Let $$[x]$$ be the greatest integer $$\leq x$$. Then the number of points in the interval $$(-2, 1)$$ where the function $$f(x) = |[x]| + \sqrt{x - [x]}$$ is discontinuous, is _____.

We need to find the number of points of discontinuity of $$f(x) = |[x]| + \sqrt{x - [x]}$$ in the interval $$(-2, 1)$$, where $$[x]$$ denotes the greatest integer function.

Note that $$x - [x] = \{x\}$$ is the fractional part function, which satisfies $$0 \leq \{x\} < 1$$. So $$f(x) = |[x]| + \sqrt{\{x\}}$$.

Let us analyze $$f(x)$$ on each sub-interval:

For $$x \in (-2, -1)$$: $$[x] = -2$$, so $$f(x) = |-2| + \sqrt{x - (-2)} = 2 + \sqrt{x + 2}$$.

For $$x \in [-1, 0)$$: $$[x] = -1$$, so $$f(x) = |-1| + \sqrt{x - (-1)} = 1 + \sqrt{x + 1}$$.

For $$x \in [0, 1)$$: $$[x] = 0$$, so $$f(x) = |0| + \sqrt{x - 0} = \sqrt{x}$$.

Checking continuity at $$x = -1$$:

Left-hand limit: $$\lim_{x \to -1^-} f(x) = 2 + \sqrt{-1 + 2} = 2 + 1 = 3$$.

Value at $$x = -1$$: $$f(-1) = 1 + \sqrt{-1 + 1} = 1 + 0 = 1$$.

Since $$3 \neq 1$$, $$f$$ is discontinuous at $$x = -1$$.

Checking continuity at $$x = 0$$:

Left-hand limit: $$\lim_{x \to 0^-} f(x) = 1 + \sqrt{0 + 1} = 1 + 1 = 2$$.

Value at $$x = 0$$: $$f(0) = \sqrt{0} = 0$$.

Since $$2 \neq 0$$, $$f$$ is discontinuous at $$x = 0$$.

Within each sub-interval $$(-2, -1)$$, $$(-1, 0)$$, and $$(0, 1)$$, the function is continuous (being a sum of a constant and a continuous radical function).

Therefore, the number of points of discontinuity in $$(-2, 1)$$ is $$\boxed{2}$$.