<p>You are given an array <code>nums</code> consisting of non-negative integers. You are also given a <code>queries</code> array, where <code>queries[i] = [x<sub>i</sub>, m<sub>i</sub>]</code>.</p>

<p>The answer to the <code>i<sup>th</sup></code> query is the maximum bitwise <code>XOR</code> value of <code>x<sub>i</sub></code> and any element of <code>nums</code> that does not exceed <code>m<sub>i</sub></code>. In other words, the answer is <code>max(nums[j] XOR x<sub>i</sub>)</code> for all <code>j</code> such that <code>nums[j] &lt;= m<sub>i</sub></code>. If all elements in <code>nums</code> are larger than <code>m<sub>i</sub></code>, then the answer is <code>-1</code>.</p>

<p>Return <em>an integer array </em><code>answer</code><em> where </em><code>answer.length == queries.length</code><em> and </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,1,2,3,4], queries = [[3,1],[1,3],[5,6]]
<strong>Output:</strong> [3,3,7]
<strong>Explanation:</strong>
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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [5,2,4,6,6,3], queries = [[12,4],[8,1],[6,3]]
<strong>Output:</strong> [15,-1,5]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, queries.length &lt;= 10<sup>5</sup></code></li>
	<li><code>queries[i].length == 2</code></li>
	<li><code>0 &lt;= nums[j], x<sub>i</sub>, m<sub>i</sub> &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution {
   public:
    struct TrieNode {
        TrieNode* child[2];
        // TrieNode() {
        //     child[0] = nullptr;
        //     child[1] = nullptr;
        // }
        // ~TrieNode() {
        //     delete child[0];
        //     delete child[1];
        // }
    };
    TrieNode* root;
    void insert(int& n) {
        TrieNode* currNode = root;
        for (int i = 31; i >= 0; i--) {
            int currBit = (n >> i) & 1;
            if (!currNode->child[currBit])
                currNode->child[currBit] = new TrieNode();
            currNode = currNode->child[currBit];
        }
    }

    int getAnsForN(int& n) {
        TrieNode* curr = root;
        int ans = 0;
        for (int i = 31; i >= 0; i--) {
            int currBit = (n >> i) & 1;
            int inverseBit = 1 - currBit;
            int availableBit = curr->child[inverseBit] ? inverseBit : currBit;
            if (availableBit == inverseBit) ans += (int)pow(2, i);
            curr = curr->child[availableBit];
        }
        return ans;
    }
    struct Query {
        int xi, ai, idx;
        Query(int xi, int ai, int idx) : xi(xi), ai(ai), idx(idx) {}
    };
    vector<int> maximizeXor(vector<int>& nums, vector<vector<int>>& queries) {
        // queries = [{xi, ai}]
        // main idea is, to only have elements which are less than xi in trie
        // while iterating queries sort nums sort queries according to ai
        // q[1].ai < q[2].ai
        // we have to itereate queries and nums in such a way that
        // for each query only nums < query.ai are present in trie
        // for each query q:
        //  while currentIdx < n && nums at currentIdx <= q.ai
        //    we add it in trie
        //    currentIdx++
        //  calculate max xor for nums[currentIdx], q.xi
        int n = (int)nums.size();
        root = new TrieNode();
        vector<int> ans((int)queries.size());
        vector<Query> offlineQueries;
        for (int i = 0; i < (int)queries.size(); i++) {
            Query q = Query(queries[i][1], queries[i][0], i);
            offlineQueries.push_back(q);
        }
        sort(nums.begin(), nums.end());
        sort(offlineQueries.begin(), offlineQueries.end(),
             [](Query x, Query y) { return x.xi < y.xi; });
        int currentIdx = 0;
        for (auto q : offlineQueries) {
            while (currentIdx < n && nums[currentIdx] <= q.xi)
                insert(nums[currentIdx++]);
            ans[q.idx] = currentIdx == 0 ? -1 : getAnsForN(q.ai);
        }
        return ans;
    }
};

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