This is *not* an answer, just a response to <a href="https://mathoverflow.net/questions/461631/number-of-real-roots-of-0-1-polynomial/461708?noredirect=1#comment1198128_461708">David's request</a> to present a construction of a polynomial $P(x)=\sum_{j=0}^d a_jx^j$ of arbitrarily high degree $d$ whose coefficients are real, bounded by $1$ in absolute value, satisfy $|a_0|=|a_d|=1$, and whose number of distinct real roots is of order $\sqrt{d}$. I believe that I've also seen the statement about pure $0,\pm 1$ coefficients somewhere, but my attempt to retrieve the proof from my memory or to reinvent it failed, so it might be a hallucination.

We will use the standard "pigeonholing with a minor correction in the end" mumbo-jumbo.

Take big $n>m,k$

Let $P(x)=exT(x)$ where $T$ is the Chebyshev's polinomial of degree $m$ adjusted to the interval $[e^{-1},1]$. Then $P$ has no free term, is of degree $m+1$, and has the alternance points $y_1,\dots,y_{m+1}$ of heights at least $1$. The coefficients of $P$ are bounded by $K^m$ for some absolute $K>0$. Let $x_j=y_j^{1/k}$

Consider all polynomials $Q$ of degree $3n$ with coefficients $\beta_s\in{0,e^{(s/(3n))^2}}$. Since $|Q(x_j)|\le 3en$ for every $Q$ and every $j$, and the total number of $Q$-s is $2^{3n}$ we can (by pigeonholing) find a set of $2^{2n}$ $Q$-s such that their values at each $x_j$ are confined to some intervals $I_j$ of length $6en2^{-n/m}$. 

Fix one such $Q_0$ in that set. Note that the number of $Q$ whose coefficients differ from those of $Q$ only within an interval of indices of length $n$ is at most $n2^n\ll 2^{2n}-1$. Thus we can find $Q_1$ in the same set such that
$$
R(x)=Q(x)-Q_1(x)=\sum_{s=a}^b \alpha_s x^s
$$
with $b>a+n$, $\alpha_s\in\{0,\pm e^{(s/(3n))^2}\}$, $\alpha_a,\alpha_b\ne 0$, $|R(x_j)|\le 6en2^{-n/m}$ for all $j$.

Dividing $R(x)$ by $e^{(a/(3n))^2}x^a$, and taking into account that $x_j\ge e^{-1/k}$ for all $j$, we get a polynomial $S(x)$ of degree $d=b-a\in[n,3n]$ with coefficients $\gamma_s\in\{0,\pm e^{\psi(s)}\}$ where $\psi(s)=[(s+a)/(3n)]^2-[a/(3n)]^2$ and such that $|S(x_j)|\le 6en 2^{-n/m}e^{3n/k}$.

The key point is that $\psi$ is strictly convex and $0$ at $0$, so
$$
e^{\psi(s)}\le e^{s\psi(d)/d}-\kappa n^{-2}
$$
for $1\le s\le d-1$ with some $\kappa>0$. This we can correct the intermediate coefficients by about $n^{-2}$ and still have them under the geometric progression joining $|\gamma_0|=1$ and $|\gamma_d|=e^{\psi(d)}$. Then, by a trivial linear change of variable, we'll turn $\gamma_d$ into $\pm 1$ and make all intermediate coefficients $\le 1$.

We will use this correction possibility to turn smallness at $x_j$ into an alternance. To this end, we will just add $ 6en 2^{-n/m}e^{3n/k}P(x^k)$. The coefficients of this correction are bounded by 
$6en 2^{-n/m}e^{3n/k}K^m$, it has no free term and the degree $(m+1)k$.

Now choose $k\approx \sqrt n$ and $m\approx \sigma\sqrt n$.
Then if $\sigma<1$, our correction stays inside the degree span of $S$, and
$$
6en 2^{-n/m}e^{3n/k}K^m\approx 6en 2^{-\sigma^{-1}\sqrt n}e^{3\sqrt n}K^{\sigma\sqrt n}\,
$$
which for small $\sigma>0$ is less than $e^{-\sqrt n}\ll \kappa n^{-2}$, so our correction has admissible size too.

That's it. You can see, however, that the alternance is minuscule, so you should exercise utmost care with your suggested replacement of negative coefficients, if it can work at all!