All the below questions are to be done in O(n) time complexity.

1>Given an array of size n-1 whose entries are integers in the range [1, n], find an integer that is missing. Use only O(1) space and treat the input as read-only.

2>Given an array of size n-2 whose entries are integers in the range [1, n], find two integer that are missing.

3>There is an array A[N] of N integers. You have to compose an array B[N] such that Output[i] will be equal to the product of all the elements of A[] except A[i].

or
Example:
INPUT:[4, 3, 2, 1, 2]
OUTPUT:[12, 16, 24, 48, 24]

Solve it without division operator and in O(n).

Solution :

1>Let the missing number be M. We know that the sum of first N natural numbers is N*(N+1)/2. Traverse through the array once and calculate the sum. This is the sum of First N natural numbers - M or S=N*(N+1)/2 - M. Therefore M = N*(N+1)/2 - S.

2>Similar approach to the first one. Traverse the array once and calculate its sum and multiplication. Let the sum be S and multiplication be M. Let the missing numbers be P and Q. From above we know that P+Q = N*(N+1)/2 - S. Also P*Q = N!/M. So we can get P and Q from these two equations.

3> Lets first see the C++ solution.

void solution(vector<int> A)
{  
    int N=A.size(),i,j;
    vector<int> L(N,0);
    vector<int> R(N,0);
    vector<int> B(N,0);
    for (i=0,j=N-1; i<N && j>=0 ;i++, j--)
    {  
        L[i] = i==0? 1 : A[i-1] * L[i-1];  
        R[j] = j==N-1? 1 : R[j+1] * A[j+1];  
    }  
   
    for (i=0; i<N ; i++)  
    {
        B[i] = L[i] * R[i];  
        printf("%d ",B[i]);
    }
}

Most is clear from the program. Anyways, through the L and R we calculate the multiplication of terms to the left and right of i-th term. then finally we multiply it to get the result.