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

Given an array of N numbers, we need to maximize the sum of selected numbers. At each step you need to select a number A_i, delete one occurrence of A_i-1 (if exists), A_i+1 (if exists) and A_i each from the array. Repeat these steps until the array gets empty. The problem is to maximize the sum of selected numbers.

Note: We have to delete A_i+1 and A_i-1 elements if they are present in the array and not A_i+1 and A_i-1.

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 and we are left with {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. 

Our aim is to maximize the sum of selected numbers. The idea is to pre-calculate the occurrence of all numbers x in the array a[] in a hash ans. Now our recurrence relation will decide either to select a number or not. If we select the number then we take the occurrences of that number and the value stored at ans[i-2] as ans[i-1] will be deleted and not be taken to count. If we do not select the number then we take ans[i-1] which have been pre-calculated while moving forward.

ans[i] = max(ans[i-1], ans[i-2] + ans[i]*i )

At the end, ans[maximum] will have the maximum sum of selected numbers.

Below is the implementation of above idea:

// 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 a[], int n) {

  // stores the occurrences of the numbers
  unordered_map<int, int> ans;

  // marks the occurrence of every number in the sequence
  for (int i = 0; i < n; i++)
    ans[a[i]]++;

  // maximum in the sequence
  int maximum = *max_element(a, a + n);

  // traverse till maximum and apply the recurrence relation
  for (int i = 2; i <= maximum; i++) 
    ans[i] = max(ans[i - 1], ans[i - 2] + ans[i] * i);  

  // return the ans stored in the index of maximum
  return ans[maximum];
}

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

Output:

4

Time Complexity: O(A_max), where A_max is the maximum element present in the array A[].

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