421. Maximum XOR of Two Numbers in an Array LeetCode Solution

In this guide, you will get 421. Maximum XOR of Two Numbers in an Array LeetCode Solution with the best time and space complexity. The solution to Maximum XOR of Two Numbers in an Array problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Maximum XOR of Two Numbers in an Array solution in C++
Maximum XOR of Two Numbers in an Array solution in Java
Maximum XOR of Two Numbers in an Array solution in Python
Additional Resources

421. Maximum XOR of Two Numbers in an Array LeetCode Solution image

Problem Statement of Maximum XOR of Two Numbers in an Array

Given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Example 1:

Input: nums = [3,10,5,25,2,8]
Output: 28
Explanation: The maximum result is 5 XOR 25 = 28.

Example 2:

Input: nums = [14,70,53,83,49,91,36,80,92,51,66,70]
Output: 127

Constraints:

1 <= nums.length <= 2 * 105
0 <= nums[i] <= 231 – 1

Complexity Analysis

Time Complexity: O(n\log_2(\max(\texttt{nums})))
Space Complexity: O(n)

421. Maximum XOR of Two Numbers in an Array LeetCode Solution in C++

class Solution {
 public:
  int findMaximumXOR(vector<int>& nums) {
    const int maxNum = ranges::max(nums);
    if (maxNum == 0)
      return 0;
    const int maxBit = static_cast<int>(log2(maxNum));
    int ans = 0;
    int prefixMask = 0;  // 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 (int i = maxBit; i >= 0; --i) {
      prefixMask |= 1 << i;
      unordered_set<int> prefixes;
      // We only care about the left parts,
      // If i = 2, nums = {1110, 1011, 0111}
      //    -> prefixes = {1100, 1000, 0100}
      for (const int num : nums)
        prefixes.insert(num & prefixMask);
      // 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.
      const int candidate = ans | 1 << i;
      for (const int prefix : prefixes)
        if (prefixes.contains(prefix ^ candidate)) {
          ans = candidate;
          break;
        }
    }

    return ans;
  }
};
/* code provided by PROGIEZ */

421. Maximum XOR of Two Numbers in an Array LeetCode Solution in Java

class Solution {
  public int findMaximumXOR(int[] nums) {
    final int maxNum = Arrays.stream(nums).max().getAsInt();
    if (maxNum == 0)
      return 0;
    final int maxBit = (int) (Math.log(maxNum) / Math.log(2));
    int ans = 0;
    int prefixMask = 0; // 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 (int i = maxBit; i >= 0; --i) {
      prefixMask |= 1 << i;
      Set<Integer> prefixes = new HashSet<>();
      // We only care about the left parts,
      // If i = 2, nums = {1110, 1011, 0111}
      //    . prefixes = {1100, 1000, 0100}
      for (final int num : nums)
        prefixes.add(num & prefixMask);
      // 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.
      final int candidate = ans | 1 << i;
      for (final int prefix : prefixes)
        if (prefixes.contains(prefix ^ candidate)) {
          ans = candidate;
          break;
        }
    }

    return ans;
  }
}
// code provided by PROGIEZ

421. Maximum XOR of Two Numbers in an Array LeetCode Solution in Python

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
 # code by PROGIEZ