Package g1901_2000.s1938_maximum_genetic_difference_query

Class Solution

java.lang.Object
g1901_2000.s1938_maximum_genetic_difference_query.Solution
public class Solution extends Object
1938 - Maximum Genetic Difference Query.<p>Hard</p> <p>There is a rooted tree consisting of <code>n</code> nodes numbered <code>0</code> to <code>n - 1</code>. Each node&rsquo;s number denotes its <strong>unique genetic value</strong> (i.e. the genetic value of node <code>x</code> is <code>x</code>). The <strong>genetic difference</strong> between two genetic values is defined as the <strong>bitwise-XOR</strong> of their values. You are given the integer array <code>parents</code>, where <code>parents[i]</code> is the parent for node <code>i</code>. If node <code>x</code> is the <strong>root</strong> of the tree, then <code>parents[x] == -1</code>.</p> <p>You are also given the array <code>queries</code> where <code>queries[i] = [node<sub>i</sub>, val<sub>i</sub>]</code>. For each query <code>i</code>, find the <strong>maximum genetic difference</strong> between <code>val<sub>i</sub></code> and <code>p<sub>i</sub></code>, where <code>p<sub>i</sub></code> is the genetic value of any node that is on the path between <code>node<sub>i</sub></code> and the root (including <code>node<sub>i</sub></code> and the root). More formally, you want to maximize <code>val<sub>i</sub> XOR p<sub>i</sub></code>.</p> <p>Return <em>an array</em> <code>ans</code> <em>where</em> <code>ans[i]</code> <em>is the answer to the</em> <code>i<sup>th</sup></code> <em>query</em>.</p> <p><strong>Example 1:</strong></p> <p><img src="https://assets.leetcode.com/uploads/2021/06/29/c1.png" alt="" /></p> <p><strong>Input:</strong> parents = [-1,0,1,1], queries = [[0,2],[3,2],[2,5]]</p> <p><strong>Output:</strong> [2,3,7]</p> <p><strong>Explanation:</strong> The queries are processed as follows:</p> <ul> <li> <p>[0,2]: The node with the maximum genetic difference is 0, with a difference of 2 XOR 0 = 2.</p> </li> <li> <p>[3,2]: The node with the maximum genetic difference is 1, with a difference of 2 XOR 1 = 3.</p> </li> <li> <p>[2,5]: The node with the maximum genetic difference is 2, with a difference of 5 XOR 2 = 7.</p> </li> </ul> <p><strong>Example 2:</strong></p> <p><img src="https://assets.leetcode.com/uploads/2021/06/29/c2.png" alt="" /></p> <p><strong>Input:</strong> parents = [3,7,-1,2,0,7,0,2], queries = [[4,6],[1,15],[0,5]]</p> <p><strong>Output:</strong> [6,14,7]</p> <p><strong>Explanation:</strong> The queries are processed as follows:</p> <ul> <li> <p>[4,6]: The node with the maximum genetic difference is 0, with a difference of 6 XOR 0 = 6.</p> </li> <li> <p>[1,15]: The node with the maximum genetic difference is 1, with a difference of 15 XOR 1 = 14.</p> </li> <li> <p>[0,5]: The node with the maximum genetic difference is 2, with a difference of 5 XOR 2 = 7.</p> </li> </ul> <p><strong>Constraints:</strong></p> <ul> <li><code>2 <= parents.length <= 10<sup>5</sup></code></li> <li><code>0 <= parents[i] <= parents.length - 1</code> for every node <code>i</code> that is <strong>not</strong> the root.</li> <li><code>parents[root] == -1</code></li> <li><code>1 <= queries.length <= 3 * 10<sup>4</sup></code></li> <li><code>0 <= node<sub>i</sub> <= parents.length - 1</code></li> <li><code>0 <= val<sub>i</sub> <= 2 * 10<sup>5</sup></code></li> </ul>

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • maxGeneticDifference

      public int[] maxGeneticDifference(int[] parents, int[][] queries)