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Queries to calculate maximum Bitwise XOR of X with any array element not exceeding M
  • Difficulty Level : Hard
  • Last Updated : 23 Mar, 2021

Given an array arr[] consisting of N non-negative integers and a 2D array queries[][] consisting of queries of the type {X, M}, the task for each query is to find the maximum Bitwise XOR of X with any array element whose value is at most M. If it is not possible to find the Bitwise XOR, then print “-1”.

Examples:

Input: arr[] = {0, 1, 2, 3, 4}, queries[][] = {{3, 1}, {1, 3}, {5, 6}}
Output: {3, 3, 7}
Explanation:
Query 1: The query is {3, 1}. Maximum Bitwise XOR = 3 ^ 0 = 3.
Query 2: The query is {1, 3}. Maximum Bitwise XOR = 1 ^ 2 = 3.
Query 3: The query is {5, 6}. Maximum Bitwise XOR = 5 ^ 2 = 7.

Input: arr[] = {5, 2, 4, 6, 6, 3}, queries[][] = {{12, 4}, {8, 1}, {6, 3}}
Output: {15, -1, 5}

Naive Approach: The simplest approach to solve the given problem is to traverse the given array for each query {X, M} and print the maximum value of Bitwise XOR of X with an array element with a value at most M. If there doesn’t exist any value less than M, then print “-1” for the query.
Time Complexity: O(N*Q)
Auxiliary Space: O(1)

Efficient Approach: The above approach can also be optimized by using the Trie data structure to store all the elements having values at most M. Therefore, the problem reduces to finding the maximum XOR of two elements in an array. Follow the steps below to solve the problem:



  • Initialize a variable, say index, to traverse the array.
  • Initialize an array, say ans[], that stores the result for each query.
  • Initialize an auxiliary array, say temp[][3], and store all the queries in it with the index of each query.
  • Sort the given array temp[] on the basis of the second parameter, i.e. temp[1](= M).
  • Sort the given array arr[] in ascending order.
  • Traverse the array temp[] and for each query {X, M, idx}, perform the following steps:
    • Iterate until the value of index is less than N and arr[index] is at most M or not. If found to be true, then insert that node as the binary representation of N and increment index.
    • After completing the above steps, if the value of index is non-zero, then find the node with value X in the Trie(say result) and update the maximum value for the current query as result. Otherwise, update the maximum value for the current query as “-1”.
  • After completing the above steps, print the array ans[] as the resultant maximum values for each query.

Below is the implementation of the above approach:

Java




// Java program for the above approach
  
import java.io.*;
import java.util.*;
  
// Trie Node Class
class TrieNode {
    TrieNode nums[] = new TrieNode[2];
    int prefixValue;
}
  
class sol {
  
    // Function to find the maximum XOR
    // of X with any array element <= M
    // for each query of the type {X, M}
    public void maximizeXor(
        int[] nums, int[][] queries)
    {
        int queriesLength = queries.length;
        int[] ans = new int[queriesLength];
        int[][] temp = new int[queriesLength][3];
  
        // Stores the queries
        for (int i = 0; i < queriesLength; i++) {
            temp[i][0] = queries[i][0];
            temp[i][1] = queries[i][1];
            temp[i][2] = i;
        }
  
        // Sort the query
        Arrays.sort(temp,
                    (a, b) -> {
                        return a[1]
                            - b[1];
                    });
        int index = 0;
  
        // Sort the array
        Arrays.sort(nums);
        TrieNode root = new TrieNode();
  
        // Traverse the given query
        for (int query[] : temp) {
  
            // Traverse the array nums[]
            while (index < nums.length
                   && nums[index]
                          <= query[1]) {
  
                // Insert the node into the Trie
                insert(root, nums[index]);
                index++;
            }
  
            // Stores the resultant
            // maximum value
            int tempAns = -1;
  
            // Find the maximum value
            if (index != 0) {
  
                // Search the node in the Trie
                tempAns = search(root,
                                 query[0]);
            }
  
            // Update the result
            // for each query
            ans[query[2]] = tempAns;
        }
  
        // Print the answer
        for (int num : ans) {
            System.out.print(num + " ");
        }
    }
  
    // Function to insert the
    // root in the trieNode
    public void insert(TrieNode root,
                       int n)
    {
        TrieNode node = root;
  
        // Iterate from 31 to 0
        for (int i = 31; i >= 0; i--) {
  
            // Find the bit at i-th position
            int bit = (n >> i) & 1;
            if (node.nums[bit] == null) {
                node.nums[bit]
                    = new TrieNode();
            }
            node = node.nums[bit];
        }
  
        // Update the value
        node.prefixValue = n;
    }
  
    // Function to search the root
    // with the value and perform
    // the Bitwise XOR with N
    public int search(TrieNode root,
                      int n)
    {
        TrieNode node = root;
  
        // Iterate from 31 to 0
        for (int i = 31; i >= 0; i--) {
  
            // Find the bit at ith
            // position
            int bit = (n >> i) & 1;
            int requiredBit = bit
                                      == 1
                                  ? 0
                                  : 1;
  
            if (node.nums[requiredBit]
                != null) {
                node = node.nums[requiredBit];
            }
            else {
                node = node.nums[bit];
            }
        }
  
        // Return the prefixvalue XORed
        // with N
        return node.prefixValue ^ n;
    }
}
  
class GFG {
  
    // Driver Code
    public static void main(String[] args)
    {
        sol tt = new sol();
        int[] nums = { 0, 1, 2, 3, 4 };
        int[][] queries = { { 3, 1 },
                            { 1, 3 },
                            { 5, 6 } };
  
        tt.maximizeXor(nums, queries);
    }
}
Output:
3 3 7

Time Complexity: O(N*log N + K*log K)
Auxiliary Space: O(N)

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