# Maximum value obtained by performing given operations in an Array

Given an array arr[], the task is to find out the maximum obtainable value. The user is allowed to add or multiply the two consecutive elements. However, there has to be at least one addition operation between two multiplication operations (i.e), two consecutive multiplication operations are not allowed.

Let the array elements be 1, 2, 3, 4 then 1 * 2 + 3 + 4 is a valid operation whereas 1 + 2 * 3 * 4 is not a a valid operation as there are consecutive multiplication operations.

Examples:

```Input : 5 -1 -5 -3 2 9 -4
Output : 33
Explanation:
The maximum value obtained by following the above conditions is 33.
The sequence of operations are given as:
5 + (-1) + (-5) * (-3) + 2 * 9 + (-4) = 33

Input : 5 -3 -5 2 3 9 4
Output : 62
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach:

This problem can be solved by using dynamic programming.

1. Assuming 2D array dp[][] of dimensions n * 2.
2. dp[i][0] represents the maximum value of the array up to ith position if the last operation is addition.
3. dp[i][1] represents the maximum value of the array up to ith position if the last operation is multiplication.

Now, since consecutive multiplication operation is not allowed, the recurrence relation can be considered as :

```dp[i][0] = max(dp[ i - 1][0], dp[ i - 1][1]) + a[ i + 1];
dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1];
```

The base cases are:

```dp[0][0] = a[0] + a[1];
dp[0][1] = a[0] * a[1];
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// A function to calculate the maximum value ` `void` `findMax(``int` `a[], ``int` `n) ` `{ ` `    ``int` `dp[n][2]; ` `    ``memset``(dp, 0, ``sizeof``(dp)); ` `     `  `    ``// basecases ` `    ``dp[0][0] = a[0] + a[1]; ` `    ``dp[0][1] = a[0] * a[1]; ` `    `  `    ``//Loop to iterate and add the max value in the dp array ` `    ``for` `(``int` `i = 1; i <= n - 2; i++) { ` `        ``dp[i][0] = max(dp[i - 1][0], dp[i - 1][1]) + a[i + 1]; ` `        ``dp[i][1] = dp[i - 1][0] - a[i] + a[i] * a[i + 1]; ` `    ``} ` ` `  `    ``cout << max(dp[n - 2][0], dp[n - 2][1]); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``int` `arr[] = { 5, -1, -5, -3, 2, 9, -4 }; ` `    ``findMax(arr, 7); ` `} `

## Java

 `// Java implementation of the above approach  ` `class` `GFG  ` `{ ` `     `  `    ``// A function to calculate the maximum value  ` `    ``static` `void` `findMax(``int` `[]a, ``int` `n)  ` `    ``{  ` `        ``int` `dp[][] = ``new` `int``[n][``2``];  ` `        ``int` `i, j; ` `         `  `        ``for` `(i = ``0``; i < n ; i++) ` `            ``for``(j = ``0``; j < ``2``; j++) ` `                ``dp[i][j] = ``0``; ` ` `  `        ``// basecases  ` `        ``dp[``0``][``0``] = a[``0``] + a[``1``];  ` `        ``dp[``0``][``1``] = a[``0``] * a[``1``];  ` `         `  `        ``// Loop to iterate and add the  ` `        ``// max value in the dp array  ` `        ``for` `(i = ``1``; i <= n - ``2``; i++) ` `        ``{  ` `            ``dp[i][``0``] = Math.max(dp[i - ``1``][``0``],  ` `                                ``dp[i - ``1``][``1``]) + a[i + ``1``];  ` `            ``dp[i][``1``] = dp[i - ``1``][``0``] - a[i] +  ` `                        ``a[i] * a[i + ``1``];  ` `        ``}  ` `     `  `        ``System.out.println(Math.max(dp[n - ``2``][``0``],  ` `                                    ``dp[n - ``2``][``1``]));  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `main (String[] args)  ` `    ``{ ` `        ``int` `arr[] = { ``5``, -``1``, -``5``, -``3``, ``2``, ``9``, -``4` `};  ` `        ``findMax(arr, ``7``);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

## Python3

 `# Python3 implementation of the above approach  ` `import` `numpy as np ` ` `  `# A function to calculate the maximum value  ` `def` `findMax(a, n) :  ` ` `  `    ``dp ``=` `np.zeros((n, ``2``)); ` `     `  `    ``# basecases  ` `    ``dp[``0``][``0``] ``=` `a[``0``] ``+` `a[``1``];  ` `    ``dp[``0``][``1``] ``=` `a[``0``] ``*` `a[``1``];  ` `     `  `    ``# Loop to iterate and add the max value in the dp array  ` `    ``for` `i ``in` `range``(``1``, n ``-` `1``) : ` `        ``dp[i][``0``] ``=` `max``(dp[i ``-` `1``][``0``], dp[i ``-` `1``][``1``]) ``+` `a[i ``+` `1``];  ` `        ``dp[i][``1``] ``=` `dp[i ``-` `1``][``0``] ``-` `a[i] ``+` `a[i] ``*` `a[i ``+` `1``];  ` ` `  `    ``print``(``max``(dp[n ``-` `2``][``0``], dp[n ``-` `2``][``1``]), end ``=``"");  ` ` `  `# Driver Code  ` `if` `__name__ ``=``=` `"__main__"` `:  ` ` `  `    ``arr ``=` `[ ``5``, ``-``1``, ``-``5``, ``-``3``, ``2``, ``9``, ``-``4` `];  ` `    ``findMax(arr, ``7``);  ` `     `  `# This code is contributed by AnkitRai01 `

## C#

 `// C# implementation of the above approach  ` `using` `System; ` ` `  `class` `GFG  ` `{ ` `     `  `    ``// A function to calculate the maximum value  ` `    ``static` `void` `findMax(``int` `[]a, ``int` `n)  ` `    ``{  ` `        ``int` `[,]dp = ``new` `int``[n, 2];  ` `        ``int` `i, j; ` `         `  `        ``for` `(i = 0; i < n ; i++) ` `            ``for``(j = 0; j < 2; j++) ` `                ``dp[i, j] = 0; ` ` `  `        ``// basecases  ` `        ``dp[0, 0] = a[0] + a[1];  ` `        ``dp[0, 1] = a[0] * a[1];  ` `         `  `        ``// Loop to iterate and add the  ` `        ``// max value in the dp array  ` `        ``for` `(i = 1; i <= n - 2; i++) ` `        ``{  ` `            ``dp[i, 0] = Math.Max(dp[i - 1, 0], dp[i - 1, 1]) + a[i + 1]; ` `                                 `  `            ``dp[i, 1] = dp[i - 1, 0] - a[i] + a[i] * a[i + 1];  ` `        ``}  ` `     `  `        ``Console.WriteLine(Math.Max(dp[n - 2, 0], dp[n - 2, 1]));  ` `    ``}  ` `     `  `    ``// Driver Code  ` `    ``public` `static` `void` `Main()  ` `    ``{ ` `        ``int` `[]arr = { 5, -1, -5, -3, 2, 9, -4 };  ` `        ``findMax(arr, 7);  ` `    ``}  ` `} ` ` `  `// This code is contributed by AnkitRai01 `

Output:

```33
```

GeeksforGeeks has prepared a complete interview preparation course with premium videos, theory, practice problems, TA support and many more features. Please refer Placement 100 for details

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : AnkitRai01