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1940. Maximum XOR for Each Query

My accepted Python solution to LeetCode problem 1940, Maximum XOR for Each Query, running in 48ms.

  • Difficulty: Medium
  • Python
  • Runtime 48ms
  • Memory 32.9MB

Read the problem on LeetCode View on GitHub

The problem statement is LeetCode’s and stays on their site. What follows is my accepted solution.

Python

Accepted on LeetCode — runtime 48ms, memory 32.9MB, accepted 2025-12-29.

python
class Solution:
    def getMaximumXor(self, nums: List[int], maximumBit: int) -> List[int]:
        # Time: O(n), Space: O(n) for result
        # Key insight: XOR all nums to get cumulative XOR
        # To maximize XOR with k where k < 2^maximumBit,
        # k should flip all bits in the result to make it (2^maximumBit - 1)
        
        n = len(nums)
        result = []
        
        # Calculate XOR of all elements
        xor_sum = 0
        for num in nums:
            xor_sum ^= num
        
        # Maximum value with maximumBit bits is (2^maximumBit - 1)
        max_val = (1 << maximumBit) - 1
        
        # For each query (starting from full array, removing from end)
        for i in range(n):
            # To maximize xor_sum XOR k, we want k = max_val XOR xor_sum
            # This gives us max_val when XOR'd with xor_sum
            k = max_val ^ xor_sum
            result.append(k)
            # Remove last element for next query
            xor_sum ^= nums[n - 1 - i]
        
        return result

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