Repeat and Missing Number Array

• Time : $O(n)$
• Space : $O(n)$

vector<int> vis;
vector<int> Solution::repeatedNumber(const vector<int> &A) {
    int n = A.size();
    vis = vector<int>(n + 1);
    
    int dup = -1;
    for(auto& a : A) if(++vis[a] == 2) dup = a;
    for(int i = 1; i <= n; i++) if(!vis[i]) return {dup, i};
    return {dup, -1};
}