Count Pairs With XOR in a Range
<p>Given a <strong>(0-indexed)</strong> integer array <code>nums</code> and two integers <code>low</code> and <code>high</code>, return <em>the number of <strong>nice pairs</strong></em>.</p>

<p>A <strong>nice pair</strong> is a pair <code>(i, j)</code> where <code>0 &lt;= i &lt; j &lt; nums.length</code> and <code>low &lt;= (nums[i] XOR nums[j]) &lt;= high</code>.</p>

<p>&nbsp;</p>

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

<pre>

<strong>Input:</strong> nums = [1,4,2,7], low = 2, high = 6

<strong>Output:</strong> 6

<strong>Explanation:</strong> All nice pairs (i, j) are as follows:

    - (0, 1): nums[0] XOR nums[1] = 5 

    - (0, 2): nums[0] XOR nums[2] = 3

    - (0, 3): nums[0] XOR nums[3] = 6

    - (1, 2): nums[1] XOR nums[2] = 6

    - (1, 3): nums[1] XOR nums[3] = 3

    - (2, 3): nums[2] XOR nums[3] = 5

</pre>

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

<pre>

<strong>Input:</strong> nums = [9,8,4,2,1], low = 5, high = 14

<strong>Output:</strong> 8

<strong>Explanation:</strong> All nice pairs (i, j) are as follows:

​​​​​    - (0, 2): nums[0] XOR nums[2] = 13

&nbsp;   - (0, 3): nums[0] XOR nums[3] = 11

&nbsp;   - (0, 4): nums[0] XOR nums[4] = 8

&nbsp;   - (1, 2): nums[1] XOR nums[2] = 12

&nbsp;   - (1, 3): nums[1] XOR nums[3] = 10

&nbsp;   - (1, 4): nums[1] XOR nums[4] = 9

&nbsp;   - (2, 3): nums[2] XOR nums[3] = 6

&nbsp;   - (2, 4): nums[2] XOR nums[4] = 5</pre>

<p>&nbsp;</p>

<p><strong>Constraints:</strong></p>

<ul>

    <li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>

    <li><code>1 &lt;= nums[i] &lt;= 2 * 10<sup>4</sup></code></li>

    <li><code>1 &lt;= low &lt;= high &lt;= 2 * 10<sup>4</sup></code></li>

</ul>
Bit Manipulation; Trie; Array
class Trie {
public:
    Trie()
        : children(2)
        , cnt(0) {}

    void insert(int x) {
        Trie* node = this;
        for (int i = 15; ~i; --i) {
            int v = x >> i & 1;
            if (!node->children[v]) {
                node->children[v] = new Trie();
            }
            node = node->children[v];
            ++node->cnt;
        }
    }

    int search(int x, int limit) {
        Trie* node = this;
        int ans = 0;
        for (int i = 15; ~i && node; --i) {
            int v = x >> i & 1;
            if (limit >> i & 1) {
                if (node->children[v]) {
                    ans += node->children[v]->cnt;
                }
                node = node->children[v ^ 1];
            } else {
                node = node->children[v];
            }
        }
        return ans;
    }

private:
    vector<Trie*> children;
    int cnt;
};

class Solution {
public:
    int countPairs(vector<int>& nums, int low, int high) {
        Trie* tree = new Trie();
        int ans = 0;
        for (int& x : nums) {
            ans += tree->search(x, high + 1) - tree->search(x, low);
            tree->insert(x);
        }
        return ans;
    }
};