Number of K’s such that the given array can be divided into two sets satisfying the given conditions
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: 2
{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: 0
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
Time Complexity: O(n * log n)
Auxiliary Space: O(1)