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:

## C++

`// 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

`// 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

`<?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. ` `?> ` |

**Output :**

2

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