Leetcode Maximum XOR of Two Numbers in an Array problem solution

In this Leetcode Maximum XOR of Two Numbers in an Array problem solution we have given an integer array nums, return the maximum result of nums[i] XOR nums[j], where 0 <= i <= j < n.

Problem solution in Python.

class Solution:
    def findMaximumXOR(self, nums: List[int]) -> int:
        mask = 0
        ans = 0
        
        for i in range(31, -1, -1):
            mask = mask | 1 << i
            s = set()
            for num in nums:
                s.add(num & mask)
            
            start = ans | 1 << i
            
            for prefix in s:
                if start ^ prefix in s:
                    ans = start
        return ans

Problem solution in Java.

class Solution {
    public int findMaximumXOR(int[] nums) {
        Node root = new Node();
        insert(root, nums[0], 30);
        int ans = 0;
        for(int i=1;i<nums.length;i++){
            ans = Math.max(ans, query(root, nums[i], 30));
            insert(root, nums[i], 30);
        }
        return ans;
    }
    public class Node{
        Node left, right;
        int lc, rc;
        public Node(){
            this.left = null;
            this.right = null;
            this.lc = 0;
            this.rc = 0;
        }
    }
    
    void insert(Node node, int val, int level){
        if(level==-1){
            return;
        }
        
        if((val&(1<<level))>0){
            if(node.right==null){
                node.right = new Node();
            }
            node.rc++;
            insert(node.right, val, level-1);
        }else{
            if(node.left==null){
                node.left = new Node();
            }
            node.lc++;
            insert(node.left, val, level-1);
        }
    }
    int query(Node node, int val, int level){
        if(level==-1){
            return 0;
        }
        
        if((val&(1<<level))>0){
            if(node.left!=null){
                return query(node.left, val, level-1)|(1<<level);
            }else{
                return query(node.right, val, level-1);
            }
        }else{
            if(node.right!=null){
                return query(node.right, val, level-1)|(1<<level);
            }else{
                return query(node.left, val, level-1);
            }
        }
    }
}

Problem solution in C++.

#define ll long long
class Solution {
public:
    
    struct Tries{
        array<Tries *, 2> mp;
    };

    void insert(Tries *root, int n){
        for(int i=31; i>=0; i--){
            int bit = (n&(1<<i));
            if(bit) bit = 1;
            if(root->mp[bit] == NULL)
                root->mp[bit] = new Tries();
            root = root->mp[bit];
        }
    }

    ll findMax(Tries *root, ll n){
        ll ans = 0;
        for(int i=31; i>=0; i--){
            int bit = (n&(1<<i));
            if(bit) bit = 1;
            if(root->mp[!bit]){
                ans += (1<<i);
                root = root->mp[!bit];
            }
            else
                root = root->mp[bit];
        }
        return ans;
    }
    
    int findMaximumXOR(vector<int>& nums) {
        int ans = 0;
        Tries *root = new Tries();
        for(int i : nums)
            insert(root, i);
        for(int i : nums){
            int x = findMax(root, i);
            ans = max(ans, x);
        }
        return ans;
    }
};

Problem solution in C.

#define MAX_BIT 30
struct tree{
    struct tree *leaf[2]; /* bit 0 --> leaf 0 ; bit 1 --> leaf 1*/
};


static void buildTree(struct tree *root, int *nums, int numsSize)
{
    int i;
    int j;
    int bit_val;
    struct tree *node = NULL;

    for (i = 0; i < numsSize; i++) {
        node = root;
        for (j = MAX_BIT; j >= 0; j--) {
            bit_val = nums[i] >> j & 1;
            if (!node->leaf[bit_val])
                node->leaf[bit_val] = calloc(1, sizeof(*root));
            node = node->leaf[bit_val];
        }
    }
}

void freeTree(struct tree *root)
{
    if (!root)
        return;
    
    if (root->leaf[0]);
        freeTree(root->leaf[0]);
    if (root->leaf[1])
        freeTree(root->leaf[1]);
    free(root);
}

int findMaximumXOR(int* nums, int numsSize)
{
    struct tree root = {0};
    int i;
    int j;
    struct tree *node = NULL;
    int max = 0;
    int cur = 0;
    int bit_val = 0;

    if (1 == numsSize)
        return 0;

    buildTree(&root, nums, numsSize);

    for (i = 0; i < numsSize; i++) {
        cur = 0;
        node = &root;
        for (j = MAX_BIT; j >= 0; j--) {
            bit_val = nums[i] >> j & 1;
            if (node->leaf[!bit_val]) {
                cur += 1 << j;
                node = node->leaf[!bit_val];
            } else {
                node = node->leaf[bit_val];
            }
        }

        if (cur > max)
            max = cur;
    }

    freeTree(root.leaf[0]);
    freeTree(root.leaf[1]);
    return max;
}