Reduce Array and Maximize sum by deleting one occurrence of A[i] and all occurrences of A[i]+1 and A[i]-1

Given an array A[] having N positive integers, the task is to perform the following operations and maximize the sum obtained while reducing the array:

• Select an array element (say A[i]) and delete one occurrence of that element and add A[i] to the sum.
• Delete all the occurrences of A[i]-1 and A[i]+1.
• Perform these operations until the array is empty.

Examples:

Input: A[] = {3,  4,  2}
Output: 6
Explanation: First, delete number 4 to add 4 to sum and consequently 3 is also deleted.
 After that the array A = [2]. 
Then we delete number 2 and add 2.
Array becomes empty i.e. array A = [].
And hence total sum = (4 + 2) = 6

Input: A[] = {2, 2, 3, 3, 3, 4}
Output: 9
Explanation: First, delete number 3 to add 3. And all 2’s and 4’s numbers are also deleted. 
After that we the array is A = [3, 3]. 
Then delete number 3 again to add 3 points. Sum = 3 + 3 = 6.
The array is now A = [3].
In last operation delete number 3 once again to add 3. Sum = 6+3 = 9.
Now array becomes empty.
Hence maximum sum obtained is 9.

Naive Approach: The naive approach is to try to reduce the array in all possible ways, i.e. for any value (say A[i] )that can be selected and one occurrence of that element can be deleted or any other element having difference of 1 with A[i] can be selected (if that is present in array) and one occurrence of that can be removed.

Time Complexity: O(2^N)
Auxiliary Space: O(N)

Efficient Approach: This problem can be solved using Dynamic Programming based on the following idea:

If an element x is deleted once then all occurrences of x-1 and x+1 will be removed from array.

• So, if all array elements having value from 0 till x is considered then maximum sum till x depends on maximum sum till x-2 and maximum sum till x-1, i.e. if x is included then x-1 cannot be included and vice versa. [No need to consider the x+1 or x+2 because here iteration is from lower value to upper value side]
• Suppose these max values for each x are stored in an array (say dp[]) then dp[x] = max(dp[x-1], dp[x-2]+x*occurrences of x).

Follow the illustration below for a better understanding.

Illustration:

Consider an array A[] = {2, 2, 3, 3, 3, 4}

So the frequency of elements will be:
freq = {{2 -> 2}, {3 -> 3}, {4 -> 1}}

Maximum of array is 4.
So the dp[] array will be of size 5.
The dp[] array will be {0, 0, 0, 0, 0}

For x = 2:
        => dp[2] = max(dp[1], dp[0] + freq[2]*2)
                       = max(0, 2*2) = max(0, 0 + 4) = 4
        => dp[] = {0, 0, 4, 0, 0}

For x = 3:
        => dp[3] = max(dp[2], dp[1] + freq[3]*3)
                       = max(4, 0 + 3*3) = max(0, 9) = 9
        => dp[] = {0, 0, 4, 9, 0}

For x = 4:
        => dp[4] = max(dp[3], dp[2] + freq[4]*4)
                       = max(9, 4 + 4*1) = max(9, 8) = 9
        => dp[] = {0, 0, 4, 9, 9}

So the answer is dp[4] = 9 which is the maximum possible sum

Follow the steps mentioned below to implement the above observation:

• Create an unordered_map mp to store the frequency of each array element.
• Calculate the maximum value of the array (say max_val).
• Create one array dp[] to store the maximum values as in the observation and initialize dp[1] as count of 1s.
• Iterate from i = 2 to max_val:
  • Calculate the dp[i] as mentioned above.
• Return the dp[max_val] after all iterations as answer because it holds the maximum possible sum.

Below is the implementation of the above approach:

C++

// C++ program to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to return Maximum number
// of points that can be earned
int MaximumPoints(int A[], int array_size)
{
    // Maximum element in array A
    int element_max = *max_element(A, A
                                          + array_size);
    unordered_map<int, int> mp;

    // Dp array for storing points
    int dp[element_max + 1] = { 0 };

    // Storing frequency of integers
    for (int i = 0; i < array_size; i++) {
        mp[A[i]]++;
    }

    dp[0] = 0;
    dp[1] = mp[1];

    // Calculation for getting maximum sum
    // in dp[] array at every steps
    for (int i = 2; i <= element_max; i++) {
        dp[i] = max(dp[i - 1],
                    dp[i - 2] + mp[i] * i);
    }

    // Returning the maximum sum
    return dp[element_max];
}

int main()
{
    int A[] = { 2, 2, 3, 3, 3, 4 };

    // Size of Array
    int array_size = sizeof(A) / sizeof(A[0]);

    // Function call
    cout << MaximumPoints(A, array_size);
    return 0;
}

Java

// Java program to implement the approach
import java.util.*;
import java.util.Arrays;

public class GFG {
  // Function to return Maximum number
  // of points that can be earned
  static int MaximumPoints(int A[], int array_size)
  {
    // Maximum element in array A
    int element_max =Arrays.stream(A).max().getAsInt();
    HashMap<Integer, Integer> mp = new HashMap<>();

    // Dp array for storing points
    int dp[] = new int[element_max + 1];

    // Storing frequency of integers
    for (int i = 0; i < array_size; i++) {
      if(mp.containsKey(A[i])){
        mp.put(A[i], mp.get(A[i]) + 1);
      }
      else{
        mp.put(A[i], 1);
      }
    }

    dp[0] = 0;
    if(mp.containsKey(1))
      dp[1] = mp.get(1);

    // Calculation for getting maximum sum
    // in dp[] array at every steps
    for (int i = 2; i <= element_max; i++) {
      dp[i] = Math.max(dp[i - 1],
                       dp[i - 2] + mp.get(i) * i);
    }

    // Returning the maximum sum
    return dp[element_max];
  }

  // Driver Code
  public static void main (String[] args) {
    int A[] = { 2, 2, 3, 3, 3, 4 };

    // Size of Array
    int array_size = A.length;

    // Function call
    System.out.print(MaximumPoints(A, array_size));
  }
}

// This code is contributed by hrithikgarg03188.

Python3

# Python program to implement the approach

# Function to return Maximum number
# of points that can be earned
def MaximumPoints(A, array_size):

    # Maximum element in array A
    element_max = max(A)
    mp = {}

    # Dp array for storing points
    dp = [0 for i in range(element_max + 1)]

    # Storing frequency of integers
    for i in range(array_size):

        if (A[i] in mp):
            mp[A[i]] = mp[A[i]] + 1
        else:
            mp[A[i]] = 1

    if(1 in mp):
        dp[1] = mp[1]

    # Calculation for getting maximum sum
    # in dp[] array at every steps
    for i in range(2,element_max+1):
        dp[i] = (max(dp[i - 1], dp[i - 2] + mp[i] * i))

    # Returning the maximum sum
    return dp[element_max]

A = [ 2, 2, 3, 3, 3, 4 ]

# Size of Array
array_size = len(A)

# Function call
print(MaximumPoints(A, array_size))

# This code is contributed by shinjanpatra

C#

// C# code to implement the approach
using System;
using System.Collections.Generic;
using System.Linq;

public class GFG
{

  // Function to return Maximum number
  // of points that can be earned
  static int MaximumPoints(int[] A, int array_size)
  {
    // Maximum element in array A
    int element_max = A.Max();
    Dictionary<int, int> mp
      = new Dictionary<int, int>();

    // Dp array for storing points
    int[] dp = new int[element_max + 1];

    // Storing frequency of integers
    for (int i = 0; i < array_size; i++) {
      if (mp.ContainsKey(A[i])) {
        mp[A[i]] += 1;
      }
      else {
        mp[A[i]] = 1;
      }
    }

    dp[0] = 0;
    if (mp.ContainsKey(1))
      dp[1] = mp[1];

    // Calculation for getting maximum sum
    // in dp[] array at every steps
    for (int i = 2; i <= element_max; i++) {
      dp[i] = Math.Max(dp[i - 1],
                       dp[i - 2] + mp[i] * i);
    }

    // Returning the maximum sum
    return dp[element_max];
  }

  // Driver Code
  public static void Main(string[] args)
  {
    int[] A = { 2, 2, 3, 3, 3, 4 };

    // Size of Array
    int array_size = A.Length;

    // Function call
    Console.Write(MaximumPoints(A, array_size));
  }
}

// This code is contributed by phasing17.

Javascript

// JavaScript program to implement the approach

// Function to return Maximum number
// of points that can be earned
function MaximumPoints(A, array_size)
{
    // Maximum element in array A
    var element_max = Math.max(... A);
    mp = {};

    // Dp array for storing points
    var dp = [];

    // Storing frequency of integers
    for (var i = 0; i < array_size; i++) {

        if (mp.hasOwnProperty(A[i]))
            mp[A[i]] = mp[A[i]] + 1;
        else {
            mp[A[i]] = 1;
        }
    }

    dp.push(0);
    if (dp.hasOwnProperty(1))
        dp.push(mp[1]);
    else
        dp.push(0);
    // Calculation for getting maximum sum
    // in dp[] array at every steps
    for (var i = 2; i <= element_max; i++) {
        dp.push(Math.max(dp[i - 1], dp[i - 2] + mp[i] * i));
    }
    // Returning the maximum sum
    return dp[element_max];
}

var A = [ 2, 2, 3, 3, 3, 4 ];

// Size of Array
var array_size = A.length;

// Function call
console.log(MaximumPoints(A, array_size));

// this code was contributed by phasing17

Output

9

Time Complexity: O(N+M) 
Auxiliary Space: O(N+M) where M is the maximum element of the array