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++
#include <bits/stdc++.h>
using namespace std;
int max_gcd(vector<pair<int, int> >& v)
{
if (v.size() == 1 || v.size() == 0)
return 0;
sort(v.begin(), v.end());
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 (b.first == a.first)
ans
= max(ans,
__gcd(a.second,
b.second));
a = b;
}
return ans;
}
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
import java.util.*;
import java.lang.*;
class GFG{
static int max_gcd(ArrayList<int[]> v)
{
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];
}
});
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 (b[0] == a[0])
ans = Math.max(ans,
gcd(a[1],
b[1]));
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
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));
}
}
|
Python3
from math import gcd
def max_gcd(v):
if (len(v) == 1 or len(v) == 0):
return 0
v.sort()
a = v[0]
ans = -10**9
for i in range(1, len(v)):
b = v[i]
if (b[0] == a[0]):
ans = max(ans, gcd(a[1], b[1]))
a = b
return ans
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))
|
C#
using System.Collections;
using System;
class GFG{
static int max_gcd(ArrayList v)
{
if (v.Count == 1 || v.Count == 0)
return 0;
v.Sort();
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 (b[0] == a[0])
ans = Math.Max(ans, gcd(a[1], b[1]));
a = b;
}
return ans;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
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));
}
}
|
Javascript
<script>
function max_gcd(v)
{
if (v.length == 1 || v.length == 0)
return 0;
v.sort((a, b) => a - b);
let a = v[0];
let b;
let ans = Number.MIN_SAFE_INTEGER;
for (let i = 1; i < v.length; i++) {
b = v[i];
if (b[0] == a[0])
ans
= Math.max(ans, gcd(a[1], b[1]));
a = b;
}
return ans;
}
function gcd(a, b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
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));
</script>
|
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:
C++
#include <bits/stdc++.h>
using namespace std;
struct TreeNode {
int val;
TreeNode* left;
TreeNode* right;
TreeNode(int x)
: val(x)
, left(NULL)
, right(NULL)
{
}
};
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;
}
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);
cout << max_gcd(root);
return 0;
}
|
Python3
class TreeNode:
def __init__(self, x):
self.val = x
self.left = None
self.right = None
def gcd(a, b):
if b == 0:
return a
return gcd(b, a % b)
def max_gcd_helper(root, ans):
if not root:
return 0
left_gcd = max_gcd_helper(root.left, ans)
right_gcd = max_gcd_helper(root.right, ans)
if left_gcd != 0 and right_gcd != 0:
siblings_gcd = gcd(left_gcd, right_gcd)
ans[0] = max(ans[0], siblings_gcd)
return root.val if not root.left else gcd(root.val, left_gcd)
def max_gcd(root):
ans = [0]
max_gcd_helper(root, ans)
return ans[0]
if __name__ == "__main__":
root = TreeNode(10)
root.left = TreeNode(5)
root.right = TreeNode(15)
root.left.left = TreeNode(4)
root.left.right = TreeNode(8)
root.right.left = TreeNode(10)
root.right.right = TreeNode(20)
print(max_gcd(root))
|
C#
using System;
public class TreeNode
{
public int val;
public TreeNode left;
public TreeNode right;
public TreeNode(int x)
{
val = x;
left = null;
right = null;
}
}
public class Program
{
public static int GCD(int a, int b)
{
if (b == 0)
return a;
return GCD(b, a % b);
}
public static int max_gcd_helper(TreeNode root, ref int ans)
{
if (root == null)
return 0;
int leftGCD = max_gcd_helper(root.left, ref ans);
int rightGCD = max_gcd_helper(root.right, ref ans);
if (leftGCD != 0 && rightGCD != 0)
{
int siblingsGCD = GCD(leftGCD, rightGCD);
ans = Math.Max(ans, siblingsGCD);
}
return (root.left != null) ? GCD(root.val, leftGCD) : root.val;
}
public static int MaxGCD(TreeNode root)
{
int ans = 0;
max_gcd_helper(root, ref ans);
return ans;
}
public static void Main(string[] args)
{
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);
Console.WriteLine(MaxGCD(root));
}
}
|
Javascript
class TreeNode {
constructor(val) {
this.val = val;
this.left = null;
this.right = null;
}
}
function gcd(a, b) {
if (b === 0)
return a;
return gcd(b, a % b);
}
function max_gcd_helper(root, ans) {
if (!root)
return 0;
const left_gcd = max_gcd_helper(root.left, ans);
const right_gcd = max_gcd_helper(root.right, ans);
if (left_gcd !== 0 && right_gcd !== 0) {
const siblings_gcd = gcd(left_gcd, right_gcd);
ans[0] = Math.max(ans[0], siblings_gcd);
}
return (root.left) ? gcd(root.val, left_gcd) : root.val;
}
function max_gcd(root) {
const ans = [0];
max_gcd_helper(root, ans);
return ans[0];
}
function main() {
const 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);
console.log(max_gcd(root));
}
main();
|
Time Complexity: O(nlogn)
Auxiliary Space: O(1)