Sqrt (or Square Root) Decomposition | Set 2 (LCA of Tree in O(sqrt(height)) time)

07 Jul, 2022
Prerequisite: Introduction and DFSThe task is to find LCA of two given nodes in a tree (not necessarily a Binary Tree). In previous posts, we have seen how to calculate LCA using the Sparse Matrix DP approach. In this post, we will see an optimization done on the Naive method by sqrt decomposition technique that works well over the Naive Approach.
Naive Approach To calculate the LCA of two nodes first we will bring both the nodes to the same height by making the node with greater depth jump one parent up the tree till both the nodes are at the same height. Once, both the nodes are at the same height we can then start jumping one parent up for both the nodes simultaneously till both the nodes become equal and that node will be the LCA of the two originally given nodes.
Consider the below n-ary Tree with depth 9 and let’s examine how the naive approach works for this sample tree.
Here in the above Tree we need to calculate the LCA of node 6 and node 30 Clearly, node 30 has greater depth than node 6. So first of all we start jumping one parent above for node 30 till we reach the depth value of node 6 i.e at depth 2.
The orange colored path in the above figure demonstrates the jumping sequence to reach depth 2. In this procedure, we just simply jump one parent above the current node.
Now both nodes are at the same depth 2.Therefore, now both the nodes will jump one parent up till both the nodes become equal. This end node at which both the nodes become equal for the first time is our LCA. The blue color path in the above figure shows the jumping route for both the nodes
Implementation:
C++
Java
Python3
C#
Javascript
// Naive C++ implementation to find LCA in a tree#include<bits/stdc++.h>using namespace std;#define MAXN 1001 int depth[MAXN];           // stores depth for each nodeint parent[MAXN];          // stores first parent for each node vector < int > adj[MAXN]; void addEdge(int u,int v){    adj[u].push_back(v);    adj[v].push_back(u);} void dfs(int cur, int prev){    // marking parent for each node    parent[cur] = prev;     // marking depth for each node    depth[cur] = depth[prev] + 1;     // propagating marking down the tree    for (int i=0; i<adj[cur].size(); i++)        if (adj[cur][i] != prev)            dfs(adj[cur][i],cur);} void preprocess(){    // a dummy node    depth[0] = -1;     // precalculating 1)depth.  2)parent.    // for each node    dfs(1,0);} // Time Complexity : O(Height of tree)// recursively jumps one node above// till both the nodes become equalint LCANaive(int u,int v){    if (u == v)  return u;    if (depth[u] > depth[v])        swap(u, v);    v = parent[v];    return LCANaive(u,v);} // Driver function to call the above functionsint main(int argc, char const *argv[]){    // adding edges to the tree    addEdge(1,2);    addEdge(1,3);    addEdge(1,4);    addEdge(2,5);    addEdge(2,6);    addEdge(3,7);    addEdge(4,8);    addEdge(4,9);    addEdge(9,10);    addEdge(9,11);    addEdge(7,12);    addEdge(7,13);     preprocess();     cout << "LCA(11,8) : " << LCANaive(11,8) << endl;    cout << "LCA(3,13) : " << LCANaive(3,13) << endl;     return 0;}
// Naive Java implementation to find LCA in a tree.import java.io.*;import java.util.*; class GFG{    static int MAXN = 1001;         // stores depth for each node    static int[] depth = new int[MAXN];          // stores first parent for each node    static int[] parent = new int[MAXN];     @SuppressWarnings("unchecked")    static Vector<Integer>[] adj = new Vector[MAXN];    static    {        for (int i = 0; i < MAXN; i++)            adj[i] = new Vector<>();    }     static void addEdge(int u, int v)    {        adj[u].add(v);        adj[v].add(u);    }     static void dfs(int cur, int prev)    {         // marking parent for each node        parent[cur] = prev;         // marking depth for each node        depth[cur] = depth[prev] + 1;         // propagating marking down the tree        for (int i = 0; i < adj[cur].size(); i++)            if (adj[cur].elementAt(i) != prev)                dfs(adj[cur].elementAt(i), cur);    }     static void preprocess()    {         // a dummy node        depth[0] = -1;         // precalculating 1)depth. 2)parent.        // for each node        dfs(1, 0);    }     // Time Complexity : O(Height of tree)    // recursively jumps one node above    // till both the nodes become equal    static int LCANaive(int u, int v)    {        if (u == v)            return u;        if (depth[u] > depth[v])        {            int temp = u;            u = v;            v = temp;        }        v = parent[v];        return LCANaive(u, v);    }     // Driver Code    public static void main(String[] args)    {         // adding edges to the tree        addEdge(1, 2);        addEdge(1, 3);        addEdge(1, 4);        addEdge(2, 5);        addEdge(2, 6);        addEdge(3, 7);        addEdge(4, 8);        addEdge(4, 9);        addEdge(9, 10);        addEdge(9, 11);        addEdge(7, 12);        addEdge(7, 13);         preprocess();         System.out.println("LCA(11,8) : " + LCANaive(11, 8));        System.out.println("LCA(3,13) : " + LCANaive(3, 13));    }} // This code is contributed by// sanjeev2552
# Python3 implementation to# find LCA in a treeMAXN = 1001  # stores depth for each nodedepth = [0 for i in range(MAXN)]; # stores first parent for each nodeparent = [0 for i in range(MAXN)];          adj = [[] for i in range(MAXN)]  def addEdge(u, v):     adj[u].append(v);    adj[v].append(u); def dfs(cur, prev):     # marking parent for    # each node    parent[cur] = prev;      # marking depth for    # each node    depth[cur] = depth[prev] + 1;      # propagating marking down    # the tree    for i in range(len(adj[cur])):           if (adj[cur][i] != prev):            dfs(adj[cur][i], cur);  def preprocess():     # a dummy node    depth[0] = -1;      # precalculating 1)depth.     # 2)parent. for each node    dfs(1, 0);  # Time Complexity : O(Height of tree)# recursively jumps one node above# till both the nodes become equaldef LCANaive(u, v):     if (u == v):        return u;             if (depth[u] > depth[v]):        u, v = v, u             v = parent[v];       return LCANaive(u, v);         # Driver codeif __name__ == "__main__":         # adding edges to the tree    addEdge(1, 2);    addEdge(1, 3);    addEdge(1, 4);    addEdge(2, 5);    addEdge(2, 6);    addEdge(3, 7);    addEdge(4, 8);    addEdge(4, 9);    addEdge(9, 10);    addEdge(9, 11);    addEdge(7, 12);    addEdge(7, 13);      preprocess();         print('LCA(11,8) : ' +           str(LCANaive(11, 8)))    print('LCA(3,13) : ' +           str(LCANaive(3, 13)))     # This code is contributed by RUtvik_56
// Naive C# implementation to find// LCA in a tree.using System;using System.Collections;  class GFG{     static int MAXN = 1001; // Stores depth for each nodestatic int[] depth = new int[MAXN]; // Stores first parent for each nodestatic int[] parent = new int[MAXN]; static ArrayList[] adj = new ArrayList[MAXN]; static void addEdge(int u, int v){    adj[u].Add(v);    adj[v].Add(u);} static void dfs(int cur, int prev){         // Marking parent for each node    parent[cur] = prev;         // Marking depth for each node    depth[cur] = depth[prev] + 1;         // Propagating marking down the tree    for(int i = 0; i < adj[cur].Count; i++)        if ((int)adj[cur][i] != prev)            dfs((int)adj[cur][i], cur);} static void preprocess(){         // A dummy node    depth[0] = -1;     // Precalculating 1)depth. 2)parent.    // for each node    dfs(1, 0);} // Time Complexity : O(Height of tree)// recursively jumps one node above// till both the nodes become equalstatic int LCANaive(int u, int v){    if (u == v)        return u;             if (depth[u] > depth[v])    {        int temp = u;        u = v;        v = temp;    }    v = parent[v];    return LCANaive(u, v);} // Driver Codepublic static void Main(string[] args){    for(int i = 0; i < MAXN; i++)        adj[i] = new ArrayList();             // Adding edges to the tree    addEdge(1, 2);    addEdge(1, 3);    addEdge(1, 4);    addEdge(2, 5);    addEdge(2, 6);    addEdge(3, 7);    addEdge(4, 8);    addEdge(4, 9);    addEdge(9, 10);    addEdge(9, 11);    addEdge(7, 12);    addEdge(7, 13);     preprocess();     Console.WriteLine("LCA(11, 8) : " +                       LCANaive(11, 8));    Console.WriteLine("LCA(3, 13) : " +                       LCANaive(3, 13));}} // This code is contributed by pratham76
<script>// Naive Javascript implementation to find// LCA in a tree.var MAXN = 1001; // Stores depth for each nodevar depth = Array(MAXN); // Stores first parent for each nodevar parent = Array(MAXN); var adj = Array.from(Array(MAXN), ()=>Array()); function addEdge(u, v){    adj[u].push(v);    adj[v].push(u);} function dfs(cur, prev){         // Marking parent for each node    parent[cur] = prev;         // Marking depth for each node    depth[cur] = depth[prev] + 1;         // Propagating marking down the tree    for(var i = 0; i < adj[cur].length; i++)        if (adj[cur][i] != prev)            dfs(adj[cur][i], cur);} function preprocess(){         // A dummy node    depth[0] = -1;     // Precalculating 1)depth. 2)parent.    // for each node    dfs(1, 0);} // Time Complexity : O(Height of tree)// recursively jumps one node above// till both the nodes become equalfunction LCANaive(u, v){    if (u == v)        return u;             if (depth[u] > depth[v])    {        var temp = u;        u = v;        v = temp;    }    v = parent[v];    return LCANaive(u, v);} // Driver Codefor(var i = 0; i < MAXN; i++)    adj[i] = [];     // Adding edges to the treeaddEdge(1, 2);addEdge(1, 3);addEdge(1, 4);addEdge(2, 5);addEdge(2, 6);addEdge(3, 7);addEdge(4, 8);addEdge(4, 9);addEdge(9, 10);addEdge(9, 11);addEdge(7, 12);addEdge(7, 13);preprocess();document.write("LCA(11, 8) : " +                   LCANaive(11, 8)+"<br>");document.write("LCA(3, 13) : " +                   LCANaive(3, 13)); // This code is contributed by famously.</script>
LCA(11,8) : 4
LCA(3,13) : 3
Time Complexity: We pre-calculate the depth for each node using one DFS traversal in O(n). Now in the worst case, the two nodes will be two bottom-most nodes on the tree in different child branches of the root node. Therefore, in this case, the root will be the LCA of both the nodes. Hence, both the nodes will have to jump exactly h height above, where h is the height of the tree. So, to answer each LCA query Time Complexity will be O(h).
The Sqrt Decomposition Trick : We categorize nodes of the tree into different groups according to their depth. Assuming the depth of the tree h is a perfect square. So once again like the general sqrt decomposition approach, we will be having sqrt(h) blocks or groups. Nodes from depth 0 to depth sqrt(h) – 1 lie in the first group; then nodes having depth sqrt(H) to 2*sqrt(h)-1 lie in second group and so on till the last node.We keep track of the corresponding group number for every node and also the depth of every node. This can be done by one single dfs on the tree (see the code for better understanding).
Sqrt trick :- In the naive approach, we were jumping one parent up the tree till both nodes aren’t on the same depth. But here we perform the group-wise jump. To perform this group-wise jump, we need two-parameter associated with each node: 1) parent and 2) jump parent Here parent for each node is defined as the first node above the current node that is directly connected to it, whereas jump_parent for each node is the node that is the first ancestor of the current node in the group just above the current node.
So, now we need to maintain 3 parameters for each node : 
depth parent jump_parent
depth 
parent 
jump_parent
All these three parameters can be maintained in one dfs(refer to the code for better understanding)
Pseudo code for the optimization process 
  LCAsqrt(u, v){

       // assuming v is at greater depth
       while (jump_parent[u]!=jump_parent[v]){       
            v = jump_parent[v]; 
       } 

       // now both nodes are in same group
       // and have same jump_parent
       return LCAnaive(u,v); 
    }
The key concept here is that first we bring both the nodes in same group and having same jump_parent by climbing decomposed blocks above the tree one by one and then when both the nodes are in same group and have same jump_parent we use our naive approach to find LCA of the nodes.
This optimized group jumping technique reduces the iterating space by a factor of sqrt(h) and hence reduces the Time Complexity(refer below for better time complexity analysis)
Lets decompose the above tree in sqrt(h) groups (h = 9) and calculate LCA for node 6 and 30.
In the above decomposed tree 
Jump_parent[6]  =  0        parent[6]  =  3
Jump_parent[5]  =  0        parent[5]  =  2
Jump_parent[1]  =  0        parent[1]  =  0 
Jump_parent[11] =  6        parent[11] =  6  
Jump_parent[15] =  6        parent[15] = 11 
Jump_parent[21] =  6        parent[21] = 15
Jump_parent[25] = 21        parent[25] = 21  
Jump_parent[26] = 21        parent[26] = 21 
Jump_parent[30] = 21        parent[30] = 25
Now at this stage Jump_parent for node 30 is 21 and Jump_parent for node 5 is 0, So we will climp to jump_parent[30] i.e to node 21Now once again Jump_parent of node 21 is not equal to Jump_parent of node 5, So once again we will climb to jump_parent[21] i.e node 6At this stage jump_parent[6] == jump_parent[5], So now we will use our naive climbing approach and climb one parent above for both the nodes till it reach node 1 and that will be the required LCA . Blue path in the above figure describes jumping path sequence for node 6 and node 5.
Implementation: 
C++
Java
C#
Javascript
// C++ program to find LCA using Sqrt decomposition#include "iostream"#include "vector"#include "math.h"using namespace std;#define MAXN 1001 int block_sz;          // block size = sqrt(height)int depth[MAXN];       // stores depth for each nodeint parent[MAXN];      // stores first parent for                       // each nodeint jump_parent[MAXN]; // stores first ancestor in                       // previous block vector < int > adj[MAXN]; void addEdge(int u,int v){    adj[u].push_back(v);    adj[v].push_back(u);} int LCANaive(int u,int v){    if (u == v)  return u;    if (depth[u] > depth[v])        swap(u,v);    v = parent[v];    return LCANaive(u,v);} // precalculating the required parameters// associated with every nodevoid dfs(int cur, int prev){    // marking depth of cur node    depth[cur] = depth[prev] + 1;     // marking parent of cur node    parent[cur] = prev;     // making jump_parent of cur node    if (depth[cur] % block_sz == 0)         /* if it is first node of the block           then its jump_parent is its cur parent */        jump_parent[cur] = parent[cur];     else         /* if it is not the first node of this block           then its jump_parent is jump_parent of           its parent */        jump_parent[cur] = jump_parent[prev];      // propagating the marking down the subtree    for (int i = 0; i<adj[cur].size(); ++i)        if (adj[cur][i] != prev)            dfs(adj[cur][i], cur);}  // using sqrt decomposition trickint LCASQRT(int u, int v){    while (jump_parent[u] != jump_parent[v])    {        if (depth[u] > depth[v])             // maintaining depth[v] > depth[u]            swap(u,v);         // climb to its jump parent        v = jump_parent[v];    }     // u and v have same jump_parent    return LCANaive(u,v);} void preprocess(int height){    block_sz = sqrt(height);    depth[0] = -1;     // precalculating 1)depth.  2)parent.  3)jump_parent    // for each node    dfs(1, 0);} // Driver function to call the above functionsint main(int argc, char const *argv[]){    // adding edges to the tree    addEdge(1,2);    addEdge(1,3);    addEdge(1,4);    addEdge(2,5);    addEdge(2,6);    addEdge(3,7);    addEdge(4,8);    addEdge(4,9);    addEdge(9,10);    addEdge(9,11);    addEdge(7,12);    addEdge(7,13);     // here we are directly taking height = 4    // according to the given tree but we can    // pre-calculate height = max depth    // in one more dfs    int height = 4;    preprocess(height);     cout << "LCA(11,8) : " << LCASQRT(11,8) << endl;    cout << "LCA(3,13) : " << LCASQRT(3,13) << endl;     return 0;}
// Java program to find LCA using Sqrt decompositionimport java.util.*;class GFG{static final int MAXN = 1001; static int block_sz;          // block size = Math.sqrt(height)static int []depth = new int[MAXN];       // stores depth for each nodestatic int []parent = new int[MAXN];      // stores first parent for                       // each nodestatic int []jump_parent = new int[MAXN]; // stores first ancestor in                       // previous block static Vector <Integer > []adj = new Vector[MAXN]; static void addEdge(int u,int v){    adj[u].add(v);    adj[v].add(u);} static int LCANaive(int u,int v){    if (u == v)  return u;    if (depth[u] > depth[v])    {        int t = u;        u = v;        v = t;    }        v = parent[v];    return LCANaive(u, v);} // precalculating the required parameters// associated with every nodestatic void dfs(int cur, int prev){       // marking depth of cur node    depth[cur] = depth[prev] + 1;     // marking parent of cur node    parent[cur] = prev;     // making jump_parent of cur node    if (depth[cur] % block_sz == 0)         /* if it is first node of the block           then its jump_parent is its cur parent */        jump_parent[cur] = parent[cur];    else         /* if it is not the first node of this block           then its jump_parent is jump_parent of           its parent */        jump_parent[cur] = jump_parent[prev];     // propagating the marking down the subtree    for (int i = 0; i < adj[cur].size(); ++i)        if (adj[cur].get(i) != prev)            dfs(adj[cur].get(i), cur);} // using sqrt decomposition trickstatic int LCASQRT(int u, int v){    while (jump_parent[u] != jump_parent[v])    {        if (depth[u] > depth[v])        {             // maintaining depth[v] > depth[u]            int t = u;            u = v;            v = t;        }         // climb to its jump parent        v = jump_parent[v];    }     // u and v have same jump_parent    return LCANaive(u, v);} static void preprocess(int height){    block_sz = (int)Math.sqrt(height);    depth[0] = -1;     // precalculating 1)depth.  2)parent.  3)jump_parent    // for each node    dfs(1, 0);} // Driver codepublic static void  main(String []args){    for (int i = 0; i < adj.length; i++)        adj[i] = new Vector<Integer>();       // adding edges to the tree    addEdge(1, 2);    addEdge(1, 3);    addEdge(1, 4);    addEdge(2, 5);    addEdge(2, 6);    addEdge(3, 7);    addEdge(4, 8);    addEdge(4, 9);    addEdge(9, 10);    addEdge(9, 11);    addEdge(7, 12);    addEdge(7, 13);     // here we are directly taking height = 4    // according to the given tree but we can    // pre-calculate height = max depth    // in one more dfs    int height = 4;    preprocess(height);    System.out.print("LCA(11,8) : " +  LCASQRT(11, 8) +"\n");    System.out.print("LCA(3,13) : " +  LCASQRT(3, 13) +"\n");}} // This code is contributed by aashish1995.
// C# program to find LCA using Sqrt decompositionusing System;using System.Collections.Generic;public class GFG{  static readonly int MAXN = 1001;   static int block_sz;          // block size = Math.Sqrt(height)  static int []depth = new int[MAXN];       // stores depth for each node  static int []parent = new int[MAXN];      // stores first parent for  // each node  static int []jump_parent = new int[MAXN]; // stores first ancestor in  // previous block   static List <int > []adj = new List<int>[MAXN];  static void addEdge(int u, int v)  {    adj[u].Add(v);    adj[v].Add(u);  }   static int LCANaive(int u, int v)  {    if (u == v)  return u;    if (depth[u] > depth[v])    {      int t = u;      u = v;      v = t;    }        v = parent[v];    return LCANaive(u, v);  }   // precalculating the required parameters  // associated with every node  static void dfs(int cur, int prev)  {     // marking depth of cur node    depth[cur] = depth[prev] + 1;     // marking parent of cur node    parent[cur] = prev;     // making jump_parent of cur node    if (depth[cur] % block_sz == 0)       /* if it is first node of the block           then its jump_parent is its cur parent */      jump_parent[cur] = parent[cur];    else       /* if it is not the first node of this block           then its jump_parent is jump_parent of           its parent */      jump_parent[cur] = jump_parent[prev];     // propagating the marking down the subtree    for (int i = 0; i < adj[cur].Count; ++i)      if (adj[cur][i] != prev)        dfs(adj[cur][i], cur);  }   // using sqrt decomposition trick  static int LCASQRT(int u, int v)  {    while (jump_parent[u] != jump_parent[v])    {      if (depth[u] > depth[v])      {         // maintaining depth[v] > depth[u]        int t = u;        u = v;        v = t;      }       // climb to its jump parent      v = jump_parent[v];    }     // u and v have same jump_parent    return LCANaive(u, v);  }   static void preprocess(int height)  {    block_sz = (int)Math.Sqrt(height);    depth[0] = -1;     // precalculating 1)depth.  2)parent.  3)jump_parent    // for each node    dfs(1, 0);  }   // Driver code  public static void  Main(String []args)  {    for (int i = 0; i < adj.Length; i++)      adj[i] = new List<int>();     // adding edges to the tree    addEdge(1, 2);    addEdge(1, 3);    addEdge(1, 4);    addEdge(2, 5);    addEdge(2, 6);    addEdge(3, 7);    addEdge(4, 8);    addEdge(4, 9);    addEdge(9, 10);    addEdge(9, 11);    addEdge(7, 12);    addEdge(7, 13);     // here we are directly taking height = 4    // according to the given tree but we can    // pre-calculate height = max depth    // in one more dfs    int height = 4;    preprocess(height);    Console.Write("LCA(11,8) : " +  LCASQRT(11, 8) +"\n");    Console.Write("LCA(3,13) : " +  LCASQRT(3, 13) +"\n");  }} // This code is contributed by Rajput-Ji
<script> // Javascript program to find LCA// using Sqrt decompositionlet MAXN = 1001; // Block size = Math.sqrt(height)let block_sz;    // Stores depth for each nodelet depth = new Array(MAXN);  // Stores first parent for// each nodelet parent = new Array(MAXN);      // Stores first ancestor in// previous blocklet jump_parent = new Array(MAXN);let adj = new Array(MAXN); function addEdge(u, v){    adj[u].push(v);    adj[v].push(u);} function LCANaive(u, v){    if (u == v)         return u;             if (depth[u] > depth[v])    {        let t = u;        u = v;        v = t;    }       v = parent[v];    return LCANaive(u, v);} // Precalculating the required parameters// associated with every nodefunction dfs(cur, prev){         // Marking depth of cur node    depth[cur] = depth[prev] + 1;     // Marking parent of cur node    parent[cur] = prev;     // Making jump_parent of cur node    if (depth[cur] % block_sz == 0)         // If it is first node of the block        // then its jump_parent is its cur parent        jump_parent[cur] = parent[cur];    else         // If it is not the first node of this block        // then its jump_parent is jump_parent of        // its parent        jump_parent[cur] = jump_parent[prev];     // Propagating the marking down the subtree    for(let i = 0; i < adj[cur].length; ++i)        if (adj[cur][i] != prev)            dfs(adj[cur][i], cur);} // Using sqrt decomposition trickfunction LCASQRT(u, v){    while (jump_parent[u] != jump_parent[v])    {        if (depth[u] > depth[v])        {                         // Maintaining depth[v] > depth[u]            let t = u;            u = v;            v = t;        }         // Climb to its jump parent        v = jump_parent[v];    }     // u and v have same jump_parent    return LCANaive(u, v);} function preprocess(height){    block_sz = parseInt(Math.sqrt(height), 10);    depth[0] = -1;     // Precalculating 1)depth.  2)parent.  3)    // jump_parent for each node    dfs(1, 0);} // Driver codefor(let i = 0; i < adj.length; i++)    adj[i] = []; // Adding edges to the treeaddEdge(1, 2);addEdge(1, 3);addEdge(1, 4);addEdge(2, 5);addEdge(2, 6);addEdge(3, 7);addEdge(4, 8);addEdge(4, 9);addEdge(9, 10);addEdge(9, 11);addEdge(7, 12);addEdge(7, 13); // Here we are directly taking height = 4// according to the given tree but we can// pre-calculate height = max depth// in one more dfslet height = 4;preprocess(height);document.write("LCA(11,8) : " +                LCASQRT(11, 8) + "</br>");document.write("LCA(3,13) : " +                LCASQRT(3, 13) + "</br>");                // This code is contributed by divyeshrabadiya07 </script>
LCA(11,8) : 4
LCA(3,13) : 3
Note : The above code works even if height is not perfect square.
Now Lets see how the Time Complexity is changed by this simple grouping technique :
Time Complexity Analysis: 
We have divided the tree into sqrt(h) groups according to their depth and each group contain nodes having max difference in their depth equal to sqrt(h). Now once again take an example of worst case, let’s say the first node ‘u’ is in first group and the node ‘v’ is in sqrt(h)th group(last group). So, first we will make group jumps(single group jumps) till we reach group 1 from last group; This will take exactly sqrt(h) – 1 iterations or jumps. So, till this step the Time Complexity is O(sqrt(h)).
Now once we are in same group, we call the LCAnaive function. The Time complexity for LCA_Naive is O(sqrt(h’)), where h’ is the height of the tree. Now, in our case value of h’ will be sqrt(h), because each group has a subtree of at max sqrt(h) height. So the complexity for this step is also O(sqrt(h)). 
Hence, the total Time Complexity will be O(sqrt(h) + sqrt(h)) ~ O(sqrt(h)).
This article is contributed by Nitish Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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