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# 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--){
Set<Integer> set = new HashSet<>();
for(int num : nums){
}
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):
for i in range(32)[::-1]:
prefixes = {num >> i for num in nums}
``````

### 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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