Maximum XOR with given value in the path from root to given node in the tree

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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 node
class 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 i
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
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 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 Code
int 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 approach
public 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 node
class 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 i
def 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 x
def 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 Code
n = 6
 
root = 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 = 6
x = 4
print( 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 node
class 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 i
function 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 x
function 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 code
let 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:

Time Complexity: O(N)

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