How do you solve Maximum XOR With an Element From Array on LeetCode?

Maximum XOR With an Element From Array (LeetCode #1707) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + q log n) time complexity and O(n) space complexity. Key insight: Maximizing XOR involves understanding how bits interact; toggling bits can lead to higher values.

What is the time complexity of Maximum XOR With an Element From Array?

The optimal solution for Maximum XOR With an Element From Array runs in O(n log n + q log n) time and uses O(n) space. The sorting step takes O(n log n), and each query involves a binary search and Trie operations, leading to a total complexity of O(n log n + q log n).

What is the brute force solution for Maximum XOR With an Element From Array?

The brute force approach for Maximum XOR With an Element From Array is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves checking each number in the array for every query. This means we calculate the XOR for each valid number in nums against the query's x value, which is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n log n + q log n).

What algorithm or data structure is used to solve Maximum XOR With an Element From Array?

Maximum XOR With an Element From Array uses the following concepts: Array, Bit Manipulation, Trie. The recommended patterns to study are: Bit Manipulation, Trie, Sorting.

What are the interview tips for Maximum XOR With an Element From Array?

Always start with a brute-force solution to understand the problem better before optimizing. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice problems involving bit manipulation and data structures like Trie to build confidence.

What are common mistakes when solving Maximum XOR With an Element From Array?

Not considering edge cases where all nums are greater than m, leading to incorrect results. Failing to optimize the search for valid nums, which can lead to timeouts in large inputs.

#1707

Maximum XOR With an Element From Array

Hard

ArrayBit ManipulationTrieBit ManipulationTrieSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n + q log n)
Space	O(1)	O(n)

💡

Intuition

Time O(n log n + q log n)Space O(n)

The optimal solution uses a Trie to efficiently store the binary representations of the numbers in nums. This allows us to quickly find the best number to maximize the XOR for each query, significantly reducing the time complexity.

⚙️

Algorithm

5 steps

1Step 1: Sort nums to facilitate efficient searching.
2Step 2: Build a Trie from the binary representations of nums.
3Step 3: For each query, filter nums to only those <= m using binary search.
4Step 4: Use the Trie to find the maximum XOR for the filtered nums with x.
5Step 5: Return the results for all queries.

solution.py45 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.value = None
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, num):
11        node = self.root
12        for i in range(31, -1, -1):
13            bit = (num >> i) & 1
14            if bit not in node.children:
15                node.children[bit] = TrieNode()
16            node = node.children[bit]
17        node.value = num
18
19    def maxXor(self, num):
20        node = self.root
21        max_xor = 0
22        for i in range(31, -1, -1):
23            bit = (num >> i) & 1
24            toggled_bit = 1 - bit
25            if toggled_bit in node.children:
26                max_xor |= (1 << i)
27                node = node.children[toggled_bit]
28            else:
29                node = node.children.get(bit, None)
30        return max_xor
31
32def maximizeXor(nums, queries):
33    nums.sort()
34    trie = Trie()
35    for num in nums:
36        trie.insert(num)
37    answer = []
38    for x, m in queries:
39        max_xor = -1
40        for num in nums:
41            if num > m:
42                break
43            max_xor = max(max_xor, trie.maxXor(x))
44        answer.append(max_xor)
45    return answer

ℹ

Complexity note: The sorting step takes O(n log n), and each query involves a binary search and Trie operations, leading to a total complexity of O(n log n + q log n).

1Maximizing XOR involves understanding how bits interact; toggling bits can lead to higher values.
2Using data structures like Trie can significantly optimize search operations for bit manipulation problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.