1. Understand the function

Given

where $[x]$ is the greatest integer function.

Let $\{x\}=x-[x]$ be the fractional part of $x$. Then $1+[x]-x=1-(x-[x])=1-\{x\}.$ So the function becomes $f(x)=\min\{\{x\},1-\{x\}\}.$

Thus on each interval $[n,n+1)$, $f(x)$ depends only on the fractional part $t=\{x\}\in[0,1)$: $f(x)=\min\{t,1-t\}.$

2. Write the function piecewise on each unit interval

For $t\in[0,1)$:

• if $0\le t\le \tfrac12$, then $t\le 1-t$, so $f(x)=t;$
• if $\tfrac12\le t<1$, then $1-t\le t$, so $f(x)=1-t.$

Hence on each interval $[n,n+1)$,

for $n=0,1,2$.

3. Find points of discontinuity: set $P$

Possible trouble points are integers, because the fractional part resets there.

At $x=0$

$f(0)=\min\{0,1\}=0.$ Right-hand limit at $0$ is also $0$, so $f$ is continuous at $0$.

At $x=1$

From the left, as $x\to1^-$, $\{x\}\to1^-$, so $f(x)=\min\{\{x\},1-\{x\}\}\to0.$ At $x=1$, $\{1\}=0$, hence $f(1)=0.$ From the right, again values are near $0$. So continuous at $1$.

Similarly, at $x=2$ and $x=3$, $f(2)=0,\quad f(3)=0,$ and the one-sided/two-sided limits match.

Within each open interval between consecutive integers, the function is piecewise linear and continuous.

Therefore, $P=\varnothing$ and number of elements in $P$ is $|P|=0.$

4. Find points where $f$ is not differentiable: set $Q$

Inside each interval, non-differentiability can occur where the two branches meet, i.e. at $x=n+\tfrac12.$ These points in $(0,3)$ are $\tfrac12,\ \tfrac32,\ \tfrac52.$

Check derivative at such a point:

• Left derivative is $+1$ (from branch $x-n$),
• Right derivative is $-1$ (from branch $n+1-x$).

Since left and right derivatives are unequal, $f$ is not differentiable at each of these three points.

Now check integers inside $(0,3)$, namely $x=1,2$.

Near $x=1$:

• for $x<1$, $f(x)=1-x$ close to $1$, so left derivative is $-1$;
• for $x>1$, $f(x)=x-1$ close to $1$, so right derivative is $+1$.

Hence not differentiable at $x=1$.

Similarly at $x=2$:

• left derivative $=-1$,
• right derivative $=+1$, so not differentiable at $x=2$.

Thus $Q=\left\{\tfrac12,1,\tfrac32,2,\tfrac52\right\}$ and $|Q|=5.$

5. Required sum

$|P|+|Q|=0+5=5.$

Therefore, the required integer is $\boxed{5}.$