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; }

**Output: **

2

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