<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>
from typing import List

class TrieNode:
    def __init__(self):
        self.children = {}
        self.count = 0

class Solution:
    def countPairs(self, nums: List[int], low: int, high: int) -> int:
        def insert(num):
            node = root
            for i in range(15, -1, -1):
                bit = (num >> i) & 1
                if bit not in node.children:
                    node.children[bit] = TrieNode()
                node = node.children[bit]
                node.count += 1
        
        def count_less_than(num, limit):
            node = root
            count = 0
            for i in range(15, -1, -1):
                if not node:
                    break
                bit_num = (num >> i) & 1
                bit_limit = (limit >> i) & 1
                if bit_limit == 1:
                    if bit_num in node.children:
                        count += node.children[bit_num].count
                    node = node.children.get(1 - bit_num, None)
                else:
                    node = node.children.get(bit_num, None)
            return count
        
        root = TrieNode()
        result = 0
        
        for num in nums:
            result += count_less_than(num, high + 1) - count_less_than(num, low)
            insert(num)
        
        return result

# Example usage:
# solution = Solution()
# print(solution.countPairs([1, 4, 2, 7], 2, 6))  # Output: 6
# print(solution.countPairs([9, 8, 4, 2, 1], 5, 14))  # Output: 8
24500