Minimum Excluded Value of subarray - c++

Question

mex here refers to the minimum excluded value of an array.
I have and array with about 300000 elements. Then I have 50000 pairs of a and b (1<=a<300000 b<=300000) and computer need to tell me the MEX of array in range from a to b.
Example:
int a[10]={0,1,2,3,0,4,5,4,8,6};
a=3;
b=8;
We start from a[2] end and at a[7] , and MEX will be 1
I made a funtion, but it takes 30 seconds for one pair which is not what I need.
int Mex=0;
vector<int> a;
// fill vector
while (find(a.begin(),a.end(), Mex )!= a.end())
{
Mex++;
}
cout << Mex;
So the quextion is: how to make fast MEX function with C++, so that computer can do 50000 various MEX operations for 300000 elements in 5 seconds? Memory limit is 900MB.

Answer 1

Here I am sorting the range, then trying to figure out mex.
int mex(std::vector<int>& num_list, int start, int end) {
std::vector<int> range_list(num_list.begin() + start, num_list.begin() + end);
std::sort(range_list.begin(), range_list.end());
int mex_val = 0;
for (auto&el : range_list) {
int diff = el - mex_val;
if (diff > 1) {
break;
}
mex_val++;
}
return mex_val;
}
int main()
{
std::vector<int>data = {0,1,2,3,0,4,5,4,8,6};
cout<<mex(data, 3, 8)<<"\n";
return 0;
}

Answer 2

Run some tests and see if this works for you.
Worst case:O((b-a)²)
Best case: O(b-a)
#include <iostream>
//Calculate mex starting at myArray[a-1] and ending at myArray[b-1]
int mex(int* myArray, int a, int b){
int mex = 0; //Lowest positive natural number
int position = a-1; //To keep track of where you are in the array
while(position != b){
if(mex == myArray[position]){
mex++;
position = a-1;
}
else
position++;
}
return mex;
}
int main(){
int myArray[10]={0,1,2,3,0,4,5,4,8,6};
int a = 3;
int b = 8;
std::cout << mex(myArray, a, b);
return 0;
}