Does Coron and May's paper for deterministically reducing finding  to factoring  work with $\lambda(n)$?
As mentioned in the other answer, the probabilistic method always works, and I've described both of these methods before.
We'll deal exclusively with $p, q$ of approximately the same size here. Suppose $l = \gcd(p-1, q-1)$, the ratio between $\phi(n) = (p-1)(q-1)$ and $\lambda(n) = \mathrm{lcm}(p-1, q-1)$. As usual, let $n = pq$, $e$ be the public exponent, $d = e^{-1} \bmod \phi(n)$, and $d' = e^{-1} \bmod \lambda(n)$.
The original Coron-May reduction observes that $f(x) = n - x$ has a &quot;small&quot; root $p+q-1$ modulo $\phi(n)$, a &quot;large&quot; divisor of $ed - 1$. More specifically, as long as there's a $\beta \in [0, 1)$ such that $p+q-1 \le (ed - 1)^{\beta^2}$ and $\phi(n) \ge (ed - 1)^{\beta}$, the method works in polynomial time. Because $\phi(n) \approx n$, this effectively translates into the requirement $ed \le n^2$.
Without any changes to the method, replacing $\phi(n)$ by $\lambda(n)$ means that $f(x) = n - x$ has a small root $p+q-1$ modulo a large factor of $ed' - 1$. Because of the common factor $l$, $d'$ may be much smaller than $d$, and in particular the size is no longer neatly related to $n$, so as to achieve a bound like $ed' \le n^2$. However, the method still works under the more elaborate constraint that there be a $\beta$ such that $p+q \le (ed' - 1)^{\beta^2}$ and $\lambda(n) \ge (ed' - 1)^{\beta}$.
For typical $e = 65537$, we still have $\lambda(n) = (ed'-1)^{\beta}$ for a $\beta$ very close to $1$. As long as $2p &lt; (ed' - 1)^{\beta^2}$, the reduction still works.
Here's a worked example using Sagemath, where $\phi(n) \approx 2^{130}\lambda(n)$.
sage: p = 30010719099306564150822479775631261956587370647517629976046542673406882291177
sage: q = 91932507744510266739810220416155764940049951037172916877150162469889273507869
sage: n = p * q
sage: l = gcd(p-1, q-1)
sage: l
1361129467683753853853498429727072846028
sage: e = 65537
sage: d = inverse_mod(e, lcm(p-1,q-1))
sage: P.&lt;x&gt; = Zmod(d*e-1)[]
sage: f = x - n
sage: beta = d.log(d*e).n()
sage: beta
0.959304860280816
sage: assert(p+q &lt; (d*e-1)^(beta^2))
sage: f.small_roots(X=2^257, beta=beta, epsilon=0.05)
[121943226843816830890632700191787026896637321684690546853196705143296155799045]
sage: p+q-1
121943226843816830890632700191787026896637321684690546853196705143296155799045

With a larger $e$, it still works. Say, $e \approx 2^{180}$:
sage: e = next_prime(2^180)
sage: d = inverse_mod(e, lcm(p-1,q-1))
sage: P.&lt;x&gt; = Zmod(d*e-1)[]
sage: f = x - n
sage: beta = d.log(d*e).n()
sage: beta
0.677720040398063
sage: f.small_roots(X=2^257, beta=beta, epsilon=0.01)
[121943226843816830890632700191787026896637321684690546853196705143296155799045]

Here we are already pushing the limit, since $(ed' - 1)^{0.6777^2} \approx 2^{256.53}$. Going much higher with $e$ will fail. The original method with $\phi(n)$ would allow $e$ to go up to $\approx 2^{512}$ in this example.