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

• Difficulty Level : Hard
• Last Updated : 12 Jun, 2021

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 â‰¥ 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 arrayint dp[101][101][101]; // Function that removes elements// from array to maximize the costint 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 elementsint 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 Codeint main(){    // Given array    int arr[] = { 1, 3, 2, 2, 2, 3, 4, 3, 1 };     int M = 3;     int N = sizeof(arr) / sizeof(arr[0]);     // Function Call    cout << maxPoints(arr, N, M);    return 0;}

## Java

 // Java program for the above approachimport java.util.*; class GFG{ // Initialize dp arraystatic int [][][]dp = new int[101][101][101]; // Function that removes elements// from array to maximize the coststatic 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 elementsstatic 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 Codepublic 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 arraydp = [[[-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 costdef 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 elementsdef maxPoints(arr, n, m):  length = n  global dp   # Function Call  return helper(arr, 0,                length - 1, 0, m)  # Driver Codeif __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 approachusing System; class GFG{ // Initialize dp arraystatic int [,,]dp = new int[101, 101, 101]; // Function that removes elements// from array to maximize the coststatic 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 elementsstatic 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 Codepublic 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

## Javascript



Output:

27

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

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