LCA for n-ary Tree | Constant Query O(1)
We have seen various methods with different Time Complexities to calculate LCA in n-ary tree:-
Method 1 : Naive Method ( by calculating root to node path) | O(n) per query
Method 2 :Using Sqrt Decomposition | O(sqrt H)
Method 3 : Using Sparse Matrix DP approach | O(logn)
Lets study another method that has faster query time than all the above methods. So, our aim will be to calculate LCA in constant time ~ O(1). Let’s see how we can achieve it.
Method 4 : Using Range Minimum Query
We have discussed LCA and RMQ for binary tree. Here we discuss LCA problem to RMQ problem conversion for n-ary tree.
Pre-requisites:- LCA in Binary Tree using RMQ RMQ using sparse table
Key Concept : In this method, we will be reducing our LCA problem to RMQ(Range Minimum Query) problem over a static array. Once, we do that then we will relate the Range minimum queries to the required LCA queries.
The first step will be to decompose the tree into a flat linear array. To do this we can apply the Euler walk. The Euler walk will give the pre-order traversal of the graph. So we will perform a Euler Walk on the tree and store the nodes in an array as we visit them. This process reduces the tree data-structure to a simple linear array.
Consider the below tree and the euler walk over it:-
Now lets think in general terms : Consider any two nodes on the tree. There will be exactly one path connecting both the nodes and the node that has the smallest depth value in the path will be the LCA of the two given nodes.
Now take any two distinct node say u and v in the Euler walk array. Now all the elements in the path from u to v will lie in between the index of nodes u and v in the Euler walk array. Therefore, we just need to calculate the node with the minimum depth between the index of node u and node v in the euler array.
For this we will maintain another array that will contain the depth of all the nodes corresponding to their position in the Euler walk array so that we can Apply our RMQ algorithm on it.
Given below is the euler walk array parallel to its depth track array.
Example :- Consider two nodes node 6 and node 7 in the euler array. To calculate the LCA of node 6 and node 7 we look the smallest depth value for all the nodes in between node 6 and node 7 .
Therefore, node 1 has the smallest depth value = 0 and hence, it is the LCA for node 6 and node 7.
Implementation:-
We will be maintaining three arrays 1)Euler Path 2)Depth array 3)First Appearance Index
Euler Path and Depth array are the same as described above
First Appearance Index FAI[] : The First Appearance index Array will store the index for the first position of every node in the Euler Path array. FAI[i] = First appearance of ith node in Euler Walk array.
The Implementation for the above method is given below:-
Implementation:
C++
// C++ program to demonstrate LCA of n-ary tree // in constant time. #include "bits/stdc++.h" using namespace std; #define sz 101 vector < int > adj[sz]; // stores the tree vector < int > euler; // tracks the eulerwalk vector < int > depthArr; // depth for each node corresponding // to eulerwalk int FAI[sz]; // stores first appearance index of every node int level[sz]; // stores depth for all nodes in the tree int ptr; // pointer to euler walk int dp[sz][18]; // sparse table int logn[sz]; // stores log values int p2[20]; // stores power of 2 void buildSparseTable( int n) { // initializing sparse table memset (dp,-1, sizeof (dp)); // filling base case values for ( int i=1; i<n; i++) dp[i-1][0] = (depthArr[i]>depthArr[i-1])?i-1:i; // dp to fill sparse table for ( int l=1; l<15; l++) for ( int i=0; i<n; i++) if (dp[i][l-1]!=-1 and dp[i+p2[l-1]][l-1]!=-1) dp[i][l] = (depthArr[dp[i][l-1]]>depthArr[dp[i+p2[l-1]][l-1]])? dp[i+p2[l-1]][l-1] : dp[i][l-1]; else break ; } int query( int l, int r) { int d = r-l; int dx = logn[d]; if (l==r) return l; if (depthArr[dp[l][dx]] > depthArr[dp[r-p2[dx]][dx]]) return dp[r-p2[dx]][dx]; else return dp[l][dx]; } void preprocess() { // memorizing powers of 2 p2[0] = 1; for ( int i=1; i<18; i++) p2[i] = p2[i-1]*2; // memorizing all log(n) values int val = 1,ptr=0; for ( int i=1; i<sz; i++) { logn[i] = ptr-1; if (val==i) { val*=2; logn[i] = ptr; ptr++; } } } /** * Euler Walk ( preorder traversal) * converting tree to linear depthArray * Time Complexity : O(n) * */ void dfs( int cur, int prev, int dep) { // marking FAI for cur node if (FAI[cur]==-1) FAI[cur] = ptr; level[cur] = dep; // pushing root to euler walk euler.push_back(cur); // incrementing euler walk pointer ptr++; for ( auto x:adj[cur]) { if (x != prev) { dfs(x,cur,dep+1); // pushing cur again in backtrack // of euler walk euler.push_back(cur); // increment euler walk pointer ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array void makeArr() { for ( auto x : euler) depthArr.push_back(level[x]); } int LCA( int u, int v) { // trivial case if (u==v) return u; if (FAI[u] > FAI[v]) swap(u,v); // doing RMQ in the required range return euler[query(FAI[u], FAI[v])]; } void addEdge( int u, int v) { adj[u].push_back(v); adj[v].push_back(u); } int main( int argc, char const *argv[]) { // constructing the described tree int numberOfNodes = 8; addEdge(1,2); addEdge(1,3); addEdge(2,4); addEdge(2,5); addEdge(2,6); addEdge(3,7); addEdge(3,8); // performing required precalculations preprocess(); // doing the Euler walk ptr = 0; memset (FAI,-1, sizeof (FAI)); dfs(1,0,0); // creating depthArray corresponding to euler[] makeArr(); // building sparse table buildSparseTable(depthArr.size()); cout << "LCA(6,7) : " << LCA(6,7) << "\n" ; cout << "LCA(6,4) : " << LCA(6,4) << "\n" ; return 0; } |
Java
// Java program to demonstrate LCA of n-ary // tree in constant time. import java.util.ArrayList; import java.util.Arrays; class GFG{ static int sz = 101 ; @SuppressWarnings ( "unchecked" ) // Stores the tree static ArrayList<Integer>[] adj = new ArrayList[sz]; // Tracks the eulerwalk static ArrayList<Integer> euler = new ArrayList<>(); // Depth for each node corresponding static ArrayList<Integer> depthArr = new ArrayList<>(); // to eulerwalk // Stores first appearance index of every node static int [] FAI = new int [sz]; // Stores depth for all nodes in the tree static int [] level = new int [sz]; // Pointer to euler walk static int ptr; // Sparse table static int [][] dp = new int [sz][ 18 ]; // Stores log values static int [] logn = new int [sz]; // Stores power of 2 static int [] p2 = new int [ 20 ]; static void buildSparseTable( int n) { // Initializing sparse table for ( int i = 0 ; i < sz; i++) { for ( int j = 0 ; j < 18 ; j++) { dp[i][j] = - 1 ; } } // Filling base case values for ( int i = 1 ; i < n; i++) dp[i - 1 ][ 0 ] = (depthArr.get(i) > depthArr.get(i - 1 )) ? i - 1 : i; // dp to fill sparse table for ( int l = 1 ; l < 15 ; l++) for ( int i = 0 ; i < n; i++) if (dp[i][l - 1 ] != - 1 && dp[i + p2[l - 1 ]][l - 1 ] != - 1 ) dp[i][l] = (depthArr.get(dp[i][l - 1 ]) > depthArr.get( dp[i + p2[l - 1 ]][l - 1 ])) ? dp[i + p2[l - 1 ]][l - 1 ] : dp[i][l - 1 ]; else break ; } static int query( int l, int r) { int d = r - l; int dx = logn[d]; if (l == r) return l; if (depthArr.get(dp[l][dx]) > depthArr.get(dp[r - p2[dx]][dx])) return dp[r - p2[dx]][dx]; else return dp[l][dx]; } static void preprocess() { // Memorizing powers of 2 p2[ 0 ] = 1 ; for ( int i = 1 ; i < 18 ; i++) p2[i] = p2[i - 1 ] * 2 ; // Memorizing all log(n) values int val = 1 , ptr = 0 ; for ( int i = 1 ; i < sz; i++) { logn[i] = ptr - 1 ; if (val == i) { val *= 2 ; logn[i] = ptr; ptr++; } } } // Euler Walk ( preorder traversal) converting // tree to linear depthArray // Time Complexity : O(n) static void dfs( int cur, int prev, int dep) { // Marking FAI for cur node if (FAI[cur] == - 1 ) FAI[cur] = ptr; level[cur] = dep; // Pushing root to euler walk euler.add(cur); // Incrementing euler walk pointer ptr++; for (Integer x : adj[cur]) { if (x != prev) { dfs(x, cur, dep + 1 ); // Pushing cur again in backtrack // of euler walk euler.add(cur); // Increment euler walk pointer ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array static void makeArr() { for (Integer x : euler) depthArr.add(level[x]); } static int LCA( int u, int v) { // Trivial case if (u == v) return u; if (FAI[u] > FAI[v]) { int temp = u; u = v; v = temp; } // Doing RMQ in the required range return euler.get(query(FAI[u], FAI[v])); } static void addEdge( int u, int v) { adj[u].add(v); adj[v].add(u); } // Driver code public static void main(String[] args) { for ( int i = 0 ; i < sz; i++) { adj[i] = new ArrayList<>(); } // Constructing the described tree int numberOfNodes = 8 ; addEdge( 1 , 2 ); addEdge( 1 , 3 ); addEdge( 2 , 4 ); addEdge( 2 , 5 ); addEdge( 2 , 6 ); addEdge( 3 , 7 ); addEdge( 3 , 8 ); // Performing required precalculations preprocess(); // Doing the Euler walk ptr = 0 ; Arrays.fill(FAI, - 1 ); dfs( 1 , 0 , 0 ); // Creating depthArray corresponding to euler[] makeArr(); // Building sparse table buildSparseTable(depthArr.size()); System.out.println( "LCA(6,7) : " + LCA( 6 , 7 )); System.out.println( "LCA(6,4) : " + LCA( 6 , 4 )); } } // This code is contributed by sanjeev2552 |
Python3
# Python program to demonstrate LCA of n-ary tree # in constant time. from typing import List # stores the tree adj = [[] for _ in range ( 101 )] # tracks the eulerwalk euler = [] # depth for each node corresponding to eulerwalk depthArr = [] # stores first appearance index of every node FAI = [ - 1 ] * 101 # stores depth for all nodes in the tree level = [ 0 ] * 101 # pointer to euler walk ptr = 0 # sparse table dp = [[ - 1 ] * 18 for _ in range ( 101 )] # stores log values logn = [ 0 ] * 101 # stores power of 2 p2 = [ 0 ] * 20 def buildSparseTable(n: int ): # initializing sparse table for i in range (n): dp[i][ 0 ] = i - 1 if depthArr[i] > depthArr[i - 1 ] else i # dp to fill sparse table for l in range ( 1 , 15 ): for i in range (n): if dp[i][l - 1 ] ! = - 1 and dp[i + p2[l - 1 ]][l - 1 ] ! = - 1 : dp[i][l] = dp[i + p2[l - 1 ]][l - 1 ] if depthArr[dp[i][l - 1 ] ] > depthArr[dp[i + p2[l - 1 ]][l - 1 ]] else dp[i][l - 1 ] else : break def query(l: int , r: int ) - > int : d = r - l dx = logn[d] if l = = r: return l if depthArr[dp[l][dx]] > depthArr[dp[r - p2[dx]][dx]]: return dp[r - p2[dx]][dx] else : return dp[l][dx] def preprocess(): global ptr # memorizing powers of 2 p2[ 0 ] = 1 for i in range ( 1 , 18 ): p2[i] = p2[i - 1 ] * 2 # memorizing all log(n) values val = 1 ptr = 0 for i in range ( 1 , 101 ): logn[i] = ptr - 1 if val = = i: val * = 2 logn[i] = ptr ptr + = 1 def dfs(cur: int , prev: int , dep: int ): global ptr # marking FAI for cur node if FAI[cur] = = - 1 : FAI[cur] = ptr level[cur] = dep # pushing root to euler walk euler.append(cur) # incrementing euler walk pointer ptr + = 1 for x in adj[cur]: if x ! = prev: dfs(x, cur, dep + 1 ) # pushing cur again in backtrack # of euler walk euler.append(cur) # increment euler walk pointer ptr + = 1 # Create Level depthArray corresponding # to the Euler walk Array def makeArr(): global depthArr for x in euler: depthArr.append(level[x]) def LCA(u: int , v: int ) - > int : # trivial case if u = = v: return u if FAI[u] > FAI[v]: u, v = v, u # doing RMQ in the required range return euler[query(FAI[u], FAI[v])] def addEdge(u, v): adj[u].append(v) adj[v].append(u) # constructing the described tree numberOfNodes = 8 addEdge( 1 , 2 ) addEdge( 1 , 3 ) addEdge( 2 , 4 ) addEdge( 2 , 5 ) addEdge( 2 , 6 ) addEdge( 3 , 7 ) addEdge( 3 , 8 ) # performing required precalculations preprocess() # doing the Euler walk ptr = 0 FAI = [ - 1 ] * (numberOfNodes + 1 ) dfs( 1 , 0 , 0 ) # creating depthArray corresponding to euler[] makeArr() # building sparse table buildSparseTable( len (depthArr)) print ( "LCA(6,7) : " , LCA( 6 , 7 )) print ( "LCA(6,4) : " , LCA( 6 , 4 )) |
C#
// C# program to demonstrate LCA of n-ary // tree in constant time. using System; using System.Collections.Generic; public class GFG { static int sz = 101; // Stores the tree static List< int >[] adj = new List< int >[sz]; // Tracks the eulerwalk static List< int > euler = new List< int >(); // Depth for each node corresponding static List< int > depthArr = new List< int >(); // to eulerwalk // Stores first appearance index of every node static int [] FAI = new int [sz]; // Stores depth for all nodes in the tree static int [] level = new int [sz]; // Pointer to euler walk static int ptr; // Sparse table static int [,] dp = new int [sz, 18]; // Stores log values static int [] logn = new int [sz]; // Stores power of 2 static int [] p2 = new int [20]; static void buildSparseTable( int n) { // Initializing sparse table for ( int i = 0; i < sz; i++) { for ( int j = 0; j < 18; j++) { dp[i,j] = -1; } } // Filling base case values for ( int i = 1; i < n; i++) dp[i - 1,0] = (depthArr[i] > depthArr[i - 1]) ? i - 1 : i; // dp to fill sparse table for ( int l = 1; l < 15; l++) for ( int i = 0; i < n; i++) if (dp[i,l - 1] != -1 && dp[i + p2[l - 1],l - 1] != -1) dp[i,l] = (depthArr[dp[i,l - 1]] > depthArr[dp[i + p2[l - 1],l - 1]]) ? dp[i + p2[l - 1],l - 1] : dp[i,l - 1]; else break ; } static int query( int l, int r) { int d = r - l; int dx = logn[d]; if (l == r) return l; if (depthArr[dp[l,dx]] > depthArr[dp[r - p2[dx],dx]]) return dp[r - p2[dx],dx]; else return dp[l,dx]; } static void preprocess() { // Memorizing powers of 2 p2[0] = 1; for ( int i = 1; i < 18; i++) p2[i] = p2[i - 1] * 2; // Memorizing all log(n) values int val = 1, ptr = 0; for ( int i = 1; i < sz; i++) { logn[i] = ptr - 1; if (val == i) { val *= 2; logn[i] = ptr; ptr++; } } } // Euler Walk ( preorder traversal) converting // tree to linear depthArray // Time Complexity : O(n) static void dfs( int cur, int prev, int dep) { // Marking FAI for cur node if (FAI[cur] == -1) FAI[cur] = ptr; level[cur] = dep; // Pushing root to euler walk euler.Add(cur); // Incrementing euler walk pointer ptr++; foreach ( int x in adj[cur]) { if (x != prev) { dfs(x, cur, dep + 1); euler.Add(cur); ptr++; } } } // Create Level depthArray corresponding // to the Euler walk Array static void makeArr() { foreach ( int x in euler) depthArr.Add(level[x]); } static int LCA( int u, int v) { // Trivial case if (u == v) return u; if (FAI[u] > FAI[v]) { int temp = u; u = v; v = temp; } // Doing RMQ in the required range return euler[query(FAI[u], FAI[v])]; } static void addEdge( int u, int v) { adj[u].Add(v); adj[v].Add(u); } // Driver Code static void Main( string [] args) { int sz = 9; adj = new List< int >[sz]; for ( int i = 0; i < sz; i++) { adj[i] = new List< int >(); } // Constructing the described tree int numberOfNodes = 8; addEdge(1, 2); addEdge(1, 3); addEdge(2, 4); addEdge(2, 5); addEdge(2, 6); addEdge(3, 7); addEdge(3, 8); // Performing required precalculations preprocess(); // Doing the Euler walk ptr = 0; Array.Fill(FAI, -1); dfs(1, 0, 0); // Creating depthArray corresponding to euler[] makeArr(); // Building sparse table buildSparseTable(depthArr.Count); Console.WriteLine( "LCA(6,7) : " + LCA(6, 7)); Console.WriteLine( "LCA(6,4) : " + LCA(6, 4)); } } // This code is contributed by Prince Kumar |
Javascript
let adj = []; for (let _ = 0; _ < 101; _++) { adj.push([]); } // tracks the eulerwalk let euler = []; // depth for each node corresponding to eulerwalk let depthArr = []; // stores first appearance index of every node let FAI = new Array(101).fill(-1); // stores depth for all nodes in the tree let level = new Array(101).fill(0); // pointer to euler walk let ptr = 0; // sparse table let dp = []; for (let _ = 0; _ < 101; _++) { dp.push( new Array(18).fill(-1)); } // stores log values let logn = new Array(101).fill(0); // stores power of 2 let p2 = new Array(20).fill(0); function buildSparseTable(n) { // initializing sparse table for (let i = 0; i < n; i++) { dp[i][0] = i - 1 >= 0 && depthArr[i] > depthArr[i - 1] ? i - 1 : i; } // dp to fill sparse table for (let l = 1; l < 15; l++) { for (let i = 0; i < n; i++) { if ( dp[i][l - 1] !== -1 && dp[i + p2[l - 1]][l - 1] !== -1 ) { dp[i][l] = depthArr[dp[i][l - 1]] > depthArr[dp[i + p2[l - 1]][l - 1]] ? dp[i + p2[l - 1]][l - 1] : dp[i][l - 1]; } else { break ; } } } } function query(l, r) { let d = r - l; let dx = logn[d]; if (l === r) { return l; } if (depthArr[dp[l][dx]] > depthArr[dp[r - p2[dx]][dx]]) { return dp[r - p2[dx]][dx]; } else { return dp[l][dx]; } } function preprocess() { // memorizing powers of 2 p2[0] = 1; for (let i = 1; i < 18; i++) { p2[i] = p2[i - 1] * 2; } // memorizing all log(n) values let val = 1; ptr = 0; for (let i = 1; i < 101; i++) { logn[i] = ptr - 1; if (val === i) { val *= 2; logn[i] = ptr; ptr += 1; } } } function dfs(cur, prev, dep) { // marking FAI for cur node if (FAI[cur] === -1) { FAI[cur] = ptr; } level[cur] = dep; // pushing root to euler walk euler.push(cur); // incrementing euler walk pointer ptr += 1; for (let x of adj[cur]) { if (x !== prev) { dfs(x, cur, dep + 1); // pushing cur again in backtrack // of euler walk euler.push(cur); // increment euler walk pointer ptr += 1; } } } // Create Level depthArray corresponding // to the Euler walk Array function makeArr() { for (let x of euler) { depthArr.push(level[x]); } } function LCA(u, v) { // trivial case if (u === v) { return u; } if (FAI[u] > FAI[v]) { [u, v] = [v, u]; } // doing RMQ in the required range return euler[query(FAI[u], FAI[v])]; } function addEdge(u, v) { adj[u].push(v); adj[v].push(u); } // constructing the described tree let numberOfNodes = 8; addEdge(1, 2); addEdge(1, 3); addEdge(2, 4); addEdge(2, 5); addEdge(2, 6); addEdge(3, 7); addEdge(3, 8); // performing required precalculations preprocess(); // doing the Euler walk ptr = 0; FAI = new Array(numberOfNodes + 1).fill(-1); dfs(1, 0, 0); // creating depthArray corresponding to euler[] makeArr(); // building sparse table buildSparseTable(depthArr.length); console.log( "LCA(6,7) : " , LCA(6, 7)); console.log( "LCA(6,4) : " , LCA(6, 4)); |
LCA(6,7) : 1 LCA(6,4) : 2
Note : We are precalculating all the required power of 2’s and also precalculating the all the required log values to ensure constant time complexity per query. Else if we did log calculation for every query operation our Time complexity would have not been constant.
Time Complexity: The Conversion process from LCA to RMQ is done by Euler Walk that takes O(n) time.
Pre-processing for the sparse table in RMQ takes O(nlogn) time and answering each Query is a Constant time process. Therefore, overall Time Complexity is O(nlogn) – preprocessing and O(1) for each query.
Auxiliary Space: O(n+sz)
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