Queries to calculate maximum Bitwise XOR of X with any array element not exceeding M
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 Classclass 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); }} |
3 3 7
Time Complexity: O(N*log N + K*log K)
Auxiliary Space: O(N)
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