6 };

    System.out.print(findMaxScore(A, n));
}
}

// This code contributed by sapnasingh4991

Python3

# python3 program to find the maximum
# Score after given operations

# Memoizing by the use of a table
dp = [[[-1 for x in range(100)]for y in range(100)]for z in range(100)]

# Function to calculate maximum score


def MaximumScoreDP(l, r, prefix_sum,
                   num):

    # Bse case
    if (l > r):
        return 0

    # If the same state has
    # already been computed
    if (dp[l][r][num] != -1):
        return dp[l][r][num]

    # Sum of array in range (l, r)
    current_sum = prefix_sum[r]
    if (l - 1 >= 0):
        current_sum -= prefix_sum[l - 1]

    # If the operation is even-numbered
    # the score is decremented
    if (num % 2 == 0):
        current_sum *= -1

    # Exploring all paths, and storing
    # maximum value in DP table to avoid
    # further repetitive recursive calls
    dp[l][r][num] = (current_sum
                     + max(
                         MaximumScoreDP(
                             l + 1, r,
                             prefix_sum,
                             num + 1),
                         MaximumScoreDP(
                             l, r - 1,
                             prefix_sum,
                             num + 1)))

    return dp[l][r][num]


# Function to find the max score
def findMaxScore(a, n):

    # Prefix sum array
    prefix_sum = [0]*n

    prefix_sum[0] = a[0]

    # Calculating prefix_sum
    for i in range(1, n):
        prefix_sum[i] = prefix_sum[i - 1] + a[i]

    # Initialising the DP table,
    # -1 represents the subproblem
    # hasn't been solved yet
    global dp

    return MaximumScoreDP(
        0, n - 1,
        prefix_sum, 1)


# Driver code
if __name__ == "__main__":

    n = 6
    A = [1, 2, 3, 4, 2, 6]

    print(findMaxScore(A, n))

C#

// C# program to find the maximum
// Score after given operations


using System;

public class GFG{

// Memoizing by the use of a table
static int [,,]dp = new int[100,100,100];

// Function to calculate maximum score
static int MaximumScoreDP(int l, int r,
                   int []prefix_sum,
                   int num)
{
    // Bse case
    if (l > r)
        return 0;

    // If the same state has
    // already been computed
    if (dp[l,r,num] != -1)
        return dp[l,r,num];

    // Sum of array in range (l, r)
    int current_sum
        = prefix_sum[r]
          - (l - 1 >= 0
                 ? prefix_sum[l - 1]
                 : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, and storing
    // maximum value in DP table to avoid
    // further repetitive recursive calls
    dp[l,r,num] = current_sum
                    + Math.Max(
                          MaximumScoreDP(
                              l + 1, r,
                              prefix_sum,
                              num + 1),
                          MaximumScoreDP(
                              l, r - 1,
                              prefix_sum,
                              num + 1));

    return dp[l,r,num];
}

// Function to find the max score
static int findMaxScore(int []a, int n)
{
    // Prefix sum array
    int []prefix_sum = new int[n];

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    // Initialising the DP table,
    // -1 represents the subproblem
    // hasn't been solved yet
    for(int i = 0;i<100;i++){
       for(int j = 0;j<100;j++){
           for(int l=0;l<100;l++)
           dp[i,j,l]=-1;
       }
   }

    return MaximumScoreDP(
        0, n - 1,
        prefix_sum, 1);
}

// Driver code
public static void Main(String[] args)
{
    int n = 6;
    int []A = { 1, 2, 3, 4, 2, 6 };

    Console.Write(findMaxScore(A, n));
}
}

// This code contributed by PrinciRaj1992

Javascript

<script>

// JavaScript program to find the maximum
// Score after given operations

// Memoizing by the use of a table
let dp = new Array(100);
// Initialising the DP table,
    // -1 represents the subproblem
    // hasn't been solved yet
for(let i=0;i<100;i++)
{
    dp[i]=new Array(100);
    for(let j=0;j<100;j++)
    {
        dp[i][j]=new Array(100);
        for(let k=0;k<100;k++)
        {
            dp[i][j][k]=-1;
        }
    }
}

// Function to calculate maximum score
function MaximumScoreDP(l,r,prefix_sum,num)
{
    // Bse case
    if (l > r)
        return 0;

    // If the same state has
    // already been computed
    if (dp[l][r][num] != -1)
        return dp[l][r][num];

    // Sum of array in range (l, r)
    let current_sum
        = prefix_sum[r]
          - (l - 1 >= 0
                 ? prefix_sum[l - 1]
                 : 0);

    // If the operation is even-numbered
    // the score is decremented
    if (num % 2 == 0)
        current_sum *= -1;

    // Exploring all paths, and storing
    // maximum value in DP table to avoid
    // further repetitive recursive calls
    dp[l][r][num] = current_sum
                    + Math.max(
                          MaximumScoreDP(
                              l + 1, r,
                              prefix_sum,
                              num + 1),
                          MaximumScoreDP(
                              l, r - 1,
                              prefix_sum,
                              num + 1));

    return dp[l][r][num];
}

// Function to find the max score
function findMaxScore(a,n)
{
    // Prefix sum array
    let prefix_sum = new Array(n);

    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (let i = 1; i < n; i++) {
        prefix_sum[i]
            = prefix_sum[i - 1] + a[i];
    }

    // Initialising the DP table,
    // -1 represents the subproblem
    // hasn't been solved yet


    return MaximumScoreDP(
        0, n - 1,
        prefix_sum, 1);
}

// Driver code
let n = 6;
let A=[1, 2, 3, 4, 2, 6 ];
document.write(findMaxScore(A, n));


// This code is contributed by rag2127

</script>

Output

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 code for above approach:

C++

// C++ program to find the maximum
// Score after given operations
#include <bits/stdc++.h>
using namespace std;

// Function to find the max score
int findMaxScore(int a[], int n)
{

    // initialize Dp to store computation of subproblems
    int dp[n][n][n] = { 0 };

    // Prefix sum array
    int prefix_sum[n] = { 0 };
    prefix_sum[0] = a[0];

    // Calculating prefix_sum
    for (int i = 1; i < n; i++) {
        prefix_sum[i] = prefix_sum[i - 1] + a[i];
    }

    // Filling the table diagonally
    for (int len = 1; len <= n; len++) {
        for (int i = 0; i + len - 1 < n; i++) {
            int j = i + len - 1;
            for (int num = 1; num <= n; num++) {
                // Sum of array in range (l, r)
                int current_sum
                    = prefix_sum[j]
                      - (i - 1 >= 0 ? prefix_sum[i - 1]
                                    : 0);
                if (num % 2 == 0) {
                    // If the operation is even-numbered
                    // the score is decremented
                    current_sum *= -1;
                }
                if (i == j) {
                    dp[i][j][num] = current_sum;
                }
                else {
                    // Exploring all paths, and storing
                    // maximum value in DP table to avoid
                    // further repetitive recursive calls
                    dp[i][j][num] = max(
                        current_sum + dp[i + 1][j][num + 1],
                        current_sum
                            + dp[i][j - 1][num + 1]);
                }
            }
        }
    }

    return dp[0][n - 1][1];
}

// Driver code
int main()
{
    int n = 6;
    int A[n] = { 1, 2, 3, 4, 2, 6 };
    cout << findMaxScore(A, n);
    return 0;
}

Javascript

// Javascript program to find the maximum
// Score after given operations

// Function to find the max score
function findMaxScore(a, n) {
    // initialize Dp to store computation of subproblems
    let dp = new Array(n).fill(0).map(() => new Array(n).fill(0).map(() => new Array(n).fill(0)));

    // Prefix sum array
      let prefix_sum = new Array(n).fill(0);
      prefix_sum[0] = a[0];

    // Calculating prefix_sum
      for (let i = 1; i < n; i++) {
        prefix_sum[i] = prefix_sum[i - 1] + a[i];
      }

     // Filling the table diagonally
      for (let len = 1; len <= n; len++) {
        for (let i = 0; i + len - 1 < n; i++) {
            let j = i + len - 1;
            for (let num = 1; num <= n; num++) {
             // Sum of array in range (l, r)
            let current_sum = prefix_sum[j] - (i - 1 >= 0 ? prefix_sum[i - 1] : 0);
            if (num % 2 === 0) {
                // If the operation is even-numbered
                // the score is decremented
              current_sum *= -1;
            }
            if (i === j) {
              dp[i][j][num] = current_sum;
            }
            else {

             // Exploring all paths, and storing
            // maximum value in DP table to avoid
            // further repetitive recursive calls
              dp[i][j][num] = Math.max(
                current_sum + dp[i + 1][j][num + 1],
                current_sum + dp[i][j - 1][num + 1]
              );
            }
          }
        }
      }
      return dp[0][n - 1][1];
}

// Driver code
let n = 6;
let A = [1, 2, 3, 4, 2, 6];
console.log(findMaxScore(A, n));

Output:

13

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