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 {
    private Trie[] children = new Trie[2];
    private int cnt;

    public void insert(int x) {
        Trie node = this;
        for (int i = 15; i >= 0; --i) {
            int v = (x >> i) & 1;
            if (node.children[v] == null) {
                node.children[v] = new Trie();
            }
            node = node.children[v];
            ++node.cnt;
        }
    }

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

class Solution {
    public int countPairs(int[] nums, int low, int high) {
        Trie trie = new Trie();
        int ans = 0;
        for (int x : nums) {
            ans += trie.search(x, high + 1) - trie.search(x, low);
            trie.insert(x);
        }
        return ans;
    }
}