Maximum XOR of Two Numbers in an Array LeetCode Solution

Problem – 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

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

public class Solution {
    public int findMaximumXOR(int[] nums) {
        int max = 0, mask = 0;
        for(int i = 31; i >= 0; i--){
            mask = mask | (1 << i);
            Set<Integer> set = new HashSet<>();
            for(int num : nums){
                set.add(num & mask);
            }
            int tmp = max | (1 << i);
            for(int prefix : set){
                if(set.contains(tmp ^ prefix)) {
                    max = tmp;
                    break;
                }
            }
        }
        return max;
    }
}

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

def findMaximumXOR(self, nums):
    answer = 0
    for i in range(32)[::-1]:
        answer <<= 1
        prefixes = {num >> i for num in nums}
        answer += any(answer^1 ^ p in prefixes for p in prefixes)
    return answer

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

int findMaximumXOR(vector<int>& nums) {
        int n = nums.size();
        
        if (n == 0 || n == 1)
            return 0;
        if (n == 2)
            return nums.at(0) ^ nums.at(1);
        
        list<int> set0;
        list<int> set1;
        int i;
        int j;
        int maxValue;
        
        for (i = 30; i >= 0; i--) {
            for (j = 0; j < n; j++) {
                if ((nums.at(j) & (1<<i)) == 0)
                    set0.push_back(nums.at(j));
                else
                    set1.push_back(nums.at(j));
            }
            
            if (set0.size() != 0 && set1.size() != 0) {
                maxValue = pow(2, i);
                break;
            }
            else {
                set0.clear();
                set1.clear();
            }
        }
        
        if (i == -1)
            return 0;
        
        maxValue += getMaxXor(set0, set1, i-1);
        
        return maxValue;
}

int getMaxXor(list<int>& set0, list<int>& set1, int pos) {
        int maxValue;
        list<int> set0list0;
        list<int> set0list1;
        list<int> set1list0;
        list<int> set1list1;
        int i;
        list<int>::iterator it;
        
        if (set0.size() == 0 || set1.size() == 0 || pos < 0)
            return 0;
        
        for (it = set0.begin(); it != set0.end(); it++) {
            int value = *it;
            if ((value & (1<<pos)) == 0)
                set0list0.push_back(value);
            else
                set0list1.push_back(value);
        }
        
        for (it = set1.begin(); it != set1.end(); it++) {
            int value = *it;
            if ((value & (1<<pos)) == 0)
                set1list0.push_back(value);
            else
                set1list1.push_back(value);
        }
        
        if (set0list0.size() == 0 && set1list0.size() == 0)
            maxValue = getMaxXor(set0, set1, pos-1);
        else if (set0list1.size() == 0 && set1list1.size() == 0)
            maxValue = getMaxXor(set0, set1, pos-1);
        else {
            int maxValue1 = getMaxXor(set0list0, set1list1, pos-1);
            int maxValue2 = getMaxXor(set0list1, set1list0, pos-1);
            maxValue = pow(2, pos) + (maxValue1 > maxValue2 ? maxValue1 : maxValue2);
        }
        
        return maxValue;
 }

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