Find LCA for K queries in Complete Binary Tree

Given an integer n. There is a complete binary tree with 2ⁿ – 1 nodes. The root of that tree is the node with the value 1, and every node with a value x has two children where the left node has the value 
2*x and the right node has the value 2*x + 1, you are given K queries of type (a_i, b_i), and the task is to return the LCA for the node pair a_i and b_i for all K queries.

Examples:

Input:  n = 5, queries = [ { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } ]

Complete binary tree for given input n=5

Output:  [ 2, 5, 7, 1, 1, 1, 2, 1 ]

Input:  n = 3, queries = [ {2, 5}, {3, 6}, {4, 1}, {7, 3} ]

Complete binary tree for given input n=3

Output: [2, 3, 1, 3]

Approach: The problem can be solved based on the following idea:

As all values on a level are smaller than values on the next level. Check which node is having greater value in a query, and divide it by 2 to reach its parent node. Repeat this step until we get common element. 

Follow the steps to solve the problem:

• In a query, we are having 2 nodes a and b, whose lowest common ancestor we have to find.
• By dividing the value of the node by 2, we will always get the parent node value.
• From a and b whichever node is having greater value divide by 2. So, as to move towards the root of the root.
• When a and b becomes equal, the common ancestor between them is got and returned.  

Below is the implementation for the approach discussed:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to find lca for
// given two nodes in tree
int helper(int a, int b)
{

    while (a != b) {

        if (a > b)
            a = a / 2;

        else
            b = b / 2;
    }

    return a;
}

// Driver code
int main()
{

    // 2^n - 1 nodes in complete
    // binary tree
    int n = 5;

    // Queries input vector
    vector<vector<int> > queries
        = { { 17, 21 }, { 23, 5 }, { 15, 7 }, { 3, 21 }, { 31, 9 }, { 5, 15 }, { 11, 2 }, { 19, 7 } };

    // Processing each query in
    // queries vector
    for (auto e : queries) {

        // Function call
        int lca = helper(e[0], e[1]);

        cout << lca << ' ';
    }

    return 0;
}

Java

// Java code for the above approach
import java.io.*;

class GFG {
    // Function to find lca for
    // given two nodes in tree
    public static int helper(int a, int b)
    {

        while (a != b) {

            if (a > b)
                a = a / 2;

            else
                b = b / 2;
        }

        return a;
    }

    // Driver Code
    public static void main(String[] args)
    {
        // 2^n - 1 nodes in complete
        // binary tree
        int n = 5;

        // Queries input vector
        int queries[][] = { { 17, 21 }, { 23, 5 },
                            { 15, 7 },  { 3, 21 },
                            { 31, 9 },  { 5, 15 },
                            { 11, 2 },  { 19, 7 } };

        // Processing each query in
        // queries vector
        for (int e[] : queries) {

            // Function call
            int lca = helper(e[0], e[1]);

            System.out.print(lca + " ");
        }
    }
}

// This code is contributed by Rohit Pradhan

C#

using System;
using System.Collections.Generic;

class Program
{
    // Function to find lca for
    // given two nodes in tree
    static int helper(int a, int b)
    {

        while (a != b)
        {

            if (a > b)
                a = a / 2;

            else
                b = b / 2;
        }

        return a;
    }

    // Driver code
    static void Main(string[] args)
    {

        // 2^n - 1 nodes in complete
        // binary tree
        int n = 5;

        // Queries input vector
        List<List<int>> queries = new List<List<int>> {
            new List<int> { 17, 21 },
            new List<int> { 23, 5 },
            new List<int> { 15, 7 },
            new List<int> { 3, 21 },
            new List<int> { 31, 9 },
            new List<int> { 5, 15 },
            new List<int> { 11, 2 },
            new List<int> { 19, 7 }
        };

        // Processing each query in
        // queries vector
        foreach (var e in queries)
        {

            // Function call
            int lca = helper(e[0], e[1]);

            Console.Write(lca + " ");
        }

    }
}
// code by ksam24000

Javascript

// Function to find lca for
// given two nodes in tree
function helper(a, b) {
    while (a != b) {
        if (a > b)
            a = Math.floor(a / 2);
        else
            b = Math.floor(b / 2);
    }
    return a;
}

// Driver code
console.log("LCA(s):");

// 2^n - 1 nodes in complete
// binary tree
let n = 5;

// Queries input array
let queries = [ [ 17, 21 ], [ 23, 5 ], [ 15, 7 ], [ 3, 21 ], [ 31, 9 ], [ 5, 15 ], [ 11, 2 ], [ 19, 7 ] ];

// Processing each query in
// queries array
for (let i = 0; i < queries.length; i++) {
    let e = queries[i];

    // Function call
    let lca = helper(e[0], e[1]);

    console.log(lca + ' ');
}

Python3

# Function to find lca for
# given two nodes in tree
def helper(a, b):
    while (a != b):
        if (a > b):
            a = a // 2
        else:
            b = b // 2
    return a

# Driver code
print("LCA(s):")

# 2^n - 1 nodes in complete
# binary tree
n = 5

# Queries input list
queries = [ [ 17, 21 ], [ 23, 5 ], [ 15, 7 ], [ 3, 21 ], [ 31, 9 ], [ 5, 15 ], [ 11, 2 ], [ 19, 7 ] ]

# Processing each query in
# queries list
for e in queries:
    # Function call
    lca = helper(e[0], e[1])

    print(lca, end=' ')

Output

2 5 7 1 1 1 2 1 

Time Complexity: O(log₂(max(a,b)), here we divide number a or b every time with 2 so it will cost log₂() complexity and here we are doing it for a and b so a max of(a,b) will be the number which takes worst time so overall time complexity will be O(log₂(max(a,b))) to find LCA of two node a&b.
Auxiliary Space: O(1)