K 的个数，使得给定的数组可以分成满足给定条件的两组
原文:https://www . geeksforgeeks . org/k-number-of-ks-so-给定数组可以被分成满足给定条件的两个集合/
给定一个大小为 N 的数组 arr[] 。任务是找到 K 的数量，这样如果一个集合中有少于 K 的元素，而其余的元素在另一个集合中，那么这个数组就可以被分成包含相同数量元素的两个集合。 注 : N 始终为偶数。
示例:
输入: arr[] = {9，1，4，4，6，7} 输出: 2 {1，4，4}和{6，7，9}是两套。 K 可以是 5 或 6。 输入: arr[] = {1，2，3，3，4，5} 输出: 0
方法:一个有效的方法是排序数组。那么如果两个中间数相同，那么答案是零，否则答案是两个数之差。 以下是上述方法的实施:
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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<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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