<p>You are an ant tasked with adding <code>n</code> new rooms numbered <code>0</code> to <code>n-1</code> to your colony. You are given the expansion plan as a <strong>0-indexed</strong> integer array of length <code>n</code>, <code>prevRoom</code>, where <code>prevRoom[i]</code> indicates that you must build room <code>prevRoom[i]</code> before building room <code>i</code>, and these two rooms must be connected <strong>directly</strong>. Room <code>0</code> is already built, so <code>prevRoom[0] = -1</code>. The expansion&nbsp;plan is given such that once all the rooms are built, every room will be reachable from room <code>0</code>.</p>

<p>You can only build <strong>one room</strong> at a time, and you can travel freely between rooms you have <strong>already built</strong> only if they are <strong>connected</strong>.&nbsp;You can choose to build <strong>any room</strong> as long as its <strong>previous room</strong>&nbsp;is already built.</p>

<p>Return <em>the <strong>number of different orders</strong> you can build all the rooms in</em>. Since the answer may be large, return it <strong>modulo</strong> <code>10<sup>9</sup> + 7</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/19/d1.JPG" style="width: 200px; height: 212px;" />
<pre>
<strong>Input:</strong> prevRoom = [-1,0,1]
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;There is only one way to build the additional rooms: 0 &rarr; 1 &rarr; 2
</pre>

<p><strong class="example">Example 2:</strong></p>
<strong><img alt="" src="https://assets.leetcode.com/uploads/2021/06/19/d2.JPG" style="width: 200px; height: 239px;" /></strong>

<pre>
<strong>Input:</strong> prevRoom = [-1,0,0,1,2]
<strong>Output:</strong> 6
<strong>Explanation:
</strong>The 6 ways are:
0 &rarr; 1 &rarr; 3 &rarr; 2 &rarr; 4
0 &rarr; 2 &rarr; 4 &rarr; 1 &rarr; 3
0 &rarr; 1 &rarr; 2 &rarr; 3 &rarr; 4
0 &rarr; 1 &rarr; 2 &rarr; 4 &rarr; 3
0 &rarr; 2 &rarr; 1 &rarr; 3 &rarr; 4
0 &rarr; 2 &rarr; 1 &rarr; 4 &rarr; 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == prevRoom.length</code></li>
	<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>
	<li><code>prevRoom[0] == -1</code></li>
	<li><code>0 &lt;= prevRoom[i] &lt; n</code> for all <code>1 &lt;= i &lt; n</code></li>
	<li>Every room is reachable from room <code>0</code> once all the rooms are built.</li>
</ul>
import numpy as np

MOD = 10 ** 9 + 7


def cumprod(A):
    L = len(A)
    Lsq = int(L**.5 + 1)
    A = np.resize(A, Lsq**2).reshape(Lsq, Lsq)
    for n in range(1, Lsq):
        A[:, n] *= A[:, n - 1]
        A[:, n] %= MOD
    for n in range(1, Lsq):
        A[n] *= A[n - 1, -1]
        A[n] %= MOD
    return A.ravel()[:L]


def make_fact(U):
    x = np.arange(U, dtype=np.int64)
    x[0] = 1
    fact = cumprod(x)
    x = np.arange(U, 0, -1, dtype=np.int64)
    x[0] = pow(int(fact[-1]), MOD - 2, MOD)
    fact_inv = cumprod(x)[::-1]
    fact.flags.writeable = False
    fact_inv.flags.writeable = False
    return fact.tolist(), fact_inv.tolist()


fac, fac_inv = make_fact(100010)


class Solution:
    def waysToBuildRooms(self, prevRoom: List[int]) -> int:
        g = collections.defaultdict(set)
        for i, u in enumerate(prevRoom):
            g[u].add(i)

        f = [1] * len(prevRoom)
        cnt = [0] * len(prevRoom)

        def dfs(u):
            tmp = 1
            for v in g[u]:
                dfs(v)
                tmp = (tmp * fac_inv[cnt[v]]) % MOD
                f[u] = (f[u] * f[v]) % MOD
                cnt[u] += cnt[v]
            f[u] = (f[u] * fac[cnt[u]] * tmp) % MOD
            cnt[u] += 1

        dfs(0)
        return f[0]

1630