leetcode-problemset/leetcode/originData/most-profitable-path-in-a-tree.json

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            "title": "Most Profitable Path in a Tree",
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            "content": "<p>There is an undirected tree with <code>n</code> nodes labeled from <code>0</code> to <code>n - 1</code>, rooted at node <code>0</code>. You are given a 2D integer array <code>edges</code> of length <code>n - 1</code> where <code>edges[i] = [a<sub>i</sub>, b<sub>i</sub>]</code> indicates that there is an edge between nodes <code>a<sub>i</sub></code> and <code>b<sub>i</sub></code> in the tree.</p>\n\n<p>At every node <code>i</code>, there is a gate. You are also given an array of even integers <code>amount</code>, where <code>amount[i]</code> represents:</p>\n\n<ul>\n\t<li>the price needed to open the gate at node <code>i</code>, if <code>amount[i]</code> is negative, or,</li>\n\t<li>the cash reward obtained on opening the gate at node <code>i</code>, otherwise.</li>\n</ul>\n\n<p>The game goes on as follows:</p>\n\n<ul>\n\t<li>Initially, Alice is at node <code>0</code> and Bob is at node <code>bob</code>.</li>\n\t<li>At every second, Alice and Bob <b>each</b> move to an adjacent node. Alice moves towards some <strong>leaf node</strong>, while Bob moves towards node <code>0</code>.</li>\n\t<li>For <strong>every</strong> node along their path, Alice and Bob either spend money to open the gate at that node, or accept the reward. Note that:\n\t<ul>\n\t\t<li>If the gate is <strong>already open</strong>, no price will be required, nor will there be any cash reward.</li>\n\t\t<li>If Alice and Bob reach the node <strong>simultaneously</strong>, they share the price/reward for opening the gate there. In other words, if the price to open the gate is <code>c</code>, then both Alice and Bob pay&nbsp;<code>c / 2</code> each. Similarly, if the reward at the gate is <code>c</code>, both of them receive <code>c / 2</code> each.</li>\n\t</ul>\n\t</li>\n\t<li>If Alice reaches a leaf node, she stops moving. Similarly, if Bob reaches node <code>0</code>, he stops moving. Note that these events are <strong>independent</strong> of each other.</li>\n</ul>\n\n<p>Return<em> the <strong>maximum</strong> net income Alice can have if she travels towards the optimal leaf node.</em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/10/29/eg1.png\" style=\"width: 275px; height: 275px;\" />\n<pre>\n<strong>Input:</strong> edges = [[0,1],[1,2],[1,3],[3,4]], bob = 3, amount = [-2,4,2,-4,6]\n<strong>Output:</strong> 6\n<strong>Explanation:</strong> \nThe above diagram represents the given tree. The game goes as follows:\n- Alice is initially on node 0, Bob on node 3. They open the gates of their respective nodes.\n  Alice&#39;s net income is now -2.\n- Both Alice and Bob move to node 1. \n&nbsp; Since they reach here simultaneously, they open the gate together and share the reward.\n&nbsp; Alice&#39;s net income becomes -2 + (4 / 2) = 0.\n- Alice moves on to node 3. Since Bob already opened its gate, Alice&#39;s income remains unchanged.\n&nbsp; Bob moves on to node 0, and stops moving.\n- Alice moves on to node 4 and opens the gate there. Her net income becomes 0 + 6 = 6.\nNow, neither Alice nor Bob can make any further moves, and the game ends.\nIt is not possible for Alice to get a higher net income.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n<img alt=\"\" src=\"https://assets.leetcode.com/uploads/2022/10/29/eg2.png\" style=\"width: 250px; height: 78px;\" />\n<pre>\n<strong>Input:</strong> edges = [[0,1]], bob = 1, amount = [-7280,2350]\n<strong>Output:</strong> -7280\n<strong>Explanation:</strong> \nAlice follows the path 0-&gt;1 whereas Bob follows the path 1-&gt;0.\nThus, Alice opens the gate at node 0 only. Hence, her net income is -7280. \n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>2 &lt;= n &lt;= 10<sup>5</sup></code></li>\n\t<li><code>edges.length == n - 1</code></li>\n\t<li><code>edges[i].length == 2</code></li>\n\t<li><code>0 &lt;= a<sub>i</sub>, b<sub>i</sub> &lt; n</code></li>\n\t<li><code>a<sub>i</sub> != b<sub>i</sub></code></li>\n\t<li><code>edges</code> represents a valid tree.</li>\n\t<li><code>1 &lt;= bob &lt; n</code></li>\n\t<li><code>amount.length == n</code></li>\n\t<li><code>amount[i]</code> is an <strong>even</strong> integer in the range <code>[-10<sup>4</sup>, 10<sup>4</sup>]</code>.</li>\n</ul>\n",
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                    "code": "class Solution {\npublic:\n    int mostProfitablePath(vector<vector<int>>& edges, int bob, vector<int>& amount) {\n        \n    }\n};",
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                    "code": "class Solution(object):\n    def mostProfitablePath(self, edges, bob, amount):\n        \"\"\"\n        :type edges: List[List[int]]\n        :type bob: int\n        :type amount: List[int]\n        :rtype: int\n        \"\"\"\n        ",
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                    "code": "/**\n * @param {number[][]} edges\n * @param {number} bob\n * @param {number[]} amount\n * @return {number}\n */\nvar mostProfitablePath = function(edges, bob, amount) {\n    \n};",
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                    "code": "# @param {Integer[][]} edges\n# @param {Integer} bob\n# @param {Integer[]} amount\n# @return {Integer}\ndef most_profitable_path(edges, bob, amount)\n    \nend",
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                "Bob travels along a fixed path (from node “bob” to node 0).",
                "Calculate Alice’s distance to each node via DFS.",
                "We can calculate Alice’s score along a path ending at some node easily using Hints 1 and 2."
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