Let's denote the number of students in the class as $n$.

Initially, the score of a student was recorded as $0$, but it was corrected to $100$. This correction increases the class average by $4$.

The increase in the total score due to this correction is $100 - 0 = 100$.

Since the average increases by $4$, we have: $\frac{100}{n} = 4$ Solving for $n$, we get: $n = \frac{100}{4} = 25$

So, there are $25$ students in the class.

Next, another student's score was corrected from $81$ to $56$. This correction decreases the total score by $81 - 56 = 25$.

The change in average due to this correction is: $\frac{-25}{25} = -1$

Thus, the final corrected average $y$ is: $y = x + 4 - 1 = x + 3$

Therefore, the difference $y - x$ is: $y - x = 3$