Maximize the sum of selected numbers from an array to make it empty

Given an array A[] of N numbers, we need to maximize the sum of selected numbers following the given operation:

• At each step, you need to select a number A_i, delete one occurrence and add it to the sum.
• Delete one occurrence of A_i-1 and A_i+1 (if they exist in the array). 
• Repeat these steps until the array gets empty.

Examples:

Input: A[] = {1, 2, 3} 
Output: 4
Explanation: At first step we select 1, so 1 and 2 are deleted from the sequence leaving us with 3. 
Then we select 3 from the sequence and delete it.
So the sum of selected numbers is 1+3 = 4. 

Input: A[] =  {1, 2, 2, 2, 3, 4}
Output: 10 
Explanation: Select one of the 2’s from the array. 
So  2, 2-1, 2+1 will be deleted. 
Now array is {2, 2, 4}, since 1 and 3 are deleted. 
Select 2 in next two steps, and then select 4 in the last step.
We get a sum of 2 + 2 + 2 + 4 = 10 which is the maximum possible. 

Approach: The idea to solve the problem is:

Pre-calculate the occurrence of all numbers ( say x ) in the array A[] and then iterate from maximum number to minimum number.
For each number x delete one occurrence of x and x-1(if exists) and add x to the sum until x is completely removed.

Follow the steps mentioned below to solve the problem

• Calculate the MAX value in the array.
• Create an array of size MAX and store the occurrences of each element in it.
• Since we want to maximize our answer, we will start iterating from the MAX value to 0.
• If the occurrence of the i^th element is greater than 0, then add it to our answer decrease the occurrences of the i-1^th element by 1, and also decrease the occurrence of i^th by 1 since we have added it to our answer.
• We don’t have to decrease the occurrence of the i+1^th element because we are already starting from the end so i+1^th is already processed.
• There might be multiple occurrences of the i^th element that’s why do not decrease i yet, to stay on the same element.

Below is the implementation of the above idea:  

C++

// CPP program to Maximize the sum of selected
// numbers by deleting three consecutive numbers.
#include <bits/stdc++.h>
using namespace std;

// function to maximize the sum of selected numbers
int maximizeSum(int arr[], int n) {

      // Largest element in the array
    int mx = -1;
    for(int i = 0; i < n; i++)
    {
        mx = max(mx, arr[i]);
    }

    // An array to count the occurrence of each element
    int freq[mx + 1];

    memset(freq, 0, sizeof(freq));

    for(int i = 0; i < n; i++)
    {
        freq[arr[i]]++;
    }

    // ans to store the result
    int ans = 0, i=mx;

    // Using the above mentioned approach
    while(i>0){

        // if occurrence is greater than 0
        if(freq[i] > 0){
            // add it to ans
            ans += i;

            // decrease i-1th element by 1
            freq[i-1]--;

            // decrease ith element by 1
            freq[i]--;
        }else{
            // decrease i
            i--;
        }
    }

    return ans;
}

// Driver code
int main()
{
  int a[] = {1, 2, 3};
  int n = sizeof(a) / sizeof(a[0]);
  cout << maximizeSum(a, n);
  return 0;
}

Java

// Java implementation of the approach
import java.util.*;
import java.math.*;

class GFG
{
      // Function to maximise the sum of selected numbers
      //by deleting occurrences of Ai-1 and Ai+1
    public static int getMaximumSum (int arr[]) {

        // Number of elements in the array
        int n = arr.length;

        // Largest element in the array
        int max = -1;
        for(int i = 0; i < n; i++)
        {
            max = Math.max(max, arr[i]);
        }

        // An array to count the occurrence of each element
        int []freq = new int[max + 1];

        for(int i = 0; i < n; i++)
        {
            freq[arr[i]]++;
        }

        // ans to store the result
        int ans = 0, i=max;

        // Using the above mentioned approach
        while(i>0){

            // if occurrence is greater than 0
            if(freq[i] > 0){
                // add it to ans
                ans += i;

                // decrease i-1th element by 1
                freq[i-1]--;

                // decrease ith element by 1
                freq[i]--;
            }else{
                // decrease i
                i--;
            }
        }

        return ans;
    }

      // Driver code
      public static void main(String[] args)
      {
          int []a = {1, 2, 3};

          System.out.println(getMaximumSum(a));
      }
}

Python3

# Python3 program to Maximize the sum of selected
# numbers by deleting three consecutive numbers.

# function to maximize the sum of
# selected numbers
def maximizeSum(a, n) :

        # maximum in the sequence
    maximum = max(a)

    # stores the occurrences of the numbers
    ans = dict.fromkeys(range(0, n + 1), 0)

    # marks the occurrence of every
    # number in the sequence
    for i in range(n) :
        ans[a[i]] += 1

    # ans to store the result
    result = 0
    i = maximum

    # Using the above mentioned approach
    while i > 0:

        # if occurrence is greater than 0
        if ans[i] > 0:
            # add it to ans
            result += i;

            # decrease i-1th element by 1
            ans[i-1] -= 1;

            # decrease ith element by 1
            ans[i] -= 1;
        else:
            # decrease i
            i -= 1;

    return result;

# Driver code
if __name__ == "__main__" :

    a = [1, 2, 3]
    n = len(a)
    print(maximizeSum(a, n))

# This code is contributed by Ryuga

C#

// C# implementation of the approach
using System;
using System.Linq;

class GFG
{

// Function to maximise the sum of selected numbers
//by deleting occurrences of Ai-1 and Ai+1
static int getMaximumSum(int []arr)
{
    // Number of elements in the array
    int n = arr.Length;

    // Largest element in the array
    int max = arr.Max();

    // An array to count the occurrence of each element
    int []freq = new int[max + 1];

    for(int j = 0; j < n; j++)
    {
      freq[arr[j]]++;
    }

    // ans to store the result
    int ans = 0, i=max;

    // Using the above mentioned approach
    while(i>0){

      // if occurrence is greater than 0
      if(freq[i] > 0){
        // add it to ans
        ans += i;

        // decrease i-1th element by 1
        freq[i-1]--;

        // decrease ith element by 1
        freq[i]--;
      }else{
        // decrease i
        i--;
      }
    }

    return ans;
}

// Driver code
public static void Main(string[] args)
{
    int []a = {1, 2, 3};

    Console.Write(getMaximumSum(a));
}
}

// This code is contributed by rock_cool

Javascript

<script>
    // Javascript implementation of the approach

    // Function to maximise the sum of selected numbers
    //by deleting occurrences of Ai-1 and Ai+1
    function getMaximumSum(arr)
    {
        // Number of elements in the array
        let n = arr.length;

        // Largest element in the array
        let max = Number.MIN_VALUE;
        for(let i = 0; i < n; i++)
        {
            max = Math.max(max, arr[i]);
        }

        // An array to count the occurrence of each element
        let freq = new Array(max + 1);
        freq.fill(0);

        for(let j = 0; j < n; j++)
        {
          freq[arr[j]]++;
        }

        // ans to store the result
        let ans = 0, i=max;

        // Using the above mentioned approach
        while(i>0){

          // if occurrence is greater than 0
          if(freq[i] > 0){
            // add it to ans
            ans += i;

            // decrease i-1th element by 1
            freq[i-1]--;

            // decrease ith element by 1
            freq[i]--;
          }else{
            // decrease i
            i--;
          }
        }

        return ans;
    }

    let a = [1, 2, 3];

    document.write(getMaximumSum(a));

    // This code is contributed by suresh07.
</script>

Output:

4

Time Complexity: (A_max + Highest occurrence of element in arr), because if the frequency is greater than 1 then we are processing that element multiple times.
Auxiliary Space: O(A_max ), where A_max is the maximum element present in array A[].