Given a tree with N distinct nodes from the range [1, n] and two integers x and val. The task is to find the maximum value of any node when XORed with x on the path from the root to val.
Examples:
Input: val = 6, x = 4
1
/ \
2 3
/ \
4 5
/
6
Output: 7
the path is 1 -> 3 -> 5 -> 6
1 ^ 4 = 5
3 ^ 4 = 7
5 ^ 4 = 1
6 ^ 4 = 2
Maximum is 7
Input: val = 4, x = 1
1
/ \
2 3
/
4
Output: 5
Approach:
- An optimized solution to the problem is to create a parent array to store the parent of each of the node.
- Start from the given node and keep on going up in the tree using the parent array (this will be helpful when answering a number of queries as only the nodes on the path will be traversed). Take the xor with x of all the nodes in the path till root.
- The maximum xor calculated for the path is the answer.
Below is the implementation of the above approach:
C++14
// CPP implementation of the approach#include <bits/stdc++.h>using namespace std;// Tree nodeclass Node{public: int data; Node *left, *right; Node(int data) { this->data = data; this->left = NULL; this->right = NULL; }};// Recursive function to update// the parent array such that parent[i]// stores the parent of ivoid updateParent(int *parent, Node *node){ // If node is null then return if (node == NULL) return; // If left child of the node is not // null then set node as the parent // of its left child if (node->left != NULL) parent[node->left->data] = node->data; // If right child of the node is not // null then set node as the parent // of its right child if (node->right != NULL) parent[node->right->data] = node->data; // Recursive call for the // children of current node updateParent(parent, node->left); updateParent(parent, node->right);}// Function to return the maximum value// of a node on the path from root to val// when Xored with xint getMaxXor(Node *root, int n, int val, int x){ // Create the parent array int *parent = new int[n + 1]; updateParent(parent, root); // Initialize max with x XOR val int maximum = x ^ val; // Get to the parent of val val = parent[val]; // 0 is the parent of the root while (val != 0) { // Update maximum maximum = max(maximum, x ^ val); // Get one level up the tree val = parent[val]; } return maximum;}// Driver Codeint main(){ int n = 6; Node *root; root = new Node(1); root->left = new Node(2); root->right = new Node(3); root->left->left = new Node(4); root->right->right = new Node(5); root->right->right->left = new Node(6); int val = 6, x = 4; cout << getMaxXor(root, n, val, x) << endl; return 0;}// This code is contributed by// sanjeev2552 |
Java
// Java implementation of the approachpublic class GFG { // Tree node static class Node { int data; Node left, right; Node(int data) { this.data = data; left = null; right = null; } } // Recursive function to update // the parent array such that parent[i] // stores the parent of i static void updateParent(int parent[], Node node) { // If node is null then return if (node == null) return; // If left child of the node is not // null then set node as the parent // of its left child if (node.left != null) parent[node.left.data] = node.data; // If right child of the node is not // null then set node as the parent // of its right child if (node.right != null) parent[node.right.data] = node.data; // Recursive call for the // children of current node updateParent(parent, node.left); updateParent(parent, node.right); } // Function to return the maximum value // of a node on the path from root to val // when Xored with x static int getMaxXor(Node root, int n, int val, int x) { // Create the parent array int parent[] = new int[n + 1]; updateParent(parent, root); // Initialize max with x XOR val int max = x ^ val; // Get to the parent of val val = parent[val]; // 0 is the parent of the root while (val != 0) { // Update maximum max = Math.max(max, x ^ val); // Get one level up the tree val = parent[val]; } return max; } // Driver code public static void main(String[] args) { int n = 6; Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.right.right = new Node(5); root.right.right.left = new Node(6); int val = 6, x = 4; System.out.println(getMaxXor(root, n, val, x)); }} |
Python3
# Python3 implementation of the approach# Tree nodeclass Node: def __init__(self, data): self.data = data self.left = None self.right = None # Recursive function to update# the parent array such that parent[i]# stores the parent of idef updateParent(parent, node): # If node is None then return if (node == None): return parent # If left child of the node is not # None then set node as the parent # of its left child if (node.left != None): parent[node.left.data] = node.data # If right child of the node is not # None then set node as the parent # of its right child if (node.right != None): parent[node.right.data] = node.data # Recursive call for the # children of current node parent = updateParent(parent, node.left) parent = updateParent(parent, node.right) return parent# Function to return the maximum value# of a node on the path from root to val# when Xored with xdef getMaxXor(root, n, val, x): # Create the parent array parent = [0]*(n + 1) parent=updateParent(parent, root) # Initialize max with x XOR val maximum = x ^ val # Get to the parent of val val = parent[val] # 0 is the parent of the root while (val != 0): # Update maximum maximum = max(maximum, x ^ val) # Get one level up the tree val = parent[val] return maximum# Driver Coden = 6root = Node(1)root.left = Node(2)root.right = Node(3)root.left.left = Node(4)root.right.right = Node(5)root.right.right.left = Node(6)val = 6x = 4print( getMaxXor(root, n, val, x) )# This code is contributed by Arnab Kundu |
C#
// C# implementation of the approach using System;class GFG{ // Tree node public class Node { public int data; public Node left, right; public Node(int data) { this.data = data; left = null; right = null; } } // Recursive function to update // the parent array such that parent[i] // stores the parent of i static void updateParent(int []parent, Node node) { // If node is null then return if (node == null) return; // If left child of the node is not // null then set node as the parent // of its left child if (node.left != null) parent[node.left.data] = node.data; // If right child of the node is not // null then set node as the parent // of its right child if (node.right != null) parent[node.right.data] = node.data; // Recursive call for the // children of current node updateParent(parent, node.left); updateParent(parent, node.right); } // Function to return the maximum value // of a node on the path from root to val // when Xored with x static int getMaxXor(Node root, int n, int val, int x) { // Create the parent array int []parent = new int[n + 1]; updateParent(parent, root); // Initialize max with x XOR val int max = x ^ val; // Get to the parent of val val = parent[val]; // 0 is the parent of the root while (val != 0) { // Update maximum max = Math.Max(max, x ^ val); // Get one level up the tree val = parent[val]; } return max; } // Driver code public static void Main(String[] args) { int n = 6; Node root = new Node(1); root.left = new Node(2); root.right = new Node(3); root.left.left = new Node(4); root.right.right = new Node(5); root.right.right.left = new Node(6); int val = 6, x = 4; Console.WriteLine(getMaxXor(root, n, val, x)); }}// This code is contributed by Princi Singh |
Javascript
<script>// Javascript implementation of above approach// Tree nodeclass Node{ constructor(data) { this.left = null; this.right = null; this.data = data; }}// Recursive function to update// the parent array such that parent[i]// stores the parent of ifunction updateParent(parent, node){ // If node is null then return if (node == null) return; // If left child of the node is not // null then set node as the parent // of its left child if (node.left != null) parent[node.left.data] = node.data; // If right child of the node is not // null then set node as the parent // of its right child if (node.right != null) parent[node.right.data] = node.data; // Recursive call for the // children of current node updateParent(parent, node.left); updateParent(parent, node.right);}// Function to return the maximum value// of a node on the path from root to val// when Xored with xfunction getMaxXor(root, n, val, x){ // Create the parent array let parent = new Array(n + 1); parent.fill(0); updateParent(parent, root); // Initialize max with x XOR val let max = x ^ val; // Get to the parent of val val = parent[val]; // 0 is the parent of the root while (val != 0) { // Update maximum max = Math.max(max, x ^ val); // Get one level up the tree val = parent[val]; } return max;}// Driver codelet n = 6;let root = new Node(1);root.left = new Node(2);root.right = new Node(3);root.left.left = new Node(4);root.right.right = new Node(5);root.right.right.left = new Node(6);let val = 6, x = 4;document.write(getMaxXor(root, n, val, x));// This code is contributed by divyeshrabadiya07</script> |
Output:
7
Time Complexity: O(N)
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