Repeat and Missing Number Array

You are given a read only array of n integers from 1 to n.

Each integer appears exactly once except A which appears twice and B which is missing.

Return A and B.

Note: Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

Note that in your output A should precede B.

Example:

Input:[3 1 2 5 3] 

Output:[3, 4] 

A = 3, B = 4


// Space Complexity = 0(n)
vector<int> Solution::repeatedNumber(const vector<int> &A) {
    vector<int> B(A.size()+1,0); 
    vector<int> result;
    for(int i=0 ; i<A.size() ; i++){
        if(B[A[i]] == 0){
        B[A[i]] = 1;            
        }
        else {
            result.push_back(A[i]);
        }
    }
    for(int i=1 ; i<B.size() ; i++){
        if(B[i] == 0){
            result.push_back(i);
        }
    }
    B.erase();
    return result;
}

// uses no extra memory
 long long int len = A.size();
    long long int sumOfN = (len * (len+1) ) /2, sumOfNsq = (len * (len +1) *(2*len +1) )/6;
    long long int missingNumber1=0, missingNumber2=0;
    
    for(int i=0;i<A.size(); i++){
       sumOfN -= (long long int)A[i];
       sumOfNsq -= (long long int)A[i]*(long long int)A[i];
    }
    
    missingNumber1 = (sumOfN + sumOfNsq/sumOfN)/2;
    missingNumber2= missingNumber1 - sumOfN;
    vector <int> ans;
    ans.push_back(missingNumber2);
    ans.push_back(missingNumber1);
    return ans;