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Problem solution · Python

Maximum Strong Pair XOR I

Maximum Strong Pair XOR I: a Python solution using direct simulation. Learn the idea, check the complexity, and read the full code, with credit to walkccc LeetCode Solutions.

Technique
Direct simulation
Source
walkccc LeetCode Solutions
Length
61 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

Direct simulation

For Maximum Strong Pair XOR I, the implementation follows the problem’s operations directly while maintaining only the state needed for the next decision.

  1. Translate each rule into one explicit state update.
  2. Maintain the invariant after every processed item.
  3. Return the accumulated state once all relevant input has been handled.

Code notes

  • 61 lines of Python from the credited upstream file 2932.py.
  • The implementation visibly relies on sequence storage.
  • No explicit loop blocks detected.

Complexity

Count the number and nesting of passes over the input, then include the maintained containers in the memory estimate.

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 Strong Pair XOR I · PythonPython
Use this to learn the idea, then write your own version.
class TrieNode:  def __init__(self):    self.children: list[TrieNode | None] = [None] * 2    self.mn = math.inf    self.mx = -math.inf  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]      node.mn = min(node.mn, num)      node.mx = max(node.mx, num)   def getMaxXor(self, x: int) -> int:    """Returns max(x ^ y) where |x - y| <= min(x, y).     If x <= y, |x - y| <= min(x, y) can be written as y - x <= x.    So, y <= 2 * x.    """    maxXor = 0    node = self.root    for i in range(self.maxBit, -1, -1):      bit = x >> i & 1      toggleBit = bit ^ 1      # If `node.children[toggleBit].mx > x`, it means there's a number in the      # node that satisfies the condition to ensure that x <= y among x and y.      # If `node.children[toggleBit].mn <= 2 * x`, it means there's a number in      # the node that satisfies the condition for a valid y.      if (node.children[toggleBit] and          node.children[toggleBit].mx > x and              node.children[toggleBit].mn <= 2 * x):        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  class Solution:  # Similar to 421. Maximum XOR of Two Numbers in an Array  def maximumStrongPairXor(self, nums: list[int]) -> int:    maxNum = max(nums)    maxBit = int(math.log2(maxNum))    bitTrie = BitTrie(maxBit)     for num in nums:      bitTrie.insert(num)     return max(bitTrie.getMaxXor(num) for num in nums)

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