You can keep an array of Boolean values, with each index representing whether or not you have seen this number in the array. It will be of size n+1, since mex of n elements is at most n+1. Fill this array in with a O(n) loop over the array, and then go over the Boolean array and find the first false value, that index will be your mex.

If there are $$$Q$$$ mex queries of ranges in an array lengthened $$$N$$$, then it can be solved in $$$O(Q\sqrt N+N)$$$ with Mo's algorithm.

in case of the given array is sorted, you can binary search on the first index that is not equal to it's value in the array, the answer will be that index. timeO(logN).

int mex(int *a, int n) {
    int l = 0, r = n - 1, ans = n;
    while (l <= r) {
        int mid = (l + r) / 2;
        if (a[mid] == mid) {
            l = mid + 1;
        } else {
            r = mid - 1;
            ans = mid;
        }
    }
    return ans;
}

If you want to just find out the mex of the whole array, write a simple loop.

int mex(vector<int> vc) {
	vector<int> cnt;
	int n=vc.size(); cnt.resize(n+2);
	for(auto id:vc) if(vc<=n) cnt[id]=1;
	for(int i=0;i<=n;i++) if(cnt[i]==0) return i;
}

If you want to query the mex of an interval for many times, this problem may help. https://www.luogu.com.cn/problem/P4137

Sorry I don't know "可持久化线段树"'s English translation.