| 10 | */ |
| 11 | |
| 12 | export class BinaryLifting { |
| 13 | constructor(root, tree) { |
| 14 | this.root = root |
| 15 | this.connections = new Map() |
| 16 | this.up = new Map() // up[node][i] stores the 2^i-th parent of node |
| 17 | for (const [i, j] of tree) { |
| 18 | this.addEdge(i, j) |
| 19 | } |
| 20 | this.log = Math.ceil(Math.log2(this.connections.size)) |
| 21 | this.dfs(root, root) |
| 22 | } |
| 23 | |
| 24 | addNode(node) { |
| 25 | // Function to add a node to the tree (connection represented by set) |
| 26 | this.connections.set(node, new Set()) |
| 27 | } |
| 28 | |
| 29 | addEdge(node1, node2) { |
| 30 | // Function to add an edge (adds the node too if they are not present in the tree) |
| 31 | if (!this.connections.has(node1)) { |
| 32 | this.addNode(node1) |
| 33 | } |
| 34 | if (!this.connections.has(node2)) { |
| 35 | this.addNode(node2) |
| 36 | } |
| 37 | this.connections.get(node1).add(node2) |
| 38 | this.connections.get(node2).add(node1) |
| 39 | } |
| 40 | |
| 41 | dfs(node, parent) { |
| 42 | // The dfs function calculates 2^i-th ancestor of all nodes for i ranging from 0 to this.log |
| 43 | // We make use of the fact the two consecutive jumps of length 2^(i-1) make the total jump length 2^i |
| 44 | this.up.set(node, new Map()) |
| 45 | this.up.get(node).set(0, parent) |
| 46 | for (let i = 1; i < this.log; i++) { |
| 47 | this.up |
| 48 | .get(node) |
| 49 | .set(i, this.up.get(this.up.get(node).get(i - 1)).get(i - 1)) |
| 50 | } |
| 51 | for (const child of this.connections.get(node)) { |
| 52 | if (child !== parent) this.dfs(child, node) |
| 53 | } |
| 54 | } |
| 55 | |
| 56 | kthAncestor(node, k) { |
| 57 | // if value of k is more than or equal to the number of total nodes, we return the root of the graph |
| 58 | if (k >= this.connections.size) { |
| 59 | return this.root |
| 60 | } |
| 61 | // if i-th bit is set in the binary representation of k, we jump from a node to its 2^i-th ancestor |
| 62 | // so after checking all bits of k, we will have made jumps of total length k, in just log k steps |
| 63 | for (let i = 0; i < this.log; i++) { |
| 64 | if (k & (1 << i)) { |
| 65 | node = this.up.get(node).get(i) |
| 66 | } |
| 67 | } |
| 68 | return node |
| 69 | } |
nothing calls this directly
no outgoing calls
no test coverage detected