LCA for general or n-ary trees (Sparse Matrix DP approach )

06 Jul, 2022
In previous posts, we have discussed how to calculate the Lowest Common Ancestor (LCA) for a binary tree and a binary search tree (this, this and this). Now let’s look at a method that can calculate LCA for any tree (not only for binary tree). We use Dynamic Programming with Sparse Matrix Approach in our method. This method is very handy and fast when you need to answer multiple queries of LCA for a tree. 
Pre-requisites:
DFS Basic DP knowledge (This and this) Range Minimum Query (Square Root Decomposition and Sparse Table)
DFS 
Basic DP knowledge (This and this) 
Range Minimum Query (Square Root Decomposition and Sparse Table)
Naive Approach:- O(n) The naive approach for this general tree LCA calculation will be the same as the naive approach for the LCA calculation of Binary Tree (this naive approach is already well described here.
The implementation for the naive approach is given below:- 
C++
Java
Python3
C#
Javascript
/* Program to find LCA of n1 and n2 using one DFS on   the Tree */#include<bits/stdc++.h>using namespace std; // Maximum number of nodes is 100000 and nodes are// numbered from 1 to 100000#define MAXN 100001 vector < int > tree[MAXN];int path[3][MAXN]; // storing root to node path // storing the path from root to nodevoid dfs(int cur, int prev, int pathNumber, int ptr,                             int node, bool &flag){    for (int i=0; i<tree[cur].size(); i++)    {        if (tree[cur][i] != prev and !flag)        {            // pushing current node into the path            path[pathNumber][ptr] = tree[cur][i];            if (tree[cur][i] == node)            {                // node found                flag = true;                 // terminating the path                path[pathNumber][ptr+1] = -1;                return;            }            dfs(tree[cur][i], cur, pathNumber, ptr+1,                                        node, flag);        }    }} // This Function compares the path from root to 'a' & root// to 'b' and returns LCA of a and b. Time Complexity : O(n)int LCA(int a, int b){    // trivial case    if (a == b)        return a;     // setting root to be first element in path    path[1][0] = path[2][0] = 1;     // calculating path from root to a    bool flag = false;    dfs(1, 0, 1, 1, a, flag);     // calculating path from root to b    flag = false;    dfs(1, 0, 2, 1, b, flag);     // runs till path 1 & path 2 matches    int i = 0;    while (path[1][i] == path[2][i])        i++;     // returns the last matching node in the paths    return path[1][i-1];} void addEdge(int a,int b){    tree[a].push_back(b);    tree[b].push_back(a);} // Driver codeint main(){    int n = 8; // Number of nodes    addEdge(1,2);    addEdge(1,3);    addEdge(2,4);    addEdge(2,5);    addEdge(2,6);    addEdge(3,7);    addEdge(3,8);     cout << "LCA(4, 7) = " << LCA(4,7) << endl;    cout << "LCA(4, 6) = " << LCA(4,6) << endl;    return 0;}
/* Program to find LCA of n1 and n2 using one DFS onthe Tree */ import java.util.*; class GFG{     // Maximum number of nodes is 100000 and nodes are    // numbered from 1 to 100000    static final int MAXN = 100001;     static Vector<Integer>[] tree = new Vector[MAXN];    static int[][] path = new int[3][MAXN]; // storing root to node path    static boolean flag;     // storing the path from root to node    static void dfs(int cur, int prev, int pathNumber, int ptr, int node)    {        for (int i = 0; i < tree[cur].size(); i++)        {            if (tree[cur].get(i) != prev && !flag)            {                // pushing current node into the path                path[pathNumber][ptr] = tree[cur].get(i);                if (tree[cur].get(i) == node)                {                    // node found                    flag = true;                     // terminating the path                    path[pathNumber][ptr + 1] = -1;                    return;                }                dfs(tree[cur].get(i), cur, pathNumber, ptr + 1, node);            }        }    }     // This Function compares the path from root to 'a' & root    // to 'b' and returns LCA of a and b. Time Complexity : O(n)    static int LCA(int a, int b)    {        // trivial case        if (a == b)            return a;         // setting root to be first element in path        path[1][0] = path[2][0] = 1;         // calculating path from root to a        flag = false;        dfs(1, 0, 1, 1, a);         // calculating path from root to b        flag = false;        dfs(1, 0, 2, 1, b);         // runs till path 1 & path 2 matches        int i = 0;        while (i < MAXN && path[1][i] == path[2][i])            i++;         // returns the last matching node in the paths        return path[1][i - 1];    }     static void addEdge(int a, int b)    {        tree[a].add(b);        tree[b].add(a);    }     // Driver code    public static void main(String[] args)    {        for (int i = 0; i < MAXN; i++)            tree[i] = new Vector<Integer>();         // Number of nodes        addEdge(1, 2);        addEdge(1, 3);        addEdge(2, 4);        addEdge(2, 5);        addEdge(2, 6);        addEdge(3, 7);        addEdge(3, 8);         System.out.print("LCA(4, 7) = " + LCA(4, 7) + "\n");        System.out.print("LCA(4, 6) = " + LCA(4, 6) + "\n");    }} // This code is contributed by 29AjayKumar
# Python3 program to find LCA of n1 and# n2 using one DFS on the Tree # Maximum number of nodes is 100000 and# nodes are numbered from 1 to 100000MAXN = 100001 tree = [0] * MAXNfor i in range(MAXN):    tree[i] = [] # Storing root to node pathpath = [0] * 3for i in range(3):    path[i] = [0] * MAXN flag = False # Storing the path from root to nodedef dfs(cur: int, prev: int, pathNumber: int,        ptr: int, node: int) -> None:                 global tree, path, flag     for i in range(len(tree[cur])):        if (tree[cur][i] != prev and not flag):             # Pushing current node into the path            path[pathNumber][ptr] = tree[cur][i]                         if (tree[cur][i] == node):                 # Node found                flag = True                 # Terminating the path                path[pathNumber][ptr + 1] = -1                return             dfs(tree[cur][i], cur, pathNumber,                ptr + 1, node) # This Function compares the path from root# to 'a' & root to 'b' and returns LCA of# a and b. Time Complexity : O(n)def LCA(a: int, b: int) -> int:         global flag         # Trivial case    if (a == b):        return a     # Setting root to be first element    # in path    path[1][0] = path[2][0] = 1     # Calculating path from root to a    flag = False    dfs(1, 0, 1, 1, a)     # Calculating path from root to b    flag = False    dfs(1, 0, 2, 1, b)     # Runs till path 1 & path 2 matches    i = 0    while (path[1][i] == path[2][i]):        i += 1     # Returns the last matching    # node in the paths    return path[1][i - 1] def addEdge(a: int, b: int) -> None:     tree[a].append(b)    tree[b].append(a) # Driver codeif __name__ == "__main__":     n = 8         # Number of nodes    addEdge(1, 2)    addEdge(1, 3)    addEdge(2, 4)    addEdge(2, 5)    addEdge(2, 6)    addEdge(3, 7)    addEdge(3, 8)     print("LCA(4, 7) = {}".format(LCA(4, 7)))    print("LCA(4, 6) = {}".format(LCA(4, 6))) # This code is contributed by sanjeev2552
/* C# Program to find LCA of n1 and n2 using one DFS onthe Tree */using System;using System.Collections.Generic; class GFG{     // Maximum number of nodes is 100000 and nodes are    // numbered from 1 to 100000    static readonly int MAXN = 100001;     static List<int>[] tree = new List<int>[MAXN];    static int[,] path = new int[3, MAXN]; // storing root to node path    static bool flag;     // storing the path from root to node    static void dfs(int cur, int prev, int pathNumber, int ptr, int node)    {        for (int i = 0; i < tree[cur].Count; i++)        {            if (tree[cur][i] != prev && !flag)            {                // pushing current node into the path                path[pathNumber,ptr] = tree[cur][i];                if (tree[cur][i] == node)                {                    // node found                    flag = true;                     // terminating the path                    path[pathNumber, ptr + 1] = -1;                    return;                }                dfs(tree[cur][i], cur, pathNumber, ptr + 1, node);            }        }    }     // This Function compares the path from root to 'a' & root    // to 'b' and returns LCA of a and b. Time Complexity : O(n)    static int LCA(int a, int b)    {        // trivial case        if (a == b)            return a;         // setting root to be first element in path        path[1, 0] = path[2, 0] = 1;         // calculating path from root to a        flag = false;        dfs(1, 0, 1, 1, a);         // calculating path from root to b        flag = false;        dfs(1, 0, 2, 1, b);         // runs till path 1 & path 2 matches        int i = 0;        while (i < MAXN && path[1, i] == path[2, i])            i++;         // returns the last matching node in the paths        return path[1, i - 1];    }     static void addEdge(int a, int b)    {        tree[a].Add(b);        tree[b].Add(a);    }     // Driver code    public static void Main(String[] args)    {        for (int i = 0; i < MAXN; i++)            tree[i] = new List<int>();         // Number of nodes        addEdge(1, 2);        addEdge(1, 3);        addEdge(2, 4);        addEdge(2, 5);        addEdge(2, 6);        addEdge(3, 7);        addEdge(3, 8);         Console.Write("LCA(4, 7) = " + LCA(4, 7) + "\n");        Console.Write("LCA(4, 6) = " + LCA(4, 6) + "\n");    }} // This code is contributed by Rajput-Ji
<script>/* Program to find LCA of n1 and n2 using one DFS onthe Tree */ // Maximum number of nodes is 100000 and nodes are    // numbered from 1 to 100000let MAXN = 100001; let tree = new Array(MAXN); let path = new Array(3); for(let i = 0; i < 3; i++){    path[i] = new Array(MAXN);    for(let j = 0; j < MAXN; j++)    {        path[i][j] = 0;    }} let flag; /// storing the path from root to nodefunction dfs(cur,prev,pathNumber,ptr,node){    for (let i = 0; i < tree[cur].length; i++)        {            if (tree[cur][i] != prev && !flag)            {                // pushing current node into the path                path[pathNumber][ptr] = tree[cur][i];                if (tree[cur][i] == node)                {                    // node found                    flag = true;                      // terminating the path                    path[pathNumber][ptr + 1] = -1;                    return;                }                dfs(tree[cur][i], cur, pathNumber, ptr + 1, node);            }        }} // This Function compares the path from root to 'a' & root    // to 'b' and returns LCA of a and b. Time Complexity : O(n)function LCA(a,b){    // trivial case        if (a == b)            return a;          // setting root to be first element in path        path[1][0] = path[2][0] = 1;          // calculating path from root to a        flag = false;        dfs(1, 0, 1, 1, a);          // calculating path from root to b        flag = false;        dfs(1, 0, 2, 1, b);          // runs till path 1 & path 2 matches        let i = 0;        while (i < MAXN && path[1][i] == path[2][i])            i++;          // returns the last matching node in the paths        return path[1][i - 1];} function addEdge(a,b){    tree[a].push(b);    tree[b].push(a);} // Driver codefor (let i = 0; i < MAXN; i++)    tree[i] = []; // Number of nodesaddEdge(1, 2);addEdge(1, 3);addEdge(2, 4);addEdge(2, 5);addEdge(2, 6);addEdge(3, 7);addEdge(3, 8); document.write("LCA(4, 7) = " + LCA(4, 7) + "<br>");document.write("LCA(4, 6) = " + LCA(4, 6) + "<br>"); // This code is contributed by rag2127</script>
LCA(4, 7) = 1
LCA(4, 6) = 2

Sparse Matrix Approach (O(nlogn) pre-processing, O(log n) – query) 
Pre-computation:
Here we store the 2^i th parent for every node, where 0 <= i < LEVEL, here “LEVEL” is a constant integer that tells the maximum number of 2^i th ancestor possible. 
Therefore, we assume the worst case to see what is the value of the constant LEVEL. In our worst case every node in our tree will have at max 1 parent and 1 child or we can say it simply reduces to a linked list. 
So, in this case LEVEL = ceil ( log(number of nodes) ). 
We also pre-compute the height for each node using one dfs in O(n) time. 
int n             // number of nodes
int parent[MAXN][LEVEL] // all initialized to -1 

parent[node][0] : contains the 2^0th(first) 
parent of all the nodes pre-computed using DFS

// Sparse matrix Approach
for node -> 1 to n :        
  for i-> 1 to LEVEL :
    if ( parent[node][i-1] != -1 ) :
      parent[node][i]  =  
         parent[ parent[node][i-1] ][i-1]
Now , as we see the above dynamic programming code runs two nested loop that runs over their complete range respectively. Hence, it can be easily be inferred that its asymptotic Time Complexity is O(number of nodes * LEVEL) ~ O(n*LEVEL) ~ O(nlogn).
Return LCA(u,v): 
First Step is to bring both the nodes at the same height. As we have already pre-computed the heights for each node. We first calculate the difference in the heights of u and v (let’s say v >=u). Now we need the node ‘v’ to jump h nodes above. This can be easily done in O(log h) time ( where h is the difference in the heights of u and v) as we have already stored the 2^i parent for each node. This process is exactly same as calculating x^y in O(log y) time. (See the code for better understanding).Now both u and v nodes are at same height. Therefore now once again we will use 2^i jumping strategy to reach the first Common Parent of u and v. 
First Step is to bring both the nodes at the same height. As we have already pre-computed the heights for each node. We first calculate the difference in the heights of u and v (let’s say v >=u). Now we need the node ‘v’ to jump h nodes above. This can be easily done in O(log h) time ( where h is the difference in the heights of u and v) as we have already stored the 2^i parent for each node. This process is exactly same as calculating x^y in O(log y) time. (See the code for better understanding).
Now both u and v nodes are at same height. Therefore now once again we will use 2^i jumping strategy to reach the first Common Parent of u and v. 
Pseudo-code:
For i->  LEVEL to 0 :
      If parent[u][i] != parent[v][i] :
           u = parent[u][i]
           v = parent[v][i]
Implementation of the above algorithm is given below: 
C++
Java
C#
Javascript
// Sparse Matrix DP approach to find LCA of two nodes#include <bits/stdc++.h>using namespace std;#define MAXN 100000#define level 18 vector <int> tree[MAXN];int depth[MAXN];int parent[MAXN][level]; // pre-compute the depth for each node and their// first parent(2^0th parent)// time complexity : O(n)void dfs(int cur, int prev){    depth[cur] = depth[prev] + 1;    parent[cur][0] = prev;    for (int i=0; i<tree[cur].size(); i++)    {        if (tree[cur][i] != prev)            dfs(tree[cur][i], cur);    }} // Dynamic Programming Sparse Matrix Approach// populating 2^i parent for each node// Time complexity : O(nlogn)void precomputeSparseMatrix(int n){    for (int i=1; i<level; i++)    {        for (int node = 1; node <= n; node++)        {            if (parent[node][i-1] != -1)                parent[node][i] =                    parent[parent[node][i-1]][i-1];        }    }} // Returning the LCA of u and v// Time complexity : O(log n)int lca(int u, int v){    if (depth[v] < depth[u])        swap(u, v);     int diff = depth[v] - depth[u];     // Step 1 of the pseudocode    for (int i=0; i<level; i++)        if ((diff>>i)&1)            v = parent[v][i];     // now depth[u] == depth[v]    if (u == v)        return u;     // Step 2 of the pseudocode    for (int i=level-1; i>=0; i--)        if (parent[u][i] != parent[v][i])        {            u = parent[u][i];            v = parent[v][i];        }     return parent[u][0];} void addEdge(int u,int v){    tree[u].push_back(v);    tree[v].push_back(u);} // driver functionint main(){    memset(parent,-1,sizeof(parent));    int n = 8;    addEdge(1,2);    addEdge(1,3);    addEdge(2,4);    addEdge(2,5);    addEdge(2,6);    addEdge(3,7);    addEdge(3,8);    depth[0] = 0;     // running dfs and precalculating depth    // of each node.    dfs(1,0);     // Precomputing the 2^i th ancestor for every node    precomputeSparseMatrix(n);     // calling the LCA function    cout << "LCA(4, 7) = " << lca(4,7) << endl;    cout << "LCA(4, 6) = " << lca(4,6) << endl;    return 0;}
// Sparse Matrix DP approach to find LCA of two nodesimport java.util.*; class GFG{    static final int MAXN = 100000;    static final int level = 18;         @SuppressWarnings("unchecked")    static Vector<Integer>[] tree = new Vector[MAXN];    static int[] depth = new int[MAXN];    static int[][] parent = new int[MAXN][level];     // pre-compute the depth for each node and their    // first parent(2^0th parent)    // time complexity : O(n)    static void dfs(int cur, int prev)    {        depth[cur] = depth[prev] + 1;        parent[cur][0] = prev;        for (int i = 0; i < tree[cur].size(); i++)        {            if (tree[cur].get(i) != prev)                dfs(tree[cur].get(i), cur);        }    }     // Dynamic Programming Sparse Matrix Approach    // populating 2^i parent for each node    // Time complexity : O(nlogn)    static void precomputeSparseMatrix(int n)    {        for (int i = 1; i < level; i++)        {            for (int node = 1; node <= n; node++)            {                if (parent[node][i - 1] != -1)                    parent[node][i] = parent[parent[node][i - 1]][i - 1];            }        }    }     // Returning the LCA of u and v    // Time complexity : O(log n)    static int lca(int u, int v)    {        if (depth[v] < depth[u])        {            u = u + v;            v = u - v;            u = u - v;        }         int diff = depth[v] - depth[u];         // Step 1 of the pseudocode        for (int i = 0; i < level; i++)            if (((diff >> i) & 1) == 1)                v = parent[v][i];         // now depth[u] == depth[v]        if (u == v)            return u;         // Step 2 of the pseudocode        for (int i = level - 1; i >= 0; i--)            if (parent[u][i] != parent[v][i])            {                u = parent[u][i];                v = parent[v][i];            }         return parent[u][0];    }     static void addEdge(int u, int v)    {        tree[u].add(v);        tree[v].add(u);    }     static void memset(int value)    {        for (int i = 0; i < MAXN; i++)        {            for (int j = 0; j < level; j++)            {                parent[i][j] = -1;            }        }    }     // driver function    public static void main(String[] args)    {        memset(-1);        for (int i = 0; i < MAXN; i++)            tree[i] = new Vector<Integer>();        int n = 8;        addEdge(1, 2);        addEdge(1, 3);        addEdge(2, 4);        addEdge(2, 5);        addEdge(2, 6);        addEdge(3, 7);        addEdge(3, 8);        depth[0] = 0;         // running dfs and precalculating depth        // of each node.        dfs(1, 0);         // Precomputing the 2^i th ancestor for every node        precomputeSparseMatrix(n);         // calling the LCA function        System.out.print("LCA(4, 7) = " + lca(4, 7) + "\n");        System.out.print("LCA(4, 6) = " + lca(4, 6) + "\n");    }} // This code is contributed by 29AjayKumar
// Sparse Matrix DP approach to find LCA of two nodesusing System;using System.Collections.Generic; class GFG{    static readonly int MAXN = 100000;    static readonly int level = 18;     static List<int>[] tree = new List<int>[MAXN];    static int[] depth = new int[MAXN];    static int[,] parent = new int[MAXN, level];     // pre-compute the depth for each node and their    // first parent(2^0th parent)    // time complexity : O(n)    static void dfs(int cur, int prev)    {        depth[cur] = depth[prev] + 1;        parent[cur,0] = prev;        for (int i = 0; i < tree[cur].Count; i++)        {            if (tree[cur][i] != prev)                dfs(tree[cur][i], cur);        }    }     // Dynamic Programming Sparse Matrix Approach    // populating 2^i parent for each node    // Time complexity : O(nlogn)    static void precomputeSparseMatrix(int n)    {        for (int i = 1; i < level; i++)        {            for (int node = 1; node <= n; node++)            {                if (parent[node, i - 1] != -1)                    parent[node, i] = parent[parent[node, i - 1], i - 1];            }        }    }     // Returning the LCA of u and v    // Time complexity : O(log n)    static int lca(int u, int v)    {        if (depth[v] < depth[u])        {            u = u + v;            v = u - v;            u = u - v;        }         int diff = depth[v] - depth[u];         // Step 1 of the pseudocode        for (int i = 0; i < level; i++)            if (((diff >> i) & 1) == 1)                v = parent[v, i];         // now depth[u] == depth[v]        if (u == v)            return u;         // Step 2 of the pseudocode        for (int i = level - 1; i >= 0; i--)            if (parent[u, i] != parent[v, i])            {                u = parent[u, i];                v = parent[v, i];            }         return parent[u, 0];    }     static void addEdge(int u, int v)    {        tree[u].Add(v);        tree[v].Add(u);    }     static void memset(int value)    {        for (int i = 0; i < MAXN; i++)        {            for (int j = 0; j < level; j++)            {                parent[i, j] = -1;            }        }    }     // Driver function    public static void Main(String[] args)    {        memset(-1);        for (int i = 0; i < MAXN; i++)            tree[i] = new List<int>();        int n = 8;        addEdge(1, 2);        addEdge(1, 3);        addEdge(2, 4);        addEdge(2, 5);        addEdge(2, 6);        addEdge(3, 7);        addEdge(3, 8);        depth[0] = 0;         // running dfs and precalculating depth        // of each node.        dfs(1, 0);         // Precomputing the 2^i th ancestor for every node        precomputeSparseMatrix(n);         // calling the LCA function        Console.Write("LCA(4, 7) = " + lca(4, 7) + "\n");        Console.Write("LCA(4, 6) = " + lca(4, 6) + "\n");    }} // This code is contributed by PrinciRaj1992
<script> // Sparse Matrix DP approach to find LCA of two nodesvar MAXN = 100000;var level = 18;var tree = Array.from(Array(MAXN), ()=>Array());var depth = Array(MAXN).fill(0);var parent = Array.from(Array(MAXN), ()=>Array(level).fill(-1));// pre-compute the depth for each node and their// first parent(2^0th parent)// time complexity : O(n)function dfs(cur, prev){    depth[cur] = depth[prev] + 1;    parent[cur][0] = prev;    for (var i = 0; i < tree[cur].length; i++)    {        if (tree[cur][i] != prev)            dfs(tree[cur][i], cur);    }}// Dynamic Programming Sparse Matrix Approach// populating 2^i parent for each node// Time complexity : O(nlogn)function precomputeSparseMatrix(n){    for (var i = 1; i < level; i++)    {        for(var node = 1; node <= n; node++)        {            if (parent[node][i - 1] != -1)                parent[node][i] = parent[parent[node][i - 1]][i - 1];        }    }}// Returning the LCA of u and v// Time complexity : O(log n)function lca(u, v){    if (depth[v] < depth[u])    {        u = u + v;        v = u - v;        u = u - v;    }    var diff = depth[v] - depth[u];    // Step 1 of the pseudocode    for (var i = 0; i < level; i++)        if (((diff >> i) & 1) == 1)            v = parent[v][i];    // now depth[u] == depth[v]    if (u == v)        return u;    // Step 2 of the pseudocode    for (var i = level - 1; i >= 0; i--)        if (parent[u][i] != parent[v][i])        {            u = parent[u][i];            v = parent[v][i];        }    return parent[u][0];}function addEdge(u, v){    tree[u].push(v);    tree[v].push(u);}function memset(value){    for (var i = 0; i < MAXN; i++)    {        for (var j = 0; j < level; j++)        {            parent[i][j] = -1;        }    }}// Driver functionmemset(-1);var n = 8;addEdge(1, 2);addEdge(1, 3);addEdge(2, 4);addEdge(2, 5);addEdge(2, 6);addEdge(3, 7);addEdge(3, 8);depth[0] = 0;// running dfs and precalculating depth// of each node.dfs(1, 0);// Precomputing the 2^i th ancestor for every nodeprecomputeSparseMatrix(n);// calling the LCA functiondocument.write("LCA(4, 7) = " + lca(4, 7) + "<br>");document.write("LCA(4, 6) = " + lca(4, 6) + "<br>");  </script>
LCA(4, 7) = 1
LCA(4, 6) = 2

Time Complexity: 
The time complexity for answering a single LCA query will be O(logn) but the overall time complexity is dominated by precalculation of the 2^i th ( 0<=i<=level ) ancestors for each node. Hence, the overall asymptotic Time Complexity will be O(n*logn) and Space Complexity will be O(nlogn), for storing the data about the ancestors of each node.
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