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

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 i^th element and other is not to pick an i^th 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 <bits/stdc++.h>

using namespace std;

// Function to find the maximum
// score of the deleting an element
// from an array
int findMaximumScore(vector<int> a, int n)
{

    // Creating a map to keep
    // the frequency of numbers
    unordered_map<int, int> 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<int> 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<int> 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<Integer,
            Integer> 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<int,
               int> freq = new Dictionary<int,
                                          int>();

    // 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

<script>
// Javascript implementation to find the
// maximum score of the deleting a
// element from an array

// Function to find the maximum
// score of the deleting an element
// from an array
function findMaximumScore(a,n)
{
    // Creating a map to keep
    // the frequency of numbers
    let freq = new Map();

    // Loop to iterate over the
    // elements of the array
    for(let i = 0; i < n; i++)
    {
        if(freq.has(a[i]))
        {
            freq.set(a[i],
                     freq.get(a[i]) + 1);
        }
        else
        {
            freq.set(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
    let dp = new Array(Math.max(...a)+1);
    dp[0] = 0;
    dp[1] = freq.get(1);

    // Loop to choose the elements of the
    // array to delete from the array
    for(let 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
let n = 3;
let a=[1, 2, 3];

// Function call
document.write(findMaximumScore(a, n));

// This code is contributed by avanitrachhadiya2155
</script>

Output:

4

Time Complexity: O(N)

Auxiliary Space: O(N)