Skip to content
Related Articles

Related Articles

Improve Article

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

  • Last Updated : 26 Apr, 2021

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:
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++




// 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

Java




// 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

Python3




# 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)

C#




// 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

Javascript




<script>
// javascript program to find the minimum
// difference possible by splitting
// all powers of 2 up to N into two 
// sets of equal size 
 
    function MinDiff(n)
    {
 
        // Store the largest
        var val = parseInt( Math.pow(2, n));
 
        // Form two separate groups
        var sep = n / 2;
 
        // Initialize the sum
        // for two groups as 0
        var grp1 = 0;
        var 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 = 1; i < sep; i++)
            grp1 = grp1 + parseInt( Math.pow(2, i));
 
        // Sum of remaining
        // n / 2 - 1 elements
        for (i = sep; i < n; i++)
            grp2 = grp2 + parseInt( Math.pow(2, i));
 
        // Min difference between
        // the two groups
        document.write(Math.abs(grp1 - grp2));
    }
 
    // Driver Code
        var n = 4;
        MinDiff(n);
 
// This code is contributed by gauravrajput1
</script>
Output: 
6

 

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

Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.




My Personal Notes arrow_drop_up
Recommended Articles
Page :