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 Ai, delete one occurrence and add it to the sum.
- Delete one occurrence of Ai-1 and Ai+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 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 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: (Amax + 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(Amax ), where Amax is the maximum element present in array A[].