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.

**Output:**

2

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.