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 it and delete all occurrences of A_i-1 and A_i+1 (if they exist) in the array. Repeat these steps until the array gets empty. The problem is to maximize the sum of the selected numbers.

Note: We have to delete all the occureneces of 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.

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

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 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; 
} 

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) : 
  
    # 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
  
    # maximum in the sequence  
    maximum = max(a)  
  
    # traverse till maximum and apply 
    # the recurrence relation  
    for i in range(2, maximum + 1) : 
        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  
if __name__ == "__main__" :  
  
    a = [1, 2, 3]  
    n = len(a) 
    print(maximizeSum(a, n)) 
  
# This code is contributed by Ryuga 

Output:

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

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Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Python3

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