#2277 - Closest Node to Path in Tree

Problem Description

You are given a positive integer n representing the number of nodes in a tree, numbered from 0 to n - 1 (inclusive). You are also given a 2D integer array edges of length n - 1, where edges[i] = [node1i, node2i] denotes that there is a bidirectional edge connecting node1i and node2i in the tree.

You are given a 0-indexed integer array query of length m where query[i] = [starti, endi, nodei] means that for the ith query, you are tasked with finding the node on the path from starti to endi that is closest to nodei.

Return an integer array answer of length m, where answer[i] is the answer to the ith query.

Solution

/**
 * @param {number} n
 * @param {number[][]} edges
 * @param {number[][]} query
 * @return {number[]}
 */
var closestNode = function(n, edges, query) {
  const graph = new Array(n).fill().map(() => []);
  for (const [a, b] of edges) {
    graph[a].push(b);
    graph[b].push(a);
  }

  const depths = new Array(n).fill(0);
  const binaryParents = new Array(n).fill().map(() => new Array(16).fill(0));

  dfs(0, [], 0);

  return query.map(([a, b, q]) => {
    const candidates = [lca(a, b), lca(a, q), lca(b, q)];
    return candidates.reduce((deepest, current) =>
      depths[current] > depths[deepest] ? current : deepest
    );
  });

  function dfs(node, parents, depth) {
    depths[node] = depth;
    for (let bit = 0; bit < 16; bit++) {
      if ((1 << bit) <= parents.length) {
        binaryParents[node][bit] = parents[parents.length - (1 << bit)];
      }
    }
    parents.push(node);
    for (const next of graph[node]) {
      if (parents.length >= 2 && next === parents[parents.length - 2]) {
        continue;
      }
      dfs(next, parents, depth + 1);
    }
    parents.pop();
  }

  function getKthParent(node, k) {
    if (!k) return node;
    let bit = 0;
    while ((1 << (bit + 1)) <= k) bit++;
    return getKthParent(binaryParents[node][bit], k & (~(1 << bit)));
  }

  function lca(a, b) {
    if (depths[a] > depths[b]) return lca(b, a);
    b = getKthParent(b, depths[b] - depths[a]);
    if (a === b) return a;
    for (let i = binaryParents[a].length - 1; i >= 0; i--) {
      if (binaryParents[a][i] !== binaryParents[b][i]) {
        a = binaryParents[a][i];
        b = binaryParents[b][i];
      }
    }
    return binaryParents[a][0];
  }
};