Leetcode #1803: Count Pairs With XOR in a Range
In this guide, we solve Leetcode #1803 Count Pairs With XOR in a Range in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
Given a (0-indexed) integer array nums and two integers low and high, return the number of nice pairs. A nice pair is a pair (i, j) where 0 <= i < j < nums.length and low <= (nums[i] XOR nums[j]) <= high.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Trie, Array
Intuition
Prefix queries are most efficient with a trie.
Each character transitions to the next node in the tree.
Approach
Insert words into the trie and traverse by characters for queries.
Track terminal markers to distinguish full words from prefixes.
Steps:
- Build the trie.
- Traverse for each query.
- Return matches or validations.
Example
Input: nums = [1,4,2,7], low = 2, high = 6
Output: 6
Explanation: 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
Python Solution
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
Complexity
The time complexity is , and the space complexity is . The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.