要添加的最小元素数，以使中位数等于x

长度为n的数组中的中位数是一个元素，在我们按非降序对元素进行排序(数组元素从1开始编号)后，它占据位置编号(n + 1)/ 2。数组(2、6、1、2、3)的中位数是数字2，数组(0、96、17、23)的中位数是数字17。

例子 ：

Input : 3 10
        10 20 30
Output : 1
In the first sample we can add number 9 
to array (10, 20, 30). The resulting array
(9, 10, 20, 30) will have a median in 
position (4+1)/2 = 2, that is, 10

Input : 3 4
        1 2 3
Output : 4
In the second sample you should add numbers 
4, 5, 5, 5. The resulting array has median
equal to 4.

第一种方法：-该方法是在数组中再加上一个数字x，直到数组的中位数等于x。下面是上述方法的实现：

C++

// C++ program to find minimum number
// of elements needs to add to the
// array so that its median equals x.
#include 
using namespace std;
 
// Returns count of elements to be
// added to make median x. This function
// assumes that a[] has enough extra space.
int minNumber(int a[], int n, int x)
{   
    // to sort the array in increasing order.
    sort(a, a + n);
 
    int k;
    for (k = 0; a[(n - 1) / 2] != x; k++) {
        a[n++] = x;
        sort(a, a + n);
    }
    return k;
}
 
// Driver code
main()
{
    int x = 10;
    int a[6] = { 10, 20, 30 };
    int n = 3;
    cout << minNumber(a, n, x) << endl;
    return 0;
}

Java

// Java program to find minimum number
// of elements needs to add to the
// array so that its median equals x.
import java.util.Arrays;
 
class GFG
{
 
// Returns count of elements to be
// added to make median x. This function
// assumes that a[] has enough extra space.
static int minNumber(int a[], int n, int x)
{
    // to sort the array in increasing order.
    Arrays.sort(a);
 
    int k;
    for (k = 0; a[(n) / 2] != x; k++)
    {
        a[n++] = x;
        Arrays.sort(a);
    }
    return k;
}
 
// Driver code
public static void main(String[] args)
{
    int x = 10;
    int a[] = { 10, 20, 30 };
    int n = 3;
    System.out.println(minNumber(a, n-1, x));
}
}
 
// This code has been contributed by 29AjayKumar

Python3

# Python3 program to find minimum number
# of elements needs to add to the
# array so that its median equals x.
 
# Returns count of elements to be added 
# to make median x. This function
# assumes that a[] has enough extra space.
def minNumber(a, n, x):
     
    # to sort the array in increasing order.
    a.sort(reverse = False)
    k = 0
    while(a[int((n - 1) / 2)] != x):
        a[n - 1] = x
        n += 1
        a.sort(reverse = False)
        k += 1
 
    return k
 
# Driver code
if __name__ == '__main__':
    x = 10
    a = [10, 20, 30]
    n = 3
    print(minNumber(a, n, x))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find minimum number
// of elements needs to add to the
// array so that its median equals x.
using System;
 
class GFG
{
 
// Returns count of elements to be
// added to make median x. This function
// assumes that a[] has enough extra space.
static int minNumber(int []a, int n, int x)
{
    // to sort the array in increasing order.
    Array.Sort(a);
 
    int k;
    for (k = 0; a[(n) / 2] != x; k++)
    {
        a[n++] = x;
        Array.Sort(a);
    }
    return k;
}
 
// Driver code
public static void Main(String[] args)
{
    int x = 10;
    int []a = { 10, 20, 30 };
    int n = 3;
    Console.WriteLine(minNumber(a, n-1, x));
}
}
 
// This code contributed by Rajput-Ji

Javascript

    // Javascript program to find minimum number
    // of elements needs to add to the
    // array so that its median equals x.
     
    // Returns count of elements to be
    // added to make median x. This function
    // assumes that a[] has enough extra space.
    function minNumber(a, n, x)
    {   
     
        // to sort the array in increasing order.
        a.sort();
 
        let k;
        for (k = 0; a[parseInt((n - 1) / 2, 10)] != x; k++) {
            a[n++] = x;
            a.sort();
        }
        return k;
    }
 
    let x = 10;
    let a = [ 10, 20, 30 ];
    let n = 3;
    document.write(minNumber(a, n, x));
     
    // This code is contributed by mukesh07.

输出 ：

1

时间复杂度： O(knLogn)

第二种方法：-更好的方法是对所有等于x(即e)，大于x(即h)且小于x(即l)的元素进行计数。然后 –
如果l大于h，则ans将为(l – h)+ 1 – e；
如果h大于l，则ans将为(h – l – 1)+ 1 – e；
我们可以使用Hoare的分区方案来计算较小，相等和较大的元素。

下面是上述方法的实现：

C++

// C++ program to find minimum number of
// elements to add so that its median
// equals x.
#include 
using namespace std;
 
int minNumber(int a[], int n, int x)
{
    int l = 0, h = 0, e = 0;
    for (int i = 0; i < n; i++) {
 
        // no. of elements equals to x,
        // that is, e.
        if (a[i] == x)
            e++;
 
        // no. of elements greater than x,
        // that is, h.
        else if (a[i] > x)
            h++;
 
        // no. of elements smaller than x,
        // that is, l.
        else if (a[i] < x)
            l++;
    }
 
    int ans = 0;
    if (l > h)
        ans = l - h;
    else if (l < h)
        ans = h - l - 1;
     
    // subtract the no. of elements
    // that are equal to x.
    return ans + 1 - e;
}
 
// Driver code
int main()
{
    int x = 10;
    int a[] = { 10, 20, 30 };
    int n = sizeof(a) / sizeof(a[0]);
    cout << minNumber(a, n, x) << endl;
    return 0;
}

Java

// Java program to find minimum number
// of elements to add so that its
// median equals x.
import java.util.*;
import java.lang.*;
 
class GFG {
 
    public static int minNumber(int a[],
                           int n, int x)
    {
        int l = 0, h = 0, e = 0;
        for (int i = 0; i < n; i++)
        {
 
            // no. of elements equals to
            // x, that is, e.
            if (a[i] == x)
                e++;
 
            // no. of elements greater
            // than x, that is, h.
            else if (a[i] > x)
                h++;
 
            // no. of elements smaller
            // than x, that is, l.
            else if (a[i] < x)
                l++;
        }
 
        int ans = 0;
        if (l > h)
            ans = l - h;
        else if (l < h)
            ans = h - l - 1;
     
        // subtract the no. of elements
        // that are equal to x.
        return ans + 1 - e;
    }
 
    // Driven Program
    public static void main(String[] args)
    {
        int x = 10;
        int a[] = { 10, 20, 30 };
        int n = a.length;
        System.out.println(
                      minNumber(a, n, x));
    }
}
 
// This code is contributed by
// Prasad Kshirsagar

Python3

# Python3 program to find minimum number
# of elements to add so that its median
# equals x.
 
def minNumber (a, n, x):
    l = 0
    h = 0
    e = 0
    for i in range(n):
     
        # no. of elements equals to x,
        # that is, e.
        if a[i] == x:
            e+=1
         
        # no. of elements greater than x,
        # that is, h.
        elif a[i] > x:
            h+=1
         
        # no. of elements smaller than x,
        # that is, l.
        elif a[i] < x:
            l+=1
     
    ans = 0;
    if l > h:
        ans = l - h
    elif l < h:
        ans = h - l - 1;
     
    # subtract the no. of elements
    # that are equal to x.
    return ans + 1 - e
 
# Driver code
x = 10
a = [10, 20, 30]
n = len(a)
print(minNumber(a, n, x))
 
# This code is contributed
# by "Abhishek Sharma 44"

C＃

// C# program to find minimum
// number of elements to add
// so that its median equals x.
using System;
 
class GFG
{
public static int minNumber(int []a,
                            int n,
                            int x)
{
    int l = 0, h = 0, e = 0;
    for (int i = 0; i < n; i++)
    {
 
        // no. of elements
        // equals to x,
        // that is, e.
        if (a[i] == x)
            e++;
 
        // no. of elements
        // greater than x,
        // that is, h.
        else if (a[i] > x)
            h++;
 
        // no. of elements smaller
        // than x, that is, l.
        else if (a[i] < x)
            l++;
    }
 
    int ans = 0;
    if (l > h)
        ans = l - h;
    else if (l < h)
        ans = h - l - 1;
 
    // subtract the no.
    // of elements that
    // are equal to x.
    return ans + 1 - e;
}
 
// Driver Code
public static void Main()
{
    int x = 10;
    int []a = {10, 20, 30};
    int n = a.Length;
    Console.WriteLine(
                minNumber(a, n, x));
}
}
 
// This code is contributed
// by anuj_67.

的PHP

 $x)
            $h++;
 
        // no. of elements smaller
        // than x, that is, l.
        else if ($a[$i] < $x)
            $l++;
    }
 
    $ans = 0;
    if ($l > $h)
        $ans = $l - $h;
    else if ($l < $h)
        $ans = $h - $l - 1;
     
    // subtract the no. of elements
    // that are equal to x.
    return $ans + 1 - $e;
}
 
// Driver code
$x = 10;
$a = array (10, 20, 30);
$n = sizeof($a) ;
echo minNumber($a, $n, $x), "\n";
 
// This code is contributed by jit_t
?>

Java脚本

 
// Javascript program to find minimum number
// of elements to add so that its median
// equals x.
function minNumber(a, n, x)
{
    let l = 0, h = 0, e = 0;
    for(let i = 0; i < n; i++)
    {
         
        // No. of elements equals to x,
        // that is, e.
        if (a[i] == x)
            e++;
 
        // No. of elements greater than x,
        // that is, h.
        else if (a[i] > x)
            h++;
 
        // No. of elements smaller than x,
        // that is, l.
        else if (a[i] < x)
            l++;
    }
 
    let ans = 0;
    if (l > h)
        ans = l - h;
    else if (l < h)
        ans = h - l - 1;
 
    // Subtract the no. of elements
    // that are equal to x.
    return ans + 1 - e;
}
 
// Driver code
let x = 10;
let a = [ 10, 20, 30 ];
let n = a.length;
 
document.write(minNumber(a, n, x));
 
// This code is contributed by suresh07
 

输出 ：

1

时间复杂度： O(n)