1707. Maximum XOR With an Element From Array

1707. Maximum XOR With an Element From Array (Hard)

You are given an array nums consisting of non-negative integers. You are also given a queries array, where queries[i] = [xi, mi].

The answer to the ith query is the maximum bitwise XOR value of xi and any element of nums that does not exceed mi. In other words, the answer is max(nums[j] XOR xi) for all j such that nums[j] <= mi. If all elements in nums are larger than mi, then the answer is -1.

Return an integer array answer where answer.length == queries.length and answer[i] is the answer to the ith query.

 

Example 1:

Input: nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
Output: [3,3,7]
Explanation:
1) 0 and 1 are the only two integers not greater than 1. 0 XOR 3 = 3 and 1 XOR 3 = 2. The larger of the two is 3.
2) 1 XOR 2 = 3.
3) 5 XOR 2 = 7.

Example 2:

Input: nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
Output: [15,-1,5]

 

Constraints:

• 1 <= nums.length, queries.length <= 105
• queries[i].length == 2
• 0 <= nums[j], xi, mi <= 109

Related Topics:
Bit Manipulation, Trie

Similar Questions:

• Maximum XOR of Two Numbers in an Array (Medium)

Solution 1. Trie

// OJ: https://leetcode.com/problems/maximum-xor-with-an-element-from-array/
// Author: github.com/lzl124631x
// Time: O(A + Q)
// Space: O(A)
struct TrieNode {
    TrieNode *next[2] = {};
    int minVal = INT_MAX;
};
class Solution {
    void add(TrieNode *node, int n) {
        for (int i = 31; i >= 0; --i) {
            int b = n >> i & 1;
            if (node->next[b] == NULL) node->next[b] = new TrieNode();
            node = node->next[b];
            node->minVal = min(node->minVal, n);
        }
    }
    int maxXor(TrieNode *node, int x, int m) {
        int ans = 0;
        for (int i = 31; i >= 0; --i) {
            int b = x >> i & 1;
            if (node->next[1 - b] && node->next[1 - b]->minVal <= m) {
                node = node->next[1 - b];
                ans |= 1 << i;
            } else if (node->next[b] && node->next[b]->minVal <= m) node = node->next[b];
            else return -1;
        }
        return ans;
    }
public:
    vector<int> maximizeXor(vector<int>& A, vector<vector<int>>& Q) {
        TrieNode root;
        vector<int> ans;
        for (int n : A) add(&root, n);
        for (auto &q : Q) {
            int x = q[0], m = q[1];
            ans.push_back(maxXor(&root, x, m));
        }
        return ans;
    }
};