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Problem solution · Python

Maximum XOR With an Element From Array

Maximum XOR With an Element From Array: a Python solution using sorting and greedy selection. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Sorting and greedy selection
Source
walkccc LeetCode Solutions
Length
69 lines
Start with the idea.

Try the problem first. If you get stuck, read the approach below, then write your own solution. The full code is at the bottom.

Approach

Sorting and greedy selection

For Maximum XOR With an Element From Array, the implementation first exposes a useful order, then scans that order while making locally justified choices.

  1. Choose the key that reveals the greedy or grouping structure.
  2. Sort the relevant records by that key.
  3. Scan in order, maintaining the invariant that makes each local choice safe.

Code notes

  • 69 lines of Python from the credited upstream file 1707.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Sorting is typically the dominant term unless the subsequent scan uses a more expensive nested operation.

Check the problem constraints before deciding whether this complexity will pass.

Source

Code and credit

This code comes from walkccc LeetCode Solutions by P.-Y. Chen (walkccc) and is used under the MIT licence.

Open problem ↗Original source ↗
Full codeMaximum XOR With an Element From Array · PythonPython
Use this to learn the idea, then write your own version.
from dataclasses import dataclass  class TrieNode:  def __init__(self):    self.children: list[TrieNode | None] = [None] * 2  class BitTrie:  def __init__(self, maxBit: int):    self.maxBit = maxBit    self.root = TrieNode()   def insert(self, num: int) -> None:    node = self.root    for i in range(self.maxBit, -1, -1):      bit = num >> i & 1      if not node.children[bit]:        node.children[bit] = TrieNode()      node = node.children[bit]   def getMaxXor(self, num: int) -> int:    maxXor = 0    node = self.root    for i in range(self.maxBit, -1, -1):      bit = num >> i & 1      toggleBit = bit ^ 1      if node.children[toggleBit]:        maxXor = maxXor | 1 << i        node = node.children[toggleBit]      elif node.children[bit]:        node = node.children[bit]      else:  # There's nothing in the Bit Trie.        return 0    return maxXor  @dataclass(frozen=True)class IndexedQuery:  queryIndex: int  x: int  m: int   def __iter__(self):    yield self.queryIndex    yield self.x    yield self.m  class Solution:  def maximizeXor(self, nums: list[int], queries: list[list[int]]) -> list[int]:    ans = [-1] * len(queries)    maxBit = int(math.log2(max(max(nums), max(x for x, _ in queries))))    bitTrie = BitTrie(maxBit)     nums.sort()     i = 0  # nums' index    for queryIndex, x, m in sorted([IndexedQuery(i, x, m)                                    for i, (x, m) in enumerate(queries)],                                   key=lambda x: x.m):      while i < len(nums) and nums[i] <= m:        bitTrie.insert(nums[i])        i += 1      if i > 0 and nums[i - 1] <= m:        ans[queryIndex] = bitTrie.getMaxXor(x)     return ans

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