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"title": "Count Ways to Build Rooms in an Ant Colony",
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"content": "<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 plan is given such that once all the rooms are built, every room will be reachable from room <code>0</code>.</p>\r\n\r\n<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>. You can choose to build <strong>any room</strong> as long as its <strong>previous room</strong> is already built.</p>\r\n\r\n<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>\r\n\r\n<p> </p>\r\n<p><strong>Example 1:</strong></p>\r\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d1.JPG\" style=\"width: 200px; height: 212px;\" />\r\n<pre>\r\n<strong>Input:</strong> prevRoom = [-1,0,1]\r\n<strong>Output:</strong> 1\r\n<strong>Explanation:</strong> There is only one way to build the additional rooms: 0 → 1 → 2\r\n</pre>\r\n\r\n<p><strong>Example 2:</strong></p>\r\n<strong><img alt=\"\" src=\"https://assets.leetcode.com/uploads/2021/06/19/d2.JPG\" style=\"width: 200px; height: 239px;\" /></strong>\r\n\r\n<pre>\r\n<strong>Input:</strong> prevRoom = [-1,0,0,1,2]\r\n<strong>Output:</strong> 6\r\n<strong>Explanation:\r\n</strong>The 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</pre>\r\n\r\n<p> </p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>n == prevRoom.length</code></li>\r\n\t<li><code>2 <= n <= 10<sup>5</sup></code></li>\r\n\t<li><code>prevRoom[0] == -1</code></li>\r\n\t<li><code>0 <= prevRoom[i] < n</code> for all <code>1 <= i < n</code></li>\r\n\t<li>Every room is reachable from room <code>0</code> once all the rooms are built.</li>\r\n</ul>",
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"code": "class Solution {\r\npublic:\r\n int waysToBuildRooms(vector<int>& prevRoom) {\r\n \r\n }\r\n};",
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"hints": [
"Use dynamic programming.",
"Let dp[i] be the number of ways to solve the problem for the subtree of node i.",
"Imagine you are trying to fill an array with the order of traversal, dp[i] equals the multiplications of the number of ways to distribute the subtrees of the children of i on the array using combinatorics, multiplied bu their dp values."
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