# Maximize cost to empty an array by removing contiguous subarrays of equal elements

Given an array arr[] consisting of N integers and an integer M, the task is to find the maximum cost that can be obtained by performing the following operation any number of times.

In one operation, choose K contiguous elements with same value(where K ≥ 1) and remove them and cost of this operation is K * M

Examples:

Input: arr[] = {1, 3, 2, 2, 2, 3, 4, 3, 1}, M = 3
Output: 27
Explanation:
Step 1: Remove three contiguous 2’s to modify arr[] = {1, 3, 3, 4, 3, 1}. Cost = 3 * 3 = 9
Step 2: Remove 4 to modify arr[] = {1, 3, 3, 3, 1}. Cost = 9 + 1 * 3 = 12
Step 3: Remove three contiguous 3’s to modify arr[] = {1, 1}. Cost = 12 + 3 * 3 = 21
Step 4: Remove two contiguous 1’s to modify arr[] = {}. Cost = 21 + 2 * 3 = 27

Input: arr[] = {1, 2, 3, 4, 5, 6, 7}, M = 2
Output: 14

Approach: This problem can be solved using Dynamic Programming. Below are the steps:

1. Initialize a 3D array dp[][][] such that dp[left][right][count], where left and right denotes operation between indices [left, right] and count is the number of elements to the left of arr[left] having same value as that of arr[left] and the count excludes arr[left].
2. Now there are the following two possible choices:
• Ending the sequence to remove the elements of the same value including the starting element (i.e., arr[left]) and then continue from the next element onward.
• Continue the sequence to search between the indices [left + 1, right] for elements having the same value as arr[left](say index i), this enables us to continue the sequence.
3. Make recursive calls from the new sequence and continue the process for the previous sequence.
4. Print the maximum cost after all the above steps.

Below is the implementation of the above approach:

## C++

 `// C++ program for the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Initialize dp array ` `int` `dp; ` ` `  `// Function that removes elements ` `// from array to maximize the cost ` `int` `helper(``int` `arr[], ``int` `left, ``int` `right, ` `           ``int` `count, ``int` `m) ` `{ ` `    ``// Base case ` `    ``if` `(left > right) ` `        ``return` `0; ` ` `  `    ``// Check if an answer is stored ` `    ``if` `(dp[left][right][count] != -1) { ` `        ``return` `dp[left][right][count]; ` `    ``} ` ` `  `    ``// Deleting count + 1 i.e. including ` `    ``// the first element and starting a ` `    ``// new sequence ` `    ``int` `ans = (count + 1) * m ` `              ``+ helper(arr, left + 1, ` `                       ``right, 0, m); ` ` `  `    ``for` `(``int` `i = left + 1; ` `         ``i <= right; ++i) { ` ` `  `        ``if` `(arr[i] == arr[left]) { ` ` `  `            ``// Removing [left + 1, i - 1] ` `            ``// elements to continue with ` `            ``// previous sequence ` `            ``ans = max( ` `                ``ans, ` `                ``helper(arr, left + 1, ` `                       ``i - 1, 0, m) ` `                    ``+ helper(arr, i, right, ` `                             ``count + 1, m)); ` `        ``} ` `    ``} ` ` `  `    ``// Store the result ` `    ``dp[left][right][count] = ans; ` ` `  `    ``// Return answer ` `    ``return` `ans; ` `} ` ` `  `// Function to remove the elements ` `int` `maxPoints(``int` `arr[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `len = n; ` `    ``memset``(dp, -1, ``sizeof``(dp)); ` ` `  `    ``// Function Call ` `    ``return` `helper(arr, 0, len - 1, 0, m); ` `} ` ` `  `// Driver Code ` `int` `main() ` `{ ` `    ``// Given array ` `    ``int` `arr[] = { 1, 3, 2, 2, 2, 3, 4, 3, 1 }; ` ` `  `    ``int` `M = 3; ` ` `  `    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr); ` ` `  `    ``// Function Call ` `    ``cout << maxPoints(arr, N, M); ` `    ``return` `0; ` `} `

## Java

 `// Java program for the above approach ` `import` `java.util.*; ` ` `  `class` `GFG{ ` ` `  `// Initialize dp array ` `static` `int` `[][][]dp = ``new` `int``[``101``][``101``][``101``]; ` ` `  `// Function that removes elements ` `// from array to maximize the cost ` `static` `int` `helper(``int` `arr[], ``int` `left, ``int` `right, ` `                  ``int` `count, ``int` `m) ` `{ ` `     `  `    ``// Base case ` `    ``if` `(left > right) ` `        ``return` `0``; ` ` `  `    ``// Check if an answer is stored ` `    ``if` `(dp[left][right][count] != -``1``)  ` `    ``{ ` `        ``return` `dp[left][right][count]; ` `    ``} ` ` `  `    ``// Deleting count + 1 i.e. including ` `    ``// the first element and starting a ` `    ``// new sequence ` `    ``int` `ans = (count + ``1``) * m +  ` `             ``helper(arr, left + ``1``, ` `                    ``right, ``0``, m); ` ` `  `    ``for``(``int` `i = left + ``1``; i <= right; ++i) ` `    ``{ ` `        ``if` `(arr[i] == arr[left])  ` `        ``{ ` `             `  `            ``// Removing [left + 1, i - 1] ` `            ``// elements to continue with ` `            ``// previous sequence ` `            ``ans = Math.max(ans, ` `                  ``helper(arr, left + ``1``, ` `                         ``i - ``1``, ``0``, m) +  ` `                  ``helper(arr, i, right, ` `                         ``count + ``1``, m)); ` `        ``} ` `    ``} ` ` `  `    ``// Store the result ` `    ``dp[left][right][count] = ans; ` ` `  `    ``// Return answer ` `    ``return` `ans; ` `} ` ` `  `// Function to remove the elements ` `static` `int` `maxPoints(``int` `arr[], ``int` `n, ``int` `m) ` `{ ` `    ``int` `len = n; ` `    ``for``(``int` `i = ``0``; i < ``101``; i++)  ` `    ``{ ` `        ``for``(``int` `j = ``0``; j < ``101``; j++) ` `        ``{ ` `            ``for``(``int` `k = ``0``; k < ``101``; k++) ` `                ``dp[i][j][k] = -``1``; ` `        ``} ` `    ``} ` ` `  `    ``// Function call ` `    ``return` `helper(arr, ``0``, len - ``1``, ``0``, m); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `main(String[] args) ` `{ ` `     `  `    ``// Given array ` `    ``int` `arr[] = { ``1``, ``3``, ``2``, ``2``, ``2``, ``3``, ``4``, ``3``, ``1` `}; ` ` `  `    ``int` `M = ``3``; ` ` `  `    ``int` `N = arr.length; ` ` `  `    ``// Function call ` `    ``System.out.print(maxPoints(arr, N, M)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar`

## Python3

 `# Python3 program for  ` `# the above approach ` ` `  `# Initialize dp array ` `dp ``=` `[[[``-``1` `for` `x ``in` `range``(``101``)] ` `            ``for` `y ``in` `range``(``101``)] ` `            ``for` `z ``in` `range``(``101``)] ` `  `  `# Function that removes elements ` `# from array to maximize the cost ` `def` `helper(arr, left,  ` `           ``right, count, m): ` ` `  `  ``# Base case ` `  ``if` `(left > right): ` `    ``return` `0` ` `  `  ``# Check if an answer is stored ` `  ``if` `(dp[left][right][count] !``=` `-``1``): ` `    ``return` `dp[left][right][count] ` ` `  `  ``# Deleting count + 1 i.e. including ` `  ``# the first element and starting a ` `  ``# new sequence ` `  ``ans ``=` `((count ``+` `1``) ``*` `m ``+`  `          ``helper(arr, left ``+` `1``, ` `                 ``right, ``0``, m)) ` ` `  `  ``for` `i ``in` `range` `(left ``+` `1``, ` `                  ``right ``+` `1``): ` ` `  `    ``if` `(arr[i] ``=``=` `arr[left]): ` ` `  `      ``# Removing [left + 1, i - 1] ` `      ``# elements to continue with ` `      ``# previous sequence ` `      ``ans ``=` `(``max``(ans, helper(arr, left ``+` `1``, ` `                             ``i ``-` `1``, ``0``, m) ``+`  `                         ``helper(arr, i, right, ` `                              ``count ``+` `1``, m))) ` ` `  `      ``# Store the result ` `      ``dp[left][right][count] ``=` `ans ` ` `  `      ``# Return answer ` `      ``return` `ans ` `  `  `# Function to remove the elements ` `def` `maxPoints(arr, n, m): ` `  ``length ``=` `n ` `  ``global` `dp ` ` `  `  ``# Function Call ` `  ``return` `helper(arr, ``0``,  ` `                ``length ``-` `1``, ``0``, m) ` `  `  `# Driver Code ` `if` `__name__ ``=``=` `"__main__"``: ` ` `  `  ``# Given array ` `  ``arr ``=` `[``1``, ``3``, ``2``, ``2``,  ` `         ``2``, ``3``, ``4``, ``3``, ``1``] ` `  ``M ``=` `3` `  ``N ``=` `len``(arr) ` ` `  `  ``# Function Call ` `  ``print``(maxPoints(arr, N, M)) ` `     `  `# This code is contributed by Chitranayal`

## C#

 `// C# program for the above approach ` `using` `System; ` ` `  `class` `GFG{ ` ` `  `// Initialize dp array ` `static` `int` `[,,]dp = ``new` `int``[101, 101, 101]; ` ` `  `// Function that removes elements ` `// from array to maximize the cost ` `static` `int` `helper(``int` `[]arr, ``int` `left, ``int` `right, ` `                  ``int` `count, ``int` `m) ` `{ ` `     `  `    ``// Base case ` `    ``if` `(left > right) ` `        ``return` `0; ` ` `  `    ``// Check if an answer is stored ` `    ``if` `(dp[left, right, count] != -1)  ` `    ``{ ` `        ``return` `dp[left, right, count]; ` `    ``} ` ` `  `    ``// Deleting count + 1 i.e. including ` `    ``// the first element and starting a ` `    ``// new sequence ` `    ``int` `ans = (count + 1) * m +  ` `              ``helper(arr, left + 1, ` `                    ``right, 0, m); ` ` `  `    ``for``(``int` `i = left + 1; i <= right; ++i) ` `    ``{ ` `        ``if` `(arr[i] == arr[left])  ` `        ``{ ` `             `  `            ``// Removing [left + 1, i - 1] ` `            ``// elements to continue with ` `            ``// previous sequence ` `            ``ans = Math.Max(ans, ` `                  ``helper(arr, left + 1, ` `                         ``i - 1, 0, m) +  ` `                  ``helper(arr, i, right, ` `                         ``count + 1, m)); ` `        ``} ` `    ``} ` ` `  `    ``// Store the result ` `    ``dp[left, right, count] = ans; ` ` `  `    ``// Return answer ` `    ``return` `ans; ` `} ` ` `  `// Function to remove the elements ` `static` `int` `maxPoints(``int` `[]arr, ``int` `n, ``int` `m) ` `{ ` `    ``int` `len = n; ` `    ``for``(``int` `i = 0; i < 101; i++)  ` `    ``{ ` `        ``for``(``int` `j = 0; j < 101; j++) ` `        ``{ ` `            ``for``(``int` `k = 0; k < 101; k++) ` `                ``dp[i, j, k] = -1; ` `        ``} ` `    ``} ` ` `  `    ``// Function call ` `    ``return` `helper(arr, 0, len - 1, 0, m); ` `} ` ` `  `// Driver Code ` `public` `static` `void` `Main(String[] args) ` `{ ` `     `  `    ``// Given array ` `    ``int` `[]arr = { 1, 3, 2, 2, 2, 3, 4, 3, 1 }; ` ` `  `    ``int` `M = 3; ` ` `  `    ``int` `N = arr.Length; ` ` `  `    ``// Function call ` `    ``Console.Write(maxPoints(arr, N, M)); ` `} ` `} ` ` `  `// This code is contributed by Amit Katiyar`

Output:

```27
```

Time Complexity: O(N4
Auxiliary Space: O(N3

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