#1707

Hard

LeetCode Problem Workspace

Maximum XOR With an Element From Array

Solve the Maximum XOR With an Element From Array problem by efficiently finding the maximum XOR for each query using bit manipulation and arrays.

array bit manipulation trie

Problem Statement

You are given an array nums of non-negative integers and a list of queries, where each query is a pair [xi, mi]. For each query, you must compute the maximum XOR value of xi with any element of nums such that the element does not exceed mi. If no element of nums meets this condition, return -1 for that query.

Return an array of results, where each element corresponds to the answer of the respective query. The answer to the ith query is the maximum value of nums[j] XOR xi for all j where nums[j] <= mi. If no such element exists, return -1.

Examples

Example 1

Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]

Output: [3,3,7]

0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
1 XOR 2 = 3.
5 XOR 2 = 7.

Example 2

Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]

Output: [15,-1,5]

Example details omitted.

Constraints

1 <= nums.length, queries.length <= 105
queries[i].length == 2
0 <= nums[j], xi, mi <= 109

Solution Approach

Bitwise XOR Evaluation

The core of the solution is to evaluate the bitwise XOR of xi with elements from nums, ensuring that the element does not exceed mi. Efficient traversal of the nums array is essential to avoid brute force and reduce complexity.

Efficient Search with Trie

Using a Trie structure to store elements of nums allows for fast prefix-based searching, enabling us to efficiently find the maximum XOR result with each query. This method takes advantage of the bitwise nature of the problem and speeds up the lookup process.

Binary Search for Optimization

By sorting the nums array and employing binary search, we can quickly determine which elements are valid candidates for XOR calculation, further optimizing the time complexity when performing multiple queries.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the method used. The Trie-based approach has a time complexity of O(N * 32) for building the Trie, where N is the length of nums. Query handling in the Trie is O(32) due to the bit-length of integers. The binary search approach can reduce the query time to O(log N) after sorting nums. Space complexity also varies: Trie uses O(N * 32) space, while binary search requires only O(N) space after sorting.

What Interviewers Usually Probe

Look for the candidate's ability to leverage data structures like Tries or binary search for optimizing time complexity.
Assess how well the candidate understands the role of bitwise operations in solving algorithmic problems efficiently.
Evaluate the candidate’s approach to handling multiple queries within large input constraints.

Common Pitfalls or Variants

Common pitfalls

Overlooking edge cases where no elements of nums satisfy the condition, leading to returning an incorrect result.
Using brute-force approaches to evaluate all elements for each query without optimization.
Failing to properly handle the constraints, especially when nums and queries are large, which can result in time limit exceeded errors.

Follow-up variants

Find the maximum XOR result with a dynamic range for each query rather than a fixed mi value.
Optimize for space by considering alternative data structures to store nums more efficiently.
Modify the problem to allow for the XOR result to be maximized across multiple queries simultaneously, such as batch processing.

FAQ

What is the main strategy for solving the Maximum XOR With an Element From Array problem?

The primary strategy is to efficiently compute the XOR for each query, using techniques like bitwise operations, Trie, or binary search to optimize the search for valid elements in nums.

How can the Trie help in solving this problem?

A Trie helps by allowing efficient prefix-based searches to quickly find the maximum XOR result for a given query, reducing the need to check every element in nums.

What is the time complexity for solving this problem with a Trie?

The time complexity for building the Trie is O(N * 32), and each query is handled in O(32) time, where N is the length of nums.

What is the space complexity of the Trie-based approach?

The space complexity for the Trie-based approach is O(N * 32), as each number is represented by its 32-bit binary representation in the Trie.

Can I optimize the solution if the array `nums` is already sorted?

Yes, sorting nums allows you to use binary search, which can significantly reduce the time complexity of each query to O(log N), making the solution more efficient.

terminal

Solution

Solution 1: Offline Query + Binary Trie

From the problem description, we know that each query is independent and the result of the query is irrelevant to the order of elements in $nums$. Therefore, we consider sorting all queries in ascending order of $m_i$, and also sorting $nums$ in ascending order.

class Trie:
    __slots__ = ["children"]

    def __init__(self):
        self.children = [None] * 2

    def insert(self, x: int):
        node = self
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v] is None:
                node.children[v] = Trie()
            node = node.children[v]

    def search(self, x: int) -> int:
        node = self
        ans = 0
        for i in range(30, -1, -1):
            v = x >> i & 1
            if node.children[v ^ 1]:
                ans |= 1 << i
                node = node.children[v ^ 1]
            elif node.children[v]:
                node = node.children[v]
            else:
                return -1
        return ans


class Solution:
    def maximizeXor(self, nums: List[int], queries: List[List[int]]) -> List[int]:
        trie = Trie()
        nums.sort()
        j, n = 0, len(queries)
        ans = [-1] * n
        for i, (x, m) in sorted(zip(range(n), queries), key=lambda x: x[1][1]):
            while j < len(nums) and nums[j] <= m:
                trie.insert(nums[j])
                j += 1
            ans[i] = trie.search(x)
        return ans

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