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.


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
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.

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)
        // if the element equals to x.
        if (arr[n] == x)
    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;



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