# Maximum score of deleting an element from an Array based on given condition

Given an array arr[], the task is to find the maximum score of deleting an element where each element of the array can be deleted with the score of the element, but the constraint is if we delete arr[i], then arr[i] + 1 and arr[i] – 1 is gets automatically deleted with 0 scores.
Examples:

Input: arr[] = {7, 2, 1, 8, 3, 3, 6, 6}
Output: 27
Explanation:
Step 0: arr[] = 7 2 1 8 3 3 6 6, Score:
Step 1: arr[] = 7 1 8 3 6 6, Score:
Step 2: arr[] = 7 1 8 6 6, Score:
Step 3: arr[] = 7 8 6 6, Score:
Step 4: arr[] = 8 6, Score: 13
Step 5: arr[] = 8 Score: 19
Step 6: arr[] = [] Score: 27
Input: arr[] = 1 2 3
Output:

Approach: The idea is to use Dynamic Programming to solve this problem. The key observation of the problem is that for removing any element from the array the occurrence of the element and the value itself is the important factor.
Let’s take an example to understand the observation if the sequence was 4 4 5. Then, we have two choices to choose from 4 to 5. Now, on choosing 4, his score would be 4*2 = 8. On the other hand, if we choose 5, his score would be 5*1 = 5. Clearly, the maximal score is 8.
Hence, for the above sequence 4 4 5, freq[4] = 2, and freq[5] = 1.
Finally, to find the optimal score, it would be easy to first break the problem down into a smaller problem. In this case, we break the sequence into smaller sequences and find an optimal solution for it. For the sequence of numbers containing only 0, the answer would be 0. Similarly, if a sequence contains only the number 0 and 1, then the solution would be count[1]*1.
Recurrence Relation:

dp[i] = max(dp[i – 1], dp[i – 2] + i*freq[i])

Basically, we have 2 cases, either to pick an ith element and other is not to pick an ith element.

Case 1: If we pick the ith element, Maximum score till the ith element will be dp[i-2] + i*freq[i] (choosing ith element means deleting (i-1)th element)
Case 2: If we don’t pick the ith element, the Maximum Score till ith element will dp[i-1]
Now as We have to maximize the score, We will take the maximum of both.
Below is the implementation of the above approach:

## C++

 // C++ implementation to find the // maximum score of the deleting a // element from an array   #include   using namespace std;   // Function to find the maximum // score of the deleting an element // from an array int findMaximumScore(vector a, int n) {       // Creating a map to keep     // the frequency of numbers     unordered_map freq;       // Loop to iterate over the     // elements of the array     for (int i = 0; i < n; i++) {         freq[a[i]]++;     }       // Creating a DP array to keep     // count of max score at ith element     // and it will be filled     // in the bottom Up manner     vector dp(*max_element(a.begin(),                                 a.end())                        + 1,                    0);     dp[0] = 0;     dp[1] = freq[1];       // Loop to choose the elements of the     // array to delete from the array     for (int i = 2; i < dp.size(); i++)         dp[i] = max(             dp[i - 1],             dp[i - 2] + freq[i] * i);       return dp[dp.size() - 1]; }   // Driver Code int main() {     int n;     n = 3;     vector a{ 1, 2, 3 };       // Function Call     cout << findMaximumScore(a, n);     return 0; }

## Java

 // Java implementation to find the // maximum score of the deleting a // element from an array import java.util.*;   class GFG{   // Function to find the maximum // score of the deleting an element // from an array static int findMaximumScore(int []a, int n) {           // Creating a map to keep     // the frequency of numbers     @SuppressWarnings("unchecked")     HashMap freq = new HashMap();       // Loop to iterate over the     // elements of the array     for(int i = 0; i < n; i++)     {         if(freq.containsKey(a[i]))         {             freq.put(a[i],                      freq.get(a[i]) + 1);         }         else         {             freq.put(a[i], 1);         }     }       // Creating a DP array to keep     // count of max score at ith element     // and it will be filled     // in the bottom Up manner     int []dp = new int[Arrays.stream(a).max().getAsInt() + 1];     dp[0] = 0;     dp[1] = freq.get(1);       // Loop to choose the elements of the     // array to delete from the array     for(int i = 2; i < dp.length; i++)         dp[i] = Math.max(dp[i - 1],                          dp[i - 2] +                        freq.get(i) * i);       return dp[dp.length - 1]; }   // Driver Code public static void main(String[] args) {     int n;     n = 3;     int []a = { 1, 2, 3 };       // Function call     System.out.print(findMaximumScore(a, n)); } }   // This code is contributed by 29AjayKumar

## Python3

 # Python3 implementation to find the # maximum score of the deleting a # element from an array from collections import defaultdict   # Function to find the maximum # score of the deleting an element # from an array def findMaximumScore(a, n):         # Creating a map to keep     # the frequency of numbers     freq = defaultdict (int)       # Loop to iterate over the     # elements of the array     for i in range (n):         freq[a[i]] += 1       # Creating a DP array to keep     # count of max score at ith element     # and it will be filled     # in the bottom Up manner     dp = [0] * (max(a) + 1)     dp[0] = 0     dp[1] = freq[1]       # Loop to choose the elements of the     # array to delete from the array     for i in range (2, len(dp)):         dp[i] = max(dp[i - 1],                     dp[i - 2] +                     freq[i] * i)       return dp[- 1]   # Driver Code if __name__ == "__main__":         n = 3     a = [1, 2, 3]       # Function Call     print(findMaximumScore(a, n))   # This code is contributed by Chitranayal

## C#

 // C# implementation to find the // maximum score of the deleting a // element from an array using System; using System.Linq; using System.Collections.Generic;   class GFG{   // Function to find the maximum // score of the deleting an element // from an array static int findMaximumScore(int []a, int n) {           // Creating a map to keep     // the frequency of numbers     Dictionary freq = new Dictionary();       // Loop to iterate over the     // elements of the array     for(int i = 0; i < n; i++)     {         if(freq.ContainsKey(a[i]))         {             freq[a[i]] = freq[a[i]] + 1;         }         else         {             freq.Add(a[i], 1);         }     }       // Creating a DP array to keep     // count of max score at ith element     // and it will be filled     // in the bottom Up manner     int []dp = new int[a.Max() + 1];     dp[0] = 0;     dp[1] = freq[1];       // Loop to choose the elements of the     // array to delete from the array     for(int i = 2; i < dp.Length; i++)         dp[i] = Math.Max(dp[i - 1],                          dp[i - 2] +                        freq[i] * i);       return dp[dp.Length - 1]; }   // Driver Code public static void Main(String[] args) {     int n;     n = 3;     int []a = { 1, 2, 3 };           // Function call     Console.Write(findMaximumScore(a, n)); } }   // This code is contributed by 29AjayKumar

## Javascript



Output

4

Time Complexity: O(N)

Auxiliary Space: O(N)

Another Approach Using Memoization:

## C++

 // C++ implementation to find the // maximum score of the deleting a // element from an array   #include   using namespace std;   // Function to find the maximum // score of the deleting an element // from an array int solve(vector &arr, int i)     {         if(i >= arr.size()) // if i is greater than size of array         {             return 0; // then simply returnn zero         }                   // current 'i' on which we are standing         int currValue = arr[i];  // current value         int currSum = arr[i]; // intial make sum as same as value         int index = i + 1; // index to take elemets, so  i + 1                   // while it is the same as the current value, include in our sum         while(index < arr.size() && arr[index] == currValue)         {             currSum += arr[i];             index++;         }                   // Now, we have to skip all the elements, whose value is equal to         // currValue + 1         while(index < arr.size() && arr[index] == currValue + 1)         {             index++;         }                   //And lastly, we have two choices-         //whether to include the sum of this current element         // in our answer         // or not include the sum of current element in our answer         // so we explore all possibility and take maximum of them                   return max(currSum + solve(arr, index), solve(arr, i + 1));                   // If we decide to take the curr element in our answer, then upto the         // elemet we skip the next value, we paas that index         // but if decided no to make this vurrent element then simply paas         // i + 1     }     int findMaximumScore(vector& arr) {         int n = arr.size(); // take the size of the array                   // sort the array to get rid of all arr[i] - 1 elements         sort(arr.begin(), arr.end());                   // solve function which give us our final answer         return solve(arr, 0);         //                â†‘         //                we start from zero index     }   // Driver Code int main() {     int n;     n = 3;     vector a{ 1, 2, 3 };       // Function Call     cout << findMaximumScore(a);     return 0; }

## Java

 import java.util.ArrayList; import java.util.Arrays; import java.util.Collections;   public class Main {       // Function to find the maximum score of deleting an     // element from an array     static int solve(ArrayList arr, int i)     {         if (i >= arr.size()) // if i is greater than size of array         {             return 0;         }                   // current 'i' on which we are standing         int currValue = arr.get(i);         int currSum = arr.get(i);         int index = i + 1;                   // while it is the same as the current value, include in our sum         while (index < arr.size()                && arr.get(index) == currValue) {             currSum += arr.get(i);             index++;         }                    // Now, we have to skip all the elements, whose value is equal to         // currValue + 1         while (index < arr.size()                && arr.get(index) == currValue + 1) {             index++;         }                   //And lastly, we have two choices-         //whether to include the sum of this current element         // in our answer         // or not include the sum of current element in our answer         // so we explore all possibility and take maximum of them         return Math.max(currSum + solve(arr, index),                         solve(arr, i + 1));     }           // If we decide to take the curr element in our answer, then upto the     // elemet we skip the next value, we paas that index     // but if decided no to make this vurrent element then simply paas     // i + 1     static int findMaximumScore(ArrayList arr)     {         int n = arr.size();                   // sort the array to get rid of all arr[i] - 1 elements         Collections.sort(arr);           return solve(arr, 0);     }           // Driver Code     public static void main(String[] args)     {         int n = 3;         ArrayList a             = new ArrayList<>(Arrays.asList(1, 2, 3));           // Function Call         System.out.println(findMaximumScore(a));     } }

## Python

 def solve(arr, i):     # Base case: if i is greater than or equal to the size of the array, return 0     if i >= len(arr):         return 0       currValue = arr[i]  # current value at index i     currSum = arr[i]  # initialize the current sum as the current value     index = i + 1  # index to take elements, starting from i + 1       # While the elements have the same value as the current value, include them in our sum     while index < len(arr) and arr[index] == currValue:         currSum += arr[i]         index += 1       # Now, we have to skip all the elements whose value is equal to currValue + 1     while index < len(arr) and arr[index] == currValue + 1:         index += 1       # Lastly, we have two choices - whether to include the sum of the current element in our answer     # or not include the sum of the current element in our answer. So, we explore all possibilities and take the maximum of them     return max(currSum + solve(arr, index), solve(arr, i + 1))     def find_maximum_score(arr):     arr.sort()  # Sort the array to get rid of all arr[i] - 1 elements     # Call the solve function to get our final answer, starting from index 0     return solve(arr, 0)     # Driver Code if __name__ == "__main__":     arr = [1, 2, 3]     print(find_maximum_score(arr))

## C#

 // C# implementation to find the // maximum score of the deleting a // element from an array using System; using System.Collections.Generic; using System.Linq;   class GFG {     // Function to find the maximum score by combining     // adjacent elements in the list     static int Solve(List arr, int i)     {         // Base case: if we reach the end of the list,         // return 0         if (i >= arr.Count) {             return 0;         }           // Initialize variables to keep track of the current         // value, current sum, and next index         int currValue = arr[i];         int currSum = arr[i];         int index = i + 1;           // Calculate the sum of adjacent elements with the         // same value         while (index < arr.Count                && arr[index] == currValue) {             currSum += arr[i];             index++;         }           // Skip elements that have a value of "currValue +         // 1" and move to the next unique value         while (index < arr.Count                && arr[index] == currValue + 1) {             index++;         }           // Return the maximum score by either choosing the         // current element and recursively calling the         // function for the next unique element or skipping         // the current element and calling the function for         // the next element.         return Math.Max(currSum + Solve(arr, index),                         Solve(arr, i + 1));     }       // Function to find the maximum score by first sorting     // the list and then calling Solve     static int FindMaximumScore(List arr)     {         arr.Sort(); // Sort the list in ascending order         return Solve(arr,                      0); // Start the recursive calculation                          // from the beginning of the list     }       static void Main()     {         List a = new List{ 1, 2, 3 };           // Find the maximum score and print the result         Console.WriteLine(FindMaximumScore(a));     } }

## Javascript

 // Function to recursively find the maximum score of // deleting an element from an array function solve(arr, i) {     if (i >= arr.length) {         return 0;     }       // current 'i' on which we are standing     let currValue = arr[i]; // current value     let currSum = arr[i]; // initialize sum as the current value     let index = i + 1; // index to take elements, so i + 1       // while the element is the same as the current value, include     // it in our sum     while (index < arr.length && arr[index] === currValue) {         currSum += arr[i];         index++;     }       // Now, we have to skip all the elements whose value is equal     // to currValue + 1     while (index < arr.length && arr[index] === currValue + 1) {         index++;     }       // Explore two choices:     // 1. Include the sum of the current element in our answer     // and move to the next unique element (index).     // 2. Exclude the current element from our answer and move     // to the next element (i + 1).     // Take the maximum of these two possibilities.     return Math.max(currSum + solve(arr, index), solve(arr, i + 1)); }   // Function to find the maximum score of deleting an element from // an array function findMaximumScore(arr) {     const n = arr.length;       // Sort the array in ascending order to get rid of all arr[i] - 1     // elements     arr.sort((a, b) => a - b);       // Call the solve function to get the final answer     return solve(arr, 0);     //        â†‘     // Start from the zero index }   // Driver Code const a = [1, 2, 3]; console.log(findMaximumScore(a));

Output

4

Time complexity: O(n)
Auxiliary Space: O(m)

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