Maximum score possible after performing given operations on an Array
Last Updated :
12 Jul, 2025
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:
- If the operation is odd-numbered, the score is incremented by the sum of all elements of the current array.
- If the operation is even-numbered, the score is decremented by the sum of all elements of the current array.
- 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
- 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.
- 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.
- 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 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 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# 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
<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>
Time complexity: O(2N)
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 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 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# 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
<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>
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:
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;
}
import java.util.*;
class GFG {
// Function to find the max score
static int findMaxScore(int[] a, int n)
{
// Initialize DP to store computation of subproblems
int[][][] dp = new int[n + 3][n + 3][n + 3];
// Prefix sum array
int[] prefixSum = new int[n];
prefixSum[0] = a[0];
// Calculating prefix_sum
for (int i = 1; i < n; i++) {
prefixSum[i] = prefixSum[i - 1] + a[i];
}
// Filling the table diagonally
for (int length = 1; length <= n; length++) {
for (int i = 0; i + length - 1 < n; i++) {
int j = i + length - 1;
for (int num = 1; num <= n; num++) {
// Sum of array in range (l, r)
int currentSum
= prefixSum[j]
- (i - 1 >= 0 ? prefixSum[i - 1]
: 0);
// If the operation is even-numbered,
// the score is decremented
if (num % 2 == 0) {
currentSum *= -1;
}
if (i == j) {
dp[i][j][num] = currentSum;
}
else {
// Exploring all paths, and storing
// maximum value in DP table
dp[i][j][num] = Math.max(
currentSum
+ dp[i + 1][j][num + 1],
currentSum
+ dp[i][j - 1][num + 1]);
}
}
}
}
return dp[0][n - 1][1];
}
// Driver code
public static void main(String[] args)
{
int n = 6;
int[] A = { 1, 2, 3, 4, 2, 6 };
System.out.println(findMaxScore(A, n));
}
}
# Python3 program to find the maximum
# Score after given operations
def findMaxScore(a, n):
# Initialize DP to store computation of subproblems
dp = [[[0] * (n + 3) for _ in range(n+3)] for _ in range(n+3)]
# 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]
# Filling the table diagonally
for length in range(1, n + 1):
for i in range(n - length + 1):
j = i + length - 1
for num in range(1, n + 1):
# Sum of array in range (l, r)
current_sum = prefix_sum[j] - (prefix_sum[i - 1] if i - 1 >= 0 else 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
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
n = 6
A = [1, 2, 3, 4, 2, 6]
print(findMaxScore(A, n))
using System;
class GFG {
// Function to find the max score
static int FindMaxScore(int[] a, int n)
{
// Initialize DP to store computation of subproblems
int[, , ] dp = new int[n + 3, n + 3, n + 3];
// Prefix sum array
int[] prefixSum = new int[n];
prefixSum[0] = a[0];
// Calculating prefix_sum
for (int i = 1; i < n; i++) {
prefixSum[i] = prefixSum[i - 1] + a[i];
}
// Filling the table diagonally
for (int length = 1; length <= n; length++) {
for (int i = 0; i + length - 1 < n; i++) {
int j = i + length - 1;
for (int num = 1; num <= n; num++) {
// Sum of array in range (l, r)
int currentSum
= prefixSum[j]
- (i - 1 >= 0 ? prefixSum[i - 1]
: 0);
// If the operation is even-numbered,
// the score is decremented
if (num % 2 == 0) {
currentSum *= -1;
}
if (i == j) {
dp[i, j, num] = currentSum;
}
else {
// Exploring all paths, and storing
// maximum value in DP table
dp[i, j, num] = Math.Max(
currentSum
+ dp[i + 1, j, num + 1],
currentSum
+ dp[i, j - 1, num + 1]);
}
}
}
}
return dp[0, n - 1, 1];
}
// Driver code
static void Main()
{
int n = 6;
int[] A = { 1, 2, 3, 4, 2, 6 };
Console.WriteLine(FindMaxScore(A, n));
}
}
// 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)
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:
C++
#include <bits/stdc++.h>
using namespace std;
int findMaxScore(int a[], int n)
{
// DP array to store maximum scores
int dp[n][n] = {0};
// Prefix sum array to calculate subarray sums
int prefix_sum[n] = {0};
// Initialize the prefix sum with the first element
prefix_sum[0] = a[0];
for (int i = 1; i < n; i++) {
// Calculate prefix sum
prefix_sum[i] = prefix_sum[i - 1] + a[i];
}
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++) {
int current_sum = prefix_sum[j] - (i - 1 >= 0 ? prefix_sum[i - 1] : 0);
// Calculate the sum of the current subarray
if (num % 2 == 0) {
// Multiply sum by -1 if num is even
current_sum *= -1;
}
if (i == j) {
// Base case: subarray has only one element
dp[i][num] = current_sum;
} else {
// Choose the maximum score among two options:
// 1. Include the current element and move to the next element
// 2. Exclude the current element and move to the next element
dp[i][num] = max(current_sum + dp[i + 1][num + 1], current_sum + dp[i][num + 1]);
}
}
}
}
return dp[0][1];
}
int main()
{
int n = 6;
int A[] = {1, 2, 3, 4, 2, 6};
cout << findMaxScore(A, n) << endl;
return 0;
}
// --by bhardwajji
public class MaxScore {
static int findMaxScore(int[] arr, int n) {
// DP array to store maximum scores
int[][] dp = new int[n][n + 1];
// Prefix sum array to calculate subarray sums
int[] prefixSum = new int[n];
// Initialize the prefix sum with the first element
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++) {
// Calculate prefix sum
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
for (int length = 1; length <= n; length++) {
for (int i = 0; i <= n - length; i++) {
int j = i + length - 1;
for (int num = 1; num <= n; num++) {
int currentSum = prefixSum[j] - ((i - 1 >= 0) ? prefixSum[i - 1] : 0);
// Calculate the sum of the current subarray
if (num % 2 == 0) {
// Multiply sum by -1 if num is even
currentSum *= -1;
}
if (i == j) {
// Base case: subarray has only one element
dp[i][num] = currentSum;
} else {
// Choose the maximum score among two options:
// 1. Include the current element and move to the next element
// 2. Exclude the current element and move to the next element
dp[i][num] = Math.max(
currentSum + ((num + 1 <= n) ? dp[i + 1][num + 1] : 0),
currentSum + ((num + 1 <= n) ? dp[i][num + 1] : 0)
);
}
}
}
}
return dp[0][1];
}
public static void main(String[] args) {
int n = 6;
int[] A = {1, 2, 3, 4, 2, 6};
System.out.println(findMaxScore(A, n));
}
}
def findMaxScore(arr, n):
# DP array to store maximum scores
dp = [[0] * (n + 1) for _ in range(n)]
# Prefix sum array to calculate subarray sums
prefix_sum = [0] * n
# Initialize the prefix sum with the first element
prefix_sum[0] = arr[0]
for i in range(1, n):
# Calculate prefix sum
prefix_sum[i] = prefix_sum[i - 1] + arr[i]
for length in range(1, n + 1):
for i in range(n - length + 1):
j = i + length - 1
for num in range(1, n + 1):
current_sum = prefix_sum[j] - (prefix_sum[i - 1] if i - 1 >= 0 else 0)
# Calculate the sum of the current subarray
if num % 2 == 0:
# Multiply sum by -1 if num is even
current_sum *= -1
if i == j:
# Base case: subarray has only one element
dp[i][num] = current_sum
else:
# Choose the maximum score among two options:
# 1. Include the current element and move to the next element
# 2. Exclude the current element and move to the next element
dp[i][num] = max(current_sum + (dp[i + 1][num + 1] if num + 1 <= n else 0),
current_sum + dp[i][num + 1] if num + 1 <= n else 0)
return dp[0][1]
def main():
n = 6
A = [1, 2, 3, 4, 2, 6]
print(findMaxScore(A, n))
if __name__ == "__main__":
main()
using System;
class MaxScoreSubarray {
static int FindMaxScore(int[] arr, int n)
{
// DP array to store maximum scores
int[, ] dp = new int[n, n + 1];
// Prefix sum array to calculate subarray sums
int[] prefixSum = new int[n];
// Initialize the prefix sum with the first element
prefixSum[0] = arr[0];
for (int i = 1; i < n; i++) {
// Calculate prefix sum
prefixSum[i] = prefixSum[i - 1] + arr[i];
}
for (int length = 1; length <= n; length++) {
for (int i = 0; i + length - 1 < n; i++) {
int j = i + length - 1;
for (int num = 1; num <= n; num++) {
int currentSum
= prefixSum[j]
- (i - 1 >= 0 ? prefixSum[i - 1]
: 0);
// Calculate the sum of the current
// subarray
if (num % 2 == 0) {
// Multiply sum by -1 if num is even
currentSum *= -1;
}
if (i == j) {
// Base case: subarray has only one
// element
dp[i, num] = currentSum;
}
else {
// Choose the maximum score among
// two options:
// 1. Include the current element
// and move to the next element
// 2. Exclude the current element
// and move to the next element
dp[i, num] = Math.Max(
currentSum
+ (num + 1 <= n
? dp[i + 1, num + 1]
: 0),
currentSum
+ (num + 1 <= n
? dp[i, num + 1]
: 0));
}
}
}
}
return dp[0, 1];
}
static void Main()
{
int n = 6;
int[] A = { 1, 2, 3, 4, 2, 6 };
Console.WriteLine(FindMaxScore(A, n));
}
}
function GFG(a, n) {
// Initialize an array to store maximum scores
const dp = new Array(n).fill(0).map(() => new Array(n).fill(0));
// Initialize an array for prefix sums to
// calculate subarray sums
const prefixSum = new Array(n).fill(0);
// Initialize the prefix sum with
// the first element
prefixSum[0] = a[0];
for (let i = 1; i < n; i++) {
// Calculate prefix sum
prefixSum[i] = prefixSum[i - 1] + a[i];
}
for (let len = 1; len <= n; len++) {
for (let i = 0; i + len - 1 < n; i++) {
const j = i + len - 1;
for (let num = 1; num <= n; num++) {
let currentSum = prefixSum[j] - (i - 1 >= 0 ? prefixSum[i - 1] : 0);
// Calculate the sum of the current subarray
if (num % 2 === 0) {
// Multiply sum by -1 if num is even
currentSum *= -1;
}
if (i === j) {
// Base case: subarray has only one element
dp[i][num] = currentSum;
} else {
dp[i][num] = Math.max(currentSum + dp[i + 1][num + 1], currentSum + dp[i][num + 1]);
}
}
}
}
return dp[0][1];
}
function main() {
const n = 6;
const A = [1, 2, 3, 4, 2, 6];
console.log(GFG(A, n));
}
main();
Time complexity: O(N^3)
Auxiliary space: O(N^2)
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