L7. Maximum XOR With an Element From Array | Queries | C++ | Java

Master the offline querying technique combined with Bitwise Tries to solve the Maximum XOR with an Element from Array problem efficiently.

1. Problem Statement & Formal Definition

We are given an array of integers named arr and a 2D array of queries named queries. Each query contains two elements: xi and mi (represented as [xi, mi]):

• xi: The number we want to perform a bitwise XOR operation with.
• mi: The upper bound threshold constraint.

The Objective: For each query i, find an element arr[j] from the array such that arr[j] <= mi, and the bitwise XOR value (arr[j] ^ xi) is maximized. If no such element exists in the array that satisfies arr[j] <= mi, the answer for that query is -1.

2. Naive Brute-Force Approach & Its Limitations

To understand why advanced data structures are needed, let's examine the straightforward brute-force strategy.

Algorithm

1. Loop through each query from 0 to Q - 1.
2. For a given query [xi, mi], iterate through every element in the given array arr.
3. Filter or check elements where arr[j] <= mi.
4. Among those valid elements, compute the bitwise XOR with xi and track the maximum value found.
5. If no valid elements meet the condition arr[j] <= mi, return -1.

Time Complexity Analysis

• Let $N$ be the size of the array arr.
• Let $Q$ be the number of queries.
• For every query, we might scan up to $N$ elements.
• Total Time Complexity: $\mathcal{O}(Q \times N)$, which results in Time Limit Exceeded (TLE) when $N$ and $Q$ are large (e.g., up to $10^5$).

3. Core Mental Model: Building Up to the Trie Solution

In a previous lesson, we learned how to find the maximum XOR of a given number x against an entire static array using a Bitwise Trie.

The New Constraint Dilemma

However, in this problem, we have an extra condition: arr[j] <= mi.

• We cannot just dump all array elements into the Trie upfront, because some elements might violate arr[j] <= mi.
• If we rebuild the Trie from scratch for every single query by filtering elements $\le mi$, the time complexity becomes disastrous: $\mathcal{O}(Q \times N \times 32)$.

We need a way to process queries dynamically or in order so we don't rebuild the Trie repeatedly.

4. The "Offline Queries" Strategy

To avoid rebuilding the Trie, we can use an algorithmic technique called Offline Queries.

Instead of processing queries in their original input order, we reorder them strategically.

Step-by-Step Offline Logic

1. Sort the input array arr in ascending order.
2. Sort the queries based on their threshold limit mi in ascending order.
3. Maintain a pointer (idx) into the sorted array arr.
4. As we iterate through our sorted queries one by one:
   • Take the current query's threshold mi.
   • Incrementally insert all elements from arr into our Bitwise Trie as long as arr[idx] <= mi.
   • Once all valid elements ($\le mi$) are inserted into the Trie, query the Trie for the maximum XOR with xi.

5. Step-by-Step Flow Walkthrough with an Example

Let's trace how the offline query approach handles sorted queries and incremental Trie insertion.

Example Setup

• Array (arr): [0, 1, 2, 3, 4] (Sorted)
• Queries ([xi, mi]):
  • Query 0: [3, 1] (xi = 3, mi = 1)
  • Query 1: [1, 5] (xi = 1, mi = 5)
  • Query 2: [3, 3] (xi = 3, mi = 3)

Execution Trace

Sort Array: [0, 1, 2, 3, 4]
Sort Queries by mi: 
  - Q_sorted[0] = [3, 1] (Original Index: 0)
  - Q_sorted[1] = [3, 3] (Original Index: 2)
  - Q_sorted[2] = [1, 5] (Original Index: 1)

1. Process Query [3, 1] (mi = 1):

   • Advance array pointer (idx) while arr[idx] <= 1.
   • Elements inserted into Trie: 0, 1.
   • Query Trie for max XOR with xi = 3.
   • Result stored for Query 0.

2. Process Query [3, 3] (mi = 3):

   • Advance array pointer (idx) while arr[idx] <= 3 (current idx moves forward to include 2 and 3).
   • Elements inserted into Trie: 0, 1, 2, 3.
   • Query Trie for max XOR with xi = 3.
   • Result stored for Query 1.

3. Process Query [1, 5] (mi = 5):

   • Advance array pointer (idx) while arr[idx] <= 5 (includes 4).
   • Elements inserted into Trie: 0, 1, 2, 3, 4.
   • Query Trie for max XOR with xi = 1.
   • Result stored for Query 2.

Finally, rearrange the answers back into their original query indices and return.

6. Complete C++ Implementation

Below is the complete C++ solution implementing the Trie structure, offline query sorting, and incremental insertion.

struct Node {
    Node* links[2];
    
    bool containsKey(int bit) {
        return (links[bit] != NULL);
    }
    
    Node* get(int bit) {
        return links[bit];
    }
    
    void put(int bit, Node* node) {
        links[bit] = node;
    }
};

class Trie {
private:
    Node* root;
public:
    Trie() {
        root = new Node();
    }
    
    void insert(int num) {
        Node* node = root;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            if (!node->containsKey(bit)) {
                node->put(bit, new Node());
            }
            node = node->get(bit);
        }
    }
    
    int getMax(int num) {
        Node* node = root;
        int maxNum = 0;
        for (int i = 31; i >= 0; i--) {
            int bit = (num >> i) & 1;
            // We want the opposite bit to maximize XOR
            if (node->containsKey(1 - bit)) {
                maxNum |= (1 << i);
                node = node->get(1 - bit);
            } else {
                node = node->get(bit);
            }
        }
        return maxNum;
    }
};

class Solution {
public:
    vector<int> maximizeXor(vector<int>& arr, vector<vector<int>>& queries) {
        // Sort the main array
        sort(arr.begin(), arr.end());
        
        // Store queries with their original indices: {mi, xi, original_index}
        vector<vector<int>> q;
        int n = queries.size();
        for (int i = 0; i < n; i++) {
            q.push_back({queries[i][1], queries[i][0], i});
        }
        
        // Sort queries based on mi (threshold)
        sort(q.begin(), q.end());
        
        vector<int> ans(n, 0);
        Trie trie;
        int idx = 0;
        int m = arr.size();
        
        for (int i = 0; i < n; i++) {
            int mi = q[i][0];
            int xi = q[i][1];
            int oriIndex = q[i][2];
            
            // Insert all elements <= mi into the Trie
            while (idx < m && arr[idx] <= mi) {
                trie.insert(arr[idx]);
                idx++;
            }
            
            // If idx == 0, no element <= mi was found
            if (idx == 0) {
                ans[oriIndex] = -1;
            } else {
                ans[oriIndex] = trie.getMax(xi);
            }
        }
        
        return ans;
    }
};

7. Complexity Analysis of the Optimized Solution

8. Summary and Key Takeaways

• Problem Extension: This problem extends standard bitwise XOR queries by adding an inequality threshold constraint (arr[j] <= mi).
• Offline Query Technique: Sorting queries by their threshold mi allows us to process them in increasing order of constraints, eliminating redundant work.
• Incremental Trie Building: Instead of rebuilding the Trie per query, we maintain a pointer in the sorted array and incrementally insert elements as their values become valid ($\le mi$).
• Bitwise Greediness: Inside the Trie traversal (getMax), always prioritize moving down the branch with the opposite bit (1 - bit) to maximize the XOR outcome.
• Handling Edge Cases: If the array pointer remains at 0 after checking the threshold condition (meaning no element in arr is less than or equal to mi), immediately return -1.
• Performance Gain: Reduces time complexity from a prohibitive $\mathcal{O}(Q \times N)$ brute-force approach down to an efficient $\mathcal{O}(N \log N + Q \log Q)$ sorting-and-Trie approach.