使给定集合的 MEX 等于 x 的最小操作

给定一组 n 个整数，执行最少数量的操作（您可以在集合中插入/删除元素）以使集合的MEX等于 x（即给定）。
注意：-一组整数的MEX是其中不存在的最小非负整数。例如，集合 {0, 2, 4} 的MEX为 1，集合 {1, 2, 3} 的MEX为 0。
例子 ：

Input : n = 5, x = 3
        0 4 5 6 7
Output : 2
The MEX of the set {0, 4, 5, 6, 7} is 1 which is 
not equal to 3. So, we should add 1 and 2 to the
set. After adding 1 and 2, the set becomes 
{0, 1, 2, 4, 5, 6, 7} and 3 is the minimum
non-negative integer that doesn't exist in it.
So, the MEX of this set is 3 which is equal to
x i.e. 3. So, the output of this example is 2 
as we inserted 1 and 2 in the set.

Input : n = 1, x = 0
        1
Output : 0
In this example, the MEX of the given set {1}
is already 0. So, we do not need to perform 
any operation. So, the output is 0.

方法：
方法是看到在最终集合中小于 x 的所有元素都应该存在，x 不应该存在并且任何大于 x 的元素都无关紧要。因此，我们将计算初始集合中不存在的小于 x 的元素的数量，并将其添加到答案中。如果 x 存在，我们将在答案中加 1，因为 x 应该被删除。
以下是上述方法的实现：

C++

// CPP program to perform minimal number
// of operations to make the MEX of the
// set equal to the given number x.
#include 
using namespace std;
 
// function to find minimum number of
// operations required
int minOpeartions(int arr[], int n, int x)
{
    int k = x, i = 0;
    while (n--) {
 
        // if the element is less than x.
        if (arr[n] < x)
            k--;
 
        // if the element equals to x.
        if (arr[n] == x)
            k++;
    }
    return k;
}
 
// driver function
int main()
{
    int arr[] = { 0, 4, 5, 6, 7 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int x = 3;
    // output
    cout << minOpeartions(arr, n, x) << endl;
}

Java

// Java program to perform minimal number
// of operations to make the MEX of the
// set equal to the given number x.
import java.io.*;
 
class GFG {
 
    // function to find minimum number of
    // operations required
    static int minOpeartions(int arr[], int n, int x)
    {
 
        int k = x, i = 0;
        n--;
 
        while (n > -1) {
 
            // if the element is less than x.
            if (arr[n] < x)
                k--;
 
            // if the element equals to x.
            if (arr[n] == x)
                k++;
 
            n--;
        }
 
        return k;
    }
 
    // driver function
    public static void main(String args[])
    {
        int arr[] = { 0, 4, 5, 6, 7 };
        int n = arr.length;
        int x = 3;
 
        // output
        System.out.println(minOpeartions(arr, n, x));
    }
}
 
/* This code is contributed by Nikita Tiwari.*/

Python

# Python 3 program to perform minimal number
# of operations to make the MEX of the
# set equal to the given number x.
 
 
# function to find minimum number of
# operations required
def minOpeartions(arr, n, x) :
     
    k = x
    i = 0
    n = n-1
    while (n>-1) :
         
        # if the element is less than x.
        if (arr[n] < x) :
            k = k - 1
     
        # if the element equals to x.
        if (arr[n] == x) :
            k = k + 1
        n = n - 1
     
    return k
 
 
# driver function
arr = [ 0, 4, 5, 6, 7 ]
n = len(arr)
x = 3
 
# output
print( minOpeartions(arr, n, x))
 
 
# This code is contributed by Nikita Tiwari.

C#

// C# program to perform minimal number
// of operations to make the MEX of the
// set equal to the given number x.
using System;
 
class GFG {
 
    // function to find minimum number
    // of operations required
    static int minOpeartions(int[] arr,
                           int n, int x)
    {
         
        int k = x;
        n--;
 
        while (n > -1) {
 
            // if the element is less
            // than x.
            if (arr[n] < x)
                k--;
 
            // if the element equals
            // to x.
            if (arr[n] == x)
                k++;
 
            n--;
        }
 
        return k;
    }
 
    // driver function
    public static void Main()
    {
        int[] arr = { 0, 4, 5, 6, 7 };
        int n = arr.Length;
        int x = 3;
 
        // output
        Console.WriteLine(
            minOpeartions(arr, n, x));
    }
}
 
// This code is contributed by vt_m.

Javascript

 
    // Javascript program to perform minimal number
    // of operations to make the MEX of the
    // set equal to the given number x.
     
    // function to find minimum number of
    // operations required
    function minOpeartions(arr, n, x)
    {
        let k = x, i = 0;
        while (n-- > 0) {
 
            // if the element is less than x.
            if (arr[n] < x)
                k--;
 
            // if the element equals to x.
            if (arr[n] == x)
                k++;
        }
        return k;
    }
     
    let arr = [ 0, 4, 5, 6, 7 ];
    let n = arr.length;
    let x = 3;
    // output
    document.write(minOpeartions(arr, n, x));
 

输出 ：

2

时间复杂度： O(n)
辅助空间： O(1)