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Problem solution · Python

Maximum XOR of Two Numbers in an Array

Maximum XOR of Two Numbers in an Array: 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
29 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 XOR of Two Numbers in an Array, 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

  • 29 lines of Python from the credited upstream file 421.py.
  • The implementation visibly relies on sequence storage, ordered lookup.
  • 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 XOR of Two Numbers in an Array · PythonPython
Use this to learn the idea, then write your own version.
class Solution:  def findMaximumXOR(self, nums: list[int]) -> int:    maxNum = max(nums)    if maxNum == 0:      return 0    maxBit = int(math.log2(maxNum))    ans = 0    prefixMask = 0  # `prefixMask` grows like: 10000 -> 11000 -> ... -> 11111.     # If ans is 11100 when i = 2, it means that before we reach the last two    # bits, 11100 is the maximum XOR we have, and we're going to explore if we    # can get another two 1s and put them into `ans`.    for i in range(maxBit, -1, -1):      prefixMask |= 1 << i      # We only care about the left parts,      # If i = 2, nums = [1110, 1011, 0111]      #    -> prefixes = [1100, 1000, 0100]      prefixes = set([num & prefixMask for num in nums])      # If i = 1 and before this iteration, the ans is 10100, it means that we      # want to grow the ans to 10100 | 1 << 1 = 10110 and we're looking for      # XOR of two prefixes = candidate.      candidate = ans | 1 << i      for prefix in prefixes:        if prefix ^ candidate in prefixes:          ans = candidate          break     return ans

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