Consider this problem: A list of distinctive integers in random order is provided. The integer is in the range of \(1\) to \(N\) and there are \(N-1\) integers provided, i.e. one integer in \([1,N]\) is not in the provided list. Find that number in \(O(1)\) space and \(O(N)\) time.

Simply, we sum up the numbers in the list and the difference between the sum and \(\sum_{k=1}^N k\) is the number in concern.

But what happen if there are two numbers missing? Obviously, we can use the same trick, for the list of integers be denoted by \(n_k\), we find

\[\begin{align} \sum_{k=1}^N k - \sum_{k=1}^{N-2} n_k &= s \\ \prod_{k=1}^N k - \prod_{k=1}^{N-2} n_k &= p \end{align}\]Then we have the two missing numbers, \(x\) and \(y\), to satisfy

\[\begin{align} x + y &= s \\ xy &= p \end{align}\]This approach works but \(\prod_k n_k\) can be a vastly large number. Can we have another simultaneous equations to find \(x\) and \(y\) without overflowing an integer variable?

I have considered the following

\[\begin{gather} \operatorname*{XOR}_{k=1}^{N-2} n_k \\ \sum_{k=1}^{N-2} \operatorname{NOT}(n_k) \\ \sum_{k=1}^{N-2} \operatorname{popcount}(n_k) \\ \sum_{k=1}^{N-2} \operatorname{reverse}(n_k) \end{gather}\]and some others. However, I find one test to prove if any of these constructs can help you solve the two unknowns. Consider \(x\) and \(y\) are different at bit \(m\), and we define \(x'\) and \(y'\) to be same as \(x\) and \(y\) except their bit \(m\) are flipped. Then we should have a function that

\[f(x,y) \ne f(x',y')\]or otherwise we cannot distinguish the solution \(x\), \(y\) from \(x'\), \(y'\). Obviously the list above does not pass this test. That’s why we need multiplication involved: We can solve for \(x\) and \(y\) in the equations below

\[\begin{align} x + y &= s \\ x^2 + y^2 &= p \end{align}\]It can also be done using a non-linear injective function other than square in the second equation, such as logarithms. However, logarithm produce floating point numbers and that may introduce noise. This is not a good solution in implementation perspective because, \(\sum_k n_k^2\) increases quicker than linear and it has a chance to overflow the variable. But this is the best solution I get so far.