Minimum operations to make the MEX of the given set equal to x

Given a set of n integers, perform minimum number of operations (you can insert/delete elements into/from the set) to make the MEX of the set equal to x (that is given). 

Note:- The MEX of a set of integers is the minimum non-negative integer that doesn't exist in it. For example, the MEX of the set {0, 2, 4} is 1 and the MEX of the set {1, 2, 3} is 0. 

Examples : 

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.

Approach: The approach is to see that in the final set all the elements less than x should exist, x shouldn't exist and any element greater than x doesn't matter. So, we will count the number of elements less than x that don't exist in the initial set and add this to the answer. If x exists we will add 1 to the answer because x should be removed.

Below is the implementation of above approach:

// CPP program to perform minimal number
// of operations to make the MEX of the
// set equal to the given number x.
#include <bits/stdc++.h>
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 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 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# 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.

<script>

    // 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));

</script>

<?php
// PHP 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)
{
    $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 Code
$arr = array(0, 4, 5, 6, 7);
$n = count($arr);
$x = 3;

echo minOpeartions($arr, $n, $x) ;

// This code is contributed by anuj_67.
?>

2

Time Complexity: O(n)
Auxiliary Space: O(1)