# 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.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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:

## 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); ` `    ``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. `

## PHP

 ` `

Output :

```2
```

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