How do you solve Count Pairs With XOR in a Range on LeetCode?

Count Pairs With XOR in a Range (LeetCode #1803) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log(max_num)) time complexity and O(n) space complexity. Key insight: Using a Trie can significantly reduce the time complexity for counting pairs based on bitwise operations.

What is the brute force solution for Count Pairs With XOR in a Range?

The brute force approach for Count Pairs With XOR in a Range is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible pairs in the array to see if their XOR falls within the specified range. This is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n log(max_num)).

What algorithm or data structure is used to solve Count Pairs With XOR in a Range?

Count Pairs With XOR in a Range uses the following concepts: Array, Bit Manipulation, Trie. The recommended patterns to study are: Bit Manipulation, Trie, Sliding Window.

What are the interview tips for Count Pairs With XOR in a Range?

Always start with a brute force solution to understand the problem better. Think about how data structures like Trie can help with problems involving bit manipulation. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Count Pairs With XOR in a Range?

Forgetting to check the range conditions correctly when counting pairs. Not handling the binary representation properly when using a Trie.

#1803

Count Pairs With XOR in a Range

Hard

ArrayBit ManipulationTrieBit ManipulationTrieSliding Window

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log(max_num))
Space	O(1)	O(n)

💡

Intuition

Time O(n log(max_num))Space O(n)

Using a Trie data structure allows us to efficiently count pairs whose XOR values fall within a specified range. This approach leverages the properties of binary representation and the Trie structure to reduce the time complexity significantly.

⚙️

Algorithm

5 steps

1Step 1: Create a Trie to store the binary representations of numbers.
2Step 2: For each number in nums, insert it into the Trie.
3Step 3: For each number, calculate the number of valid pairs by traversing the Trie to count how many numbers produce an XOR within the range [low, high].
4Step 4: Use a helper function to count the valid pairs for a given number based on the Trie.
5Step 5: Return the total count of nice pairs.

solution.py39 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.count = 0
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, num):
11        node = self.root
12        for i in range(15, -1, -1):
13            bit = (num >> i) & 1
14            if bit not in node.children:
15                node.children[bit] = TrieNode()
16            node = node.children[bit]
17            node.count += 1
18
19    def countPairs(self, num, low, high):
20        return self._count(num, low, 0, 15) - self._count(num, high + 1, 0, 15)
21
22    def _count(self, num, limit, node, bit):
23        if node is None or bit < 0:
24            return 0
25        if limit < 0:
26            return 0
27        bit_num = (num >> bit) & 1
28        if (limit >> bit) & 1:
29            return node.count + self._count(num, limit, node.children.get(bit_num), bit - 1) + self._count(num, limit, node.children.get(1 - bit_num), bit - 1)
30        else:
31            return self._count(num, limit, node.children.get(bit_num), bit - 1)
32
33def countPairs(nums, low, high):
34    trie = Trie()
35    count = 0
36    for num in nums:
37        count += trie.countPairs(num, low, high)
38        trie.insert(num)
39    return count

ℹ

Complexity note: The time complexity is O(n log(max_num)) because we insert each number into the Trie and count pairs based on the binary representation, where max_num is the maximum value in nums. The space complexity is O(n) due to the storage of the Trie nodes.

1Using a Trie can significantly reduce the time complexity for counting pairs based on bitwise operations.
2Understanding the properties of XOR and its binary representation is crucial for optimizing the solution.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.