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

Namachila October 13, 2025

0 0 10 minutes read

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:

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].
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.
Make recursive calls from the new sequence and continue the process for the previous sequence.
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 <bits/stdc++.h>
using namespace std;
 
// Initialize dp array
int dp[101][101][101];
 
// 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[0]);
 
    // 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

Javascript

<script>
 
// Javascript program for the above approach
 
// Initialize dp array
let dp = new Array(101);
for(let i = 0; i < 101; i++)
{
    dp[i] = new Array(101);
    for(let j = 0; j < 101; j++)
    {
        dp[i][j] = new Array(101);
        for(let k = 0; k < 101; k++)
        {
            dp[i][j][k] = -1;
        }
    }
}
 
// Function that removes elements
// from array to maximize the cost
function 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
    let ans = (count + 1) * m + 
              helper(arr, left + 1,
                     right, 0, m);
 
    for(let 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
function maxPoints(arr, n, m)
{
    let len = arr.length;
     
    // Function call
    return helper(arr, 0, len - 1, 0, m);
}
 
// Driver Code
 
// Given array
let arr = [ 1, 3, 2, 2, 2, 3, 4, 3, 1 ];
let M = 3;
let N = arr.length;
 
// Function call
document.write(maxPoints(arr, N, M));
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output:

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

Efficient approach : Using DP Tabulation method ( Iterative approach )

The approach to solve this problem is same but DP tabulation(bottom-up) method is better then Dp + memoization(top-down) because memoization method needs extra stack space of recursion calls.

Steps to solve this problem :

Create a 3D DP table to store the solution of the subproblems.
Initialize the table with base cases dp[i][i][1] = m + 1.
Now Iterate over subproblems to get the value of current problem form previous computation of subproblems stored in DP.
Return the final solution stored in dp[0][n-1][0].

Implementation :

C++

// C++ program for above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to remove the elements
int maxPoints(int arr[], int n, int m)
{
    int dp[n][n][n+1];
    memset(dp, 0, sizeof(dp));
     
    // Initializing diagonal elements
    for (int i = 0; i < n; i++) {
        dp[i][i][1] = m + 1;
    }
     
    // Filling dp table
      // get the current value from the previous
      // computations of subproblems
    for (int len = 2; len <= n; len++) {
        for (int i = 0; i <= n-len; i++) {
            int j = i + len - 1;
             
            // Deleting count + 1 i.e. including
            // the first element and starting a
            // new sequence
            dp[i][j][0] = (len * m);
             
            for (int k = 1; k <= len; k++) {
                if (arr[i] == arr[i+k-1]) {
                    dp[i][j][k] = max(dp[i][j][k], dp[i+1][i+k-2][0] + dp[i+k][j][k+1]);
                }
                for (int l = 0; l < k; l++) {
                    dp[i][j][k] = max(dp[i][j][k], dp[i][i+l][k] + dp[i+l+1][j][0]);
                }
            }
        }
    }
     
    // Return the maximum points
    return dp[0][n-1][0];
}
 
// 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[0]);
 
    // Function Call
    cout << maxPoints(arr, N, M);
    return 0;
}
// this code is contributed by bhardwajji

Java

import java.util.*;
 
public class Main {
 
    public static int maxPoints(int[] arr, int n, int m)
    {
        int[][][] dp = new int[n][n][n + 1];
        for (int[][] layer : dp) {
            for (int[] row : layer) {
                Arrays.fill(row, 0);
            }
        }
 
        for (int i = 0; i < n; i++) {
            dp[i][i][1] = m + 1;
        }
 
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                int j = i + len - 1;
                dp[i][j][0] = (len * m);
 
                for (int k = 1; k <= len; k++) {
                    if (arr[i] == arr[i + k - 1]) {
                        if (i + k - 2 >= i && i + k < n
                            && k + 1 <= n) {
                            dp[i][j][k] = Math.max(
                                dp[i][j][k],
                                dp[i + 1][i + k - 2][0]
                                    + dp[i + k][j][k + 1]);
                        }
                    }
                    for (int l = 0; l < k; l++) {
                        if (i + l <= j && i + l >= i
                            && i + l + 1 <= j) {
                            dp[i][j][k] = Math.max(
                                dp[i][j][k],
                                dp[i][i + l][k]
                                    + dp[i + l + 1][j][0]);
                        }
                    }
                }
            }
        }
 
        return dp[0][n - 1][0];
    }
 
    public static void main(String[] args)
    {
        int[] arr = { 1, 3, 2, 2, 2, 3, 4, 3, 1 };
        int m = 3;
        int n = arr.length;
        int maxPoints = maxPoints(arr, n, m);
        System.out.println(maxPoints);
    }
}

Python3

def maxPoints(arr, n, m):
    dp = [[[0 for i in range(n + 1)] for j in range(n)] for k in range(n)]
     
    for layer in dp:
        for row in layer:
            row[1] = m + 1
     
    for i in range(n):
        dp[i][i][1] = m + 1
     
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            j = i + length - 1
            dp[i][j][0] = length * m
             
            for k in range(1, length + 1):
                if arr[i] == arr[i + k - 1]:
                    if i + k - 2 >= i and i + k < n and k + 1 <= n:
                        dp[i][j][k] = max(dp[i][j][k], 
                        dp[i + 1][i + k - 2][0] + dp[i + k][j][k + 1])
                 
                for l in range(k):
                    if i + l <= j and i + l >= i and i + l + 1 <= j:
                        dp[i][j][k] = max(dp[i][j][k], dp[i][i +
                                      l][k] + dp[i + l + 1][j][0])
     
    return dp[0][n - 1][0]
 
 
arr = [1, 3, 2, 2, 2, 3, 4, 3, 1]
m = 3
n = len(arr)
max_points = maxPoints(arr, n, m)
print(max_points)

Output

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

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