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:
    def __init__(self):
        self.children = [None] * 2
        self.cnt = 0

    def insert(self, x):
        node = self
        for i in range(15, -1, -1):
            v = x >> i & 1
            if node.children[v] is None:
                node.children[v] = Trie()
            node = node.children[v]
            node.cnt += 1

    def search(self, x, limit):
        node = self
        ans = 0
        for i in range(15, -1, -1):
            if node is None:
                return ans
            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


class Solution:
    def countPairs(self, nums: List[int], low: int, high: int) -> int:
        ans = 0
        tree = Trie()
        for x in nums:
            ans += tree.search(x, high + 1) - tree.search(x, low)
            tree.insert(x)
        return ans
