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Number of K’s such that the given array can be divided into two sets satisfying the given conditions

  • Last Updated : 27 Apr, 2021

Given an array arr[] of size N. The task is to find the number of K‘s such that the array can be divided into two sets containing equal number of elements if elements less than K are in one set and the rest of them are in the other set. 
Note: N is always even.
 

Examples:  

Input: arr[] = {9, 1, 4, 4, 6, 7} 
Output:
{1, 4, 4} and {6, 7, 9} are the two sets. 
K can be 5 or 6.
Input: arr[] = {1, 2, 3, 3, 4, 5} 
Output:

Approach: An efficient approach is to sort the array. Then if two middle numbers are the same then the answer is zero otherwise answer is the difference between the two numbers.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of K's
// such that the array can be divided
// into two sets containing equal number
// of elements when all the elements less
// than K are in one set and the rest
// of the elements are in the other set
int two_sets(int a[], int n)
{
    // Sort the given array
    sort(a, a + n);
 
    // Return number of such Ks
    return a[n / 2] - a[(n / 2) - 1];
}
 
// Driver code
int main()
{
    int a[] = { 1, 4, 4, 6, 7, 9 };
    int n = sizeof(a) / sizeof(a[0]);
 
    cout << two_sets(a, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the count of K's
// such that the array can be divided
// into two sets containing equal number
// of elements when all the elements less
// than K are in one set and the rest
// of the elements are in the other set
static int two_sets(int a[], int n)
{
    // Sort the given array
    Arrays.sort(a);
 
    // Return number of such Ks
    return a[n / 2] - a[(n / 2) - 1];
}
 
// Driver code
public static void main(String []args)
{
    int a[] = { 1, 4, 4, 6, 7, 9 };
    int n = a.length;
 
    System.out.println(two_sets(a, n));
}
}
 
// This code is contributed by Rajput-Ji

Python3




# Python3 implementation of the approach
 
# Function to return the count of K's
# such that the array can be divided
# into two sets containing equal number
# of elements when all the elements less
# than K are in one set and the rest
# of the elements are in the other set
def two_sets(a, n) :
 
    # Sort the given array
    a.sort();
 
    # Return number of such Ks
    return (a[n // 2] - a[(n // 2) - 1]);
 
# Driver code
if __name__ == "__main__" :
 
    a = [ 1, 4, 4, 6, 7, 9 ];
    n = len(a);
 
    print(two_sets(a, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
// Function to return the count of K's
// such that the array can be divided
// into two sets containing equal number
// of elements when all the elements less
// than K are in one set and the rest
// of the elements are in the other set
static int two_sets(int []a, int n)
{
    // Sort the given array
    Array.Sort(a);
 
    // Return number of such Ks
    return a[n / 2] - a[(n / 2) - 1];
}
 
// Driver code
public static void Main(String []args)
{
    int []a = { 1, 4, 4, 6, 7, 9 };
    int n = a.Length;
 
    Console.WriteLine(two_sets(a, n));
}
}
 
// This code is contributed by PrinciRaj1992

Javascript




<script>
// javascript implementation of the approach
 
    // Function to return the count of K's
    // such that the array can be divided
    // into two sets containing equal number
    // of elements when all the elements less
    // than K are in one set and the rest
    // of the elements are in the other set
    function two_sets(a , n) {
        // Sort the given array
        a.sort();
 
        // Return number of such Ks
        return a[n / 2] - a[(n / 2) - 1];
    }
 
    // Driver code
     
        var a = [ 1, 4, 4, 6, 7, 9 ];
        var n = a.length;
 
        document.write(two_sets(a, n));
 
// This code contributed by umadevi9616
</script>
Output: 
2

 

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