Find the maximum GCD of the siblings of a Binary Tree

Given a 2d-array arr[][] which represents the nodes of a Binary tree, the task is to find the maximum GCD of the siblings of this tree without actually constructing it.

Example:

Input: arr[][] = {{4, 5}, {4, 2}, {2, 3}, {2, 1}, {3, 6}, {3, 12}} 
Output: 6 
Explanation:

For the above tree, the maximum GCD for the siblings is formed for the nodes 6 and 12 for the children of node 3.

Input: arr[][] = {{5, 4}, {5, 8}, {4, 6}, {4, 9}, {8, 10}, {10, 20}, {10, 30}} 
Output: 10 

Approach: The idea is to form a vector and store the tree in the form of the vector. After storing the tree in the form of a vector, the following cases occur: 

• If the vector size is 0 or 1, then print 0 as GCD could not be found.
• For all other cases, since we store the tree in the form of a pair, we consider the first values of two pairs and compare them. But first, you need to Sort the pair of edges.
  For example, let’s assume there are two pairs in the vector A and B. We check if:

A.first == B.first

• If both of them match, then both of them belongs to the same parent. Therefore, we compute the GCD of the second values in the pairs and finally print the maximum of all such GCD’s.

Below is the implementation of the above approach:  

C++

// C++ program to find the maximum
// GCD of the siblings of a binary tree

#include <bits/stdc++.h>
using namespace std;

// Function to find maximum GCD
int max_gcd(vector<pair<int, int> >& v)
{
    // No child or Single child
    if (v.size() == 1 || v.size() == 0)
        return 0;

    sort(v.begin(), v.end());

    // To get the first pair
    pair<int, int> a = v[0];
    pair<int, int> b;
    int ans = INT_MIN;
    for (int i = 1; i < v.size(); i++) {
        b = v[i];

        // If both the pairs belongs to
        // the same parent
        if (b.first == a.first)

            // Update ans with the max
            // of current gcd and
            // gcd of both children
            ans
                = max(ans,
                      __gcd(a.second,
                            b.second));

        // Update previous
        // for next iteration
        a = b;
    }
    return ans;
}

// Driver function
int main()
{
    vector<pair<int, int> > v;
    v.push_back(make_pair(5, 4));
    v.push_back(make_pair(5, 8));
    v.push_back(make_pair(4, 6));
    v.push_back(make_pair(4, 9));
    v.push_back(make_pair(8, 10));
    v.push_back(make_pair(10, 20));
    v.push_back(make_pair(10, 30));

    cout << max_gcd(v);
    return 0;
}

Java

// Java program to find the maximum
// GCD of the siblings of a binary tree
import java.util.*;
import java.lang.*;

class GFG{

// Function to find maximum GCD
static int max_gcd(ArrayList<int[]> v)
{

    // No child or Single child
    if (v.size() == 1 || v.size() == 0)
        return 0;

      Collections.sort(v, new Comparator<int[]>() {
        public int compare(int[] a, int[] b) {
            return a[0]-b[0];
        }
    });

    // To get the first pair
    int[] a = v.get(0);
    int[] b = new int[2];
    int ans = Integer.MIN_VALUE;

    for(int i = 1; i < v.size(); i++)
    {
        b = v.get(i);

        // If both the pairs belongs to
        // the same parent
        if (b[0] == a[0])

            // Update ans with the max
            // of current gcd and
            // gcd of both children
            ans = Math.max(ans,
                           gcd(a[1],
                               b[1]));

        // Update previous
        // for next iteration
        a = b;
    }
    return ans;
}

static int gcd(int a, int b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

// Driver code
public static void main(String[] args)
{
    ArrayList<int[]> v = new ArrayList<>();

    v.add(new int[]{5, 4});
    v.add(new int[]{5, 8});
    v.add(new int[]{4, 6});
    v.add(new int[]{4, 9});
    v.add(new int[]{8, 10});
    v.add(new int[]{10, 20});
    v.add(new int[]{10, 30});

    System.out.println(max_gcd(v));
}
}

// This code is contributed by offbeat

Python3

# Python3 program to find the maximum
# GCD of the siblings of a binary tree
from math import gcd

# Function to find maximum GCD
def max_gcd(v):

    # No child or Single child
    if (len(v) == 1 or len(v) == 0):
        return 0

    v.sort()

    # To get the first pair
    a = v[0]
    ans = -10**9
    for i in range(1, len(v)):
        b = v[i]

        # If both the pairs belongs to
        # the same parent
        if (b[0] == a[0]):

            # Update ans with the max
            # of current gcd and
            # gcd of both children
            ans = max(ans, gcd(a[1], b[1]))

        # Update previous
        # for next iteration
        a = b
    return ans

# Driver function
if __name__ == '__main__':
    v=[]
    v.append([5, 4])
    v.append([5, 8])
    v.append([4, 6])
    v.append([4, 9])
    v.append([8, 10])
    v.append([10, 20])
    v.append([10, 30])

    print(max_gcd(v))

# This code is contributed by mohit kumar 29

C#

// C# program to find the maximum
// GCD of the siblings of a binary tree
using System.Collections;
using System;

class GFG{

// Function to find maximum GCD
static int max_gcd(ArrayList v)
{

    // No child or Single child
    if (v.Count == 1 || v.Count == 0)
        return 0;

    v.Sort();

    // To get the first pair
    int[] a = (int [])v[0];
    int[] b = new int[2];
    int ans = -10000000;

    for(int i = 1; i < v.Count; i++)
    {
      b = (int[])v[i];

      // If both the pairs belongs to
      // the same parent
      if (b[0] == a[0])

        // Update ans with the max
        // of current gcd and
        // gcd of both children
        ans = Math.Max(ans, gcd(a[1], b[1]));

      // Update previous
      // for next iteration
      a = b;
    }
    return ans;
}

static int gcd(int a, int b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}

// Driver code
public static void Main(string[] args)
{
    ArrayList v = new ArrayList();

    v.Add(new int[]{5, 4});
    v.Add(new int[]{5, 8});
    v.Add(new int[]{4, 6});
    v.Add(new int[]{4, 9});
    v.Add(new int[]{8, 10});
    v.Add(new int[]{10, 20});
    v.Add(new int[]{10, 30});

    Console.Write(max_gcd(v));
}
}

// This code is contributed by rutvik_56

Javascript

<script>

// JavaScript program to find the maximum
// GCD of the siblings of a binary tree

// Function to find maximum GCD
function max_gcd(v)
{
    // No child or Single child
    if (v.length == 1 || v.length == 0)
        return 0;

    v.sort((a, b) => a - b);

    // To get the first pair
    let a = v[0];
    let b;
    let ans = Number.MIN_SAFE_INTEGER;
    for (let i = 1; i < v.length; i++) {
        b = v[i];

        // If both the pairs belongs to
        // the same parent
        if (b[0] == a[0])

            // Update ans with the max
            // of current gcd and
            // gcd of both children
            ans
                = Math.max(ans, gcd(a[1], b[1]));

        // Update previous
        // for next iteration
        a = b;
    }
    return ans;
}

function gcd(a, b)
{
    if (b == 0)
        return a;

    return gcd(b, a % b);
}


// Driver function

    let v = new Array();
    v.push([5, 4]);
    v.push([5, 8]);
    v.push([4, 6]);
    v.push([4, 9]);
    v.push([8, 10]);
    v.push([10, 20]);
    v.push([10, 30]);

    document.write(max_gcd(v));

    // This code is contributed by gfgking

</script>

Output

10

Time Complexity: O(nlogn), as sort() takes O(nlogn) time.
Auxiliary Space: O(1)

Another Approach:

The approach is to traverse the binary tree in a postorder manner and for each node, compute the greatest common divisor (GCD) of the values of its left and right children. If both children exist and their GCD is greater than the current maximum GCD, update the maximum GCD. Then return the GCD of the current node value and the maximum of its children’s GCDs.

Here is an implementation of above approach in c++:

C++

// C++ Implementation
#include <bits/stdc++.h>
using namespace std;

// Structure of Tree Node
struct TreeNode {
    int val;
    TreeNode* left;
    TreeNode* right;
    TreeNode(int x)
        : val(x)
        , left(NULL)
        , right(NULL)
    {
    }
};

// Find out gcd
int gcd(int a, int b)
{
    if (b == 0)
        return a;
    return gcd(b, a % b);
}

int max_gcd_helper(TreeNode* root, int& ans)
{
    if (!root)
        return 0;

    int left_gcd = max_gcd_helper(root->left, ans);
    int right_gcd = max_gcd_helper(root->right, ans);

    if (left_gcd != 0 && right_gcd != 0) {
        int siblings_gcd = gcd(left_gcd, right_gcd);
        ans = max(ans, siblings_gcd);
    }

    return (root->left) ? gcd(root->val, left_gcd)
                        : root->val;
}

int max_gcd(TreeNode* root)
{
    int ans = 0;
    max_gcd_helper(root, ans);
    return ans;
}

// Driver code
int main()
{
    TreeNode* root = new TreeNode(10);
    root->left = new TreeNode(5);
    root->right = new TreeNode(15);
    root->left->left = new TreeNode(4);
    root->left->right = new TreeNode(8);
    root->right->left = new TreeNode(10);
    root->right->right = new TreeNode(20);

    // Function call
    cout << max_gcd(root); // Output: 10
    return 0;
}

Output

10

Time Complexity: O(nlogn)
Auxiliary Space: O(1)