421: Solution with step by step explanation
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n1. We first find the maximum number in the input list nums using the buil
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n1. We first find the maximum number in the input list nums using the built-in max function.\n\n2. If the maximum number is 0, it means that all the numbers in the list are also 0, and so the maximum XOR value will also be 0. So we return 0.\n\n3. Otherwise, we calculate the maximum bit position of the maximum number using the math.log2 function.\n\n4. We initialize the ans variable to 0, which will eventually hold the maximum XOR value we find.\n\n5. We also initialize a mask variable to 0, which we will use to generate a set of prefixes for each bit position.\n\n6. We loop through the bit positions in descending order from the maximum bit position to 0.\n\n7. In each iteration, we set the mask variable to the current bit position by left-shifting 1 by the bit position and subtracting 1. For example, if the bit position is 2, mask will be set to 0b00000100.\n\n8. We generate a set of prefixes for the current bit position by applying the mask to each number in the input list using the bitwise AND operator (&). This results in a set of numbers where the rightmost bits up to the current bit position are all 0.\n\n9. We initialize a candidate variable to the current ans value with the bit at the current bit position set to 1 using the bitwise OR operator (|).\n\n10. We loop through each prefix in the set of prefixes we generated in step 8.\n\n11. For each prefix, we check if there exists another prefix in the set that, when XORed with the candidate variable, results in another prefix in the set. If such a prefix exists, it means that we can use the candidate variable as the new ans value, since it indicates that we can form a larger XOR value by setting the bit at the current bit position to 1.\n\n12. If we find such a prefix, we set the ans variable to the candidate value and break out of the loop.\n\n13. We return the ans variable at the end of the loop, which should contain the maximum XOR value we found.\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution:\n  def findMaximumXOR(self, nums: List[int]) -> int:\n    maxNum = max(nums)\n    if maxNum == 0:\n      return 0\n    maxBit = int(math.log2(maxNum))\n    ans = 0\n    mask = 0\n\n    # If ans is 11100 when i = 2, it means that before we reach the last two\n    # bits, 11100 is the maximum XOR we have, and we\'re going to explore if we\n    # can get another two \'1\'s and put them into ans.\n    for i in range(maxBit, -1, -1):\n      # Mask grows like: 100...000, 110...000, 111...000, ..., 111...111.\n      mask |= 1 << i\n      # We only care about the left parts,\n      # If i = 2, nums = [1110, 1011, 0111]\n      #    -> prefixes = [1100, 1000, 0100]\n      prefixes = set([num & mask for num in nums])\n      # If i = 1 and before this iteration, the ans is 10100, it means that we\n      # want to grow the ans to 10100 | 1 << 1 = 10110 and we\'re looking for\n      # XOR of two prefixes = candidate.\n      candidate = ans | 1 << i\n      for prefix in prefixes:\n        if prefix ^ candidate in prefixes:\n          ans = candidate\n          break\n\n    return ans\n\n```
['Array', 'Hash Table', 'Bit Manipulation', 'Python', 'Python3']