Split N powers of 2 into two subsets such that their difference of sum is minimum

Given an even number N, the task is to split all N powers of 2 into two sets such that the difference of their sum is minimum.

Examples:

Input: n = 4
Output: 6
Explanation:
Here n = 4 which means we have 21, 22, 23, 24. The most optimal way to divide it into two groups with equal element is 24 + 21 in one group and 22 + 23 in another group giving a minimum possible difference of 6.

Input: n = 8
Output: 30
Explanation:
Here n = 8 which means we have 21, 22, 23, 24, 25, 26, 27, 28. The most optimal way to divide it into two groups with equal element is 28 + 21 + 22 + 23 in one group and 24 + 25 + 26 + 27 in another group giving a minimum possible difference of 30.

Approach: To solve the problem mentioned above we have to follow the steps given below:



  • In the first group add the largest element that is 2N.
  • After adding the first number in one group, add N/2 – 1 more elements to this group, where the elements has to start from the least power of two or from the beginning of the sequence.

    For example: if N = 4 then add 24 to the first group and add N/2 – 1, i.e. 1 more element to this group which is the least which means is 21.

  • The remaining element of the sequence forms the elements of the second group.
  • Calculate the sum for both the groups and then find the absolute difference between the groups which will eventually be the minimum.

Below is the implementation of the above approach:

C++

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// C++ program to find the minimum
// difference possible by splitting
// all powers of 2 up to N into two 
// sets of equal size 
#include<bits/stdc++.h>
using namespace std;
  
void MinDiff(int n)
{
  
    // Store the largest
    int val = pow(2, n);
      
    // Form two separate groups
    int sep = n / 2;
      
    // Initialize the sum 
    // for two groups as 0
    int grp1 = 0;
    int grp2 = 0;
      
    // Insert 2 ^ n in the
    // first group
    grp1 = grp1 + val;
      
    // Calculate the sum of
    // least n / 2 -1 elements
    // added to the first set 
    for(int i = 1; i < sep; i++)
       grp1 = grp1 + pow(2, i);
          
    // Sum of remaining
    // n / 2 - 1 elements
    for(int i = sep; i < n; i++)
       grp2 = grp2 + pow(2, i);
          
    // Min Difference between
    // the two groups
    cout << (abs(grp1 - grp2));
  
// Driver code
int main()
{
    int n = 4;
    MinDiff(n); 
}
  
// This code is contributed by Bhupendra_Singh

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Java

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// Java program to find the minimum 
// difference possible by splitting 
// all powers of 2 up to N into two  
// sets of equal size  
import java.lang.Math; 
  
class GFG{ 
  
public static void MinDiff(int n) 
  
    // Store the largest 
    int val = (int)Math.pow(2, n); 
      
    // Form two separate groups 
    int sep = n / 2
      
    // Initialize the sum 
    // for two groups as 0 
    int grp1 = 0
    int grp2 = 0
      
    // Insert 2 ^ n in the 
    // first group 
    grp1 = grp1 + val; 
      
    // Calculate the sum of 
    // least n / 2 -1 elements 
    // added to the first set 
    for(int i = 1; i < sep; i++) 
       grp1 = grp1 + (int)Math.pow(2, i); 
          
    // Sum of remaining 
    // n / 2 - 1 elements 
    for(int i = sep; i < n; i++) 
       grp2 = grp2 + (int)Math.pow(2, i); 
          
    // Min difference between 
    // the two groups 
    System.out.println(Math.abs(grp1 - grp2)); 
  
// Driver Code 
public static void main(String args[]) 
    int n = 4
      
    MinDiff(n); 
  
// This code is contributed by grand_master

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Python3

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# Python3 program to find the minimum
# difference possible by splitting
# all powers of 2 up to N into two 
# sets of equal size  
  
def MinDiff(n):
      
    # Store the largest
    val = 2 ** n
      
    # Form two separate groups
    sep = n // 2
      
    # Initialize the sum 
    # for two groups as 0
    grp1 = 0
    grp2 = 0
      
    # Insert 2 ^ n in the
    # first group
    grp1 = grp1 + val
      
    # Calculate the sum of
    # least n / 2 -1 elements
    # added to the first set 
    for i in range(1, sep):
        grp1 = grp1 + 2 ** i
          
      
    # sum of remaining
    # n / 2 - 1 elements
    for i in range(sep, n):
        grp2 = grp2 + 2 ** i
          
    # Min Difference between
    # the two groups
    print(abs(grp1 - grp2))     
      
# Driver code
if __name__=='__main__':
    n = 4
    MinDiff(n)

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C#

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// C# program to find the minimum 
// difference possible by splitting 
// all powers of 2 up to N into two 
// sets of equal size 
using System;
class GFG{ 
  
public static void MinDiff(int n) 
  
    // Store the largest 
    int val = (int)Math.Pow(2, n); 
      
    // Form two separate groups 
    int sep = n / 2; 
      
    // Initialize the sum 
    // for two groups as 0 
    int grp1 = 0; 
    int grp2 = 0; 
      
    // Insert 2 ^ n in the 
    // first group 
    grp1 = grp1 + val; 
      
    // Calculate the sum of 
    // least n / 2 -1 elements 
    // added to the first set 
    for(int i = 1; i < sep; i++) 
    grp1 = grp1 + (int)Math.Pow(2, i); 
          
    // Sum of remaining 
    // n / 2 - 1 elements 
    for(int i = sep; i < n; i++) 
    grp2 = grp2 + (int)Math.Pow(2, i); 
          
    // Min difference between 
    // the two groups 
    Console.Write(Math.Abs(grp1 - grp2)); 
  
// Driver Code 
public static void Main() 
    int n = 4; 
      
    MinDiff(n); 
  
// This code is contributed by Akanksha_Rai

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Output:

6

Time complexity: O(N)
Auxiliary Space: O(1)

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