<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>
class TrieNode:
    def __init__(self):
        self.childs = [None] * 2
        self.cnt = 0
class Trie:
    def __init__(self):
        self.root = TrieNode()
    def add(self, num):
        p = self.root
        for i in range(15, -1, -1):
            bit = (num >> i) & 1
            if not p.childs[bit]:
                p.childs[bit] = TrieNode()
            p.childs[bit].cnt += 1
            p = p.childs[bit]

    def query(self, num, limit): # count number of x ^ num < limit
        p = self.root
        count = 0
        for i in range(15, -1, -1):
            if not p:
                break
            nb = (num >> i) & 1
            lb = (limit >> i) & 1
            if lb == 1:
                if p.childs[nb]:
                    count += p.childs[nb].cnt
                p = p.childs[nb ^ 1]
            else:
                p = p.childs[nb]
        return count

class Solution:
    def countPairs(self, nums: List[int], low: int, high: int) -> int:
        res = 0
        tree = Trie()
        for num in nums:
            res += tree.query(num, high + 1) - tree.query(num, low)
            tree.add(num)
        return res

24600