CW complex structure for $\mathbb{R}P^n$ or a space homotopy equivalent I'm doing a project about Topological Complexity (it doesn't matter what it is for the questions I will ask) and I have proofs for a few results about the bounds of the topological complexity of spaces which are homotopy equivalent to finite CW complexes.
It is known that one CW complex structure for the projective space $\mathbb{R}P^n$  consists of one cell in each dimension and the attaching maps being the projection maps from $\mathbb{S}^n$ to $\mathbb{R}P^n$  (See Hatcher Example 0.4). It is also known that a space could be given different CW complex structures. My first question is:
Does there exist a more "economic" CW complex structure for $\mathbb{R}P^n$? I mean, could we give a CW complex structure with less cells or using more cells but without using cells of all dimensions? I will put an example of what I want:
For example, we can give a CW complex structure for $\mathbb{S}^n$ using only two cells, one in dimension 0 and other in dimension n. 
Assuming that the answer was negative, this lead us to my second question:
Are there spaces homotopy equivalent to $\mathbb{R}P^n$ which satisfy the property I ask for in my first question?
Thanks in advance and any help would be appreciated.
 A: Your questions can be answered by using cellular homology. I have given a quick summary below. For the details, see Hatcher's Algebraic Topology, pages $137 - 140$. Although Hatcher uses $\mathbb{Z}$ coefficients, you can define cellular homology for any coefficients in the same way. See the discussion at the end for why I used $\mathbb{Z}_2$ coefficients rather than $\mathbb{Z}$ coefficients.
If $X_k$ denotes the $k$-skeleton of a CW complex $X$, then $H_k(X_k, X_{k-1}; \mathbb{Z}_2)$ is a vector space over $\mathbb{Z}_2$ with dimension equal to the number of $k$-cells of $X$. In addition, there are maps $d_k : H_k(X_k, X_{k-1}; \mathbb{Z}_2) \to H_{k-1}(X_{k-1}, X_{k-2}; \mathbb{Z}_2)$ which give rise to a complex $(H_k(X_k, X_{k-1}; \mathbb{Z}_2), d_k)$. The homology of this complex is called the cellular homology of $X$, denoted $H^{\text{CW}}_k(X; \mathbb{Z}_2)$, and it is isomorphic to the singular homology of $X$.
If $X$ has no $k$-cells, $H_k(X_k, X_{k-1}; \mathbb{Z}_2) = 0$, so 
$$H_k^{\text{CW}}(X; \mathbb{Z}_2) = \frac{\ker d_k : H_k(X_k, X_{k-1}; \mathbb{Z}_2) \to H_{k-1}(X_{k-1}, X_{k-2}; \mathbb{Z}_2)}{\operatorname{im}d_{k+1} : H_{k+1} : H_{k+1}(X_{k+1}, X_k; \mathbb{Z}_2) \to H_k(X_k, X_{k-1}; \mathbb{Z}_2)} = 0$$
and hence $H_k(X; \mathbb{Z}_2) = 0$. Therefore, if $H_k(X; \mathbb{Z}_2) \neq 0$, $X$ must have at least one $k$-cell. More generally, 
\begin{align*}
\operatorname{dim}H_k^{\text{CW}}(X; \mathbb{Z}_2) &\leq \operatorname{dim}\ker d_k\\ 
&\leq \operatorname{dim}H_k(X_k, X_{k-1}; \mathbb{Z}_2)\\ 
&= \text{number of $k$-cells in the CW complex.}
\end{align*}
As 
$$H_k(\mathbb{RP}^n ; \mathbb{Z}_2) = \begin{cases}
\mathbb{Z}_2 & 0 \leq k \leq n\\
0 & \text{otherwise},
\end{cases}$$
any CW structure for $\mathbb{RP}^n$ must have at least one cell in each dimension between $0$ and $n$. The same is true for any space homotopy equivalent to $\mathbb{RP}^n$ as it will have isomorphic homology.

Initially, I wrote this answer using $\mathbb{Z}$ coefficients. However, as Anubhav. K rightly points out in the comments, you can't quite get away with this because if $n$ is even, $H_n(\mathbb{RP}^n; \mathbb{Z}) = 0$ (it isn't orientable), so you can't conclude from the argument that the CW complex must contain an $n$-cell.
A: Answer is NO
Observe that $H_i(\mathbb RP^n , \mathbb Z_2) = \mathbb Z_2, 0\leq i\leq n$, so if we missed out one cell , lets say if there is no $k$ cell, then consider the cellular homology sequence, there $H_i(X_i,X_{i-1};\mathbb Z_2)=0$, in particlar $k$ th homology group will vanish. Which is not possible. 
