You are an ant tasked with adding n new rooms numbered 0 to n-1 to your colony. You are given the expansion plan as a 0-indexed integer array of length n, prevRoom, where prevRoom[i] indicates that you must build room prevRoom[i] before building room i, and these two rooms must be connected directly. Room 0 is already built, so prevRoom[0] = -1. The expansion plan is given such that once all the rooms are built, every room will be reachable from room 0.
You can only build one room at a time, and you can travel freely between rooms you have already built only if they are connected. You can choose to build any room as long as its previous room is already built.
\r\n\r\nReturn the number of different orders you can build all the rooms in. Since the answer may be large, return it modulo 109 + 7.
\r\n
Example 1:
\r\n![\"\"](\"https://assets.leetcode.com/uploads/2021/06/19/d1.JPG\")
\r\n\r\nInput: prevRoom = [-1,0,1]\r\nOutput: 1\r\nExplanation: There is only one way to build the additional rooms: 0 → 1 → 2\r\n\r\n\r\n
Example 2:
\r\n![\"\"](\"https://assets.leetcode.com/uploads/2021/06/19/d2.JPG\")
\r\n\r\n\r\nInput: prevRoom = [-1,0,0,1,2]\r\nOutput: 6\r\nExplanation:\r\nThe 6 ways are:\r\n0 → 1 → 3 → 2 → 4\r\n0 → 2 → 4 → 1 → 3\r\n0 → 1 → 2 → 3 → 4\r\n0 → 1 → 2 → 4 → 3\r\n0 → 2 → 1 → 3 → 4\r\n0 → 2 → 1 → 4 → 3\r\n\r\n\r\n
\r\n
Constraints:
\r\n\r\nn == prevRoom.length2 <= n <= 105prevRoom[0] == -10 <= prevRoom[i] < n for all 1 <= i < n0 once all the rooms are built.Compiled with clang 11 using the latest C++ 20 standard.
Your code is compiled with level two optimization (-O2). AddressSanitizer is also enabled to help detect out-of-bounds and use-after-free bugs.
Most standard library headers are already included automatically for your convenience.
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