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Maximize the sum of selected numbers from an array to make it empty
  • Difficulty Level : Hard
  • Last Updated : 19 Apr, 2021

Given an array of N numbers, we need to maximize the sum of selected numbers. At each step, you need to select a number Ai, delete one occurrence of it and delete all occurrences of Ai-1 and Ai+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 occurrences of Ai+1 and Ai-1 elements if they are present in the array and not Ai+1 and Ai-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[]. 

Approach:

  • 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 ith element is greater than 0, then add it to our answer decrease the occurrences of the i-1th element by 1, and also decrease the occurrence of ith by 1 since we have added it to our answer.
  • We don’t have to decrease the occurrence of the i+1th element because we are already starting from the end so i+1th is already processed.
  • There might be multiple occurrences of the ith element that’s why do not decrease i yet, to stay on the same element.

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 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 occurence 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 occurence 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 nummbers
      //by deleting occurences 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 occurence 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 occurence 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 occurence 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 nummbers
//by deleting occurences 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 occurence 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 occurence 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

Output: 

4

Time Complexity: Time Complexity will be the sum of (Amax + Highest occurrence of element in arr), because if the frequency is greater than 1 then we are processing that element multiple times.

-where Amax is the maximum element present in array A[].

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

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