Leetcode #2935: Maximum Strong Pair XOR II
In this guide, we solve Leetcode #2935 Maximum Strong Pair XOR II 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
You are given a 0-indexed integer array nums. A pair of integers x and y is called a strong pair if it satisfies the condition: |x - y| <= min(x, y) You need to select two integers from nums such that they form a strong pair and their bitwise XOR is the maximum among all strong pairs in the array.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Bit Manipulation, Trie, Array, Hash Table, Sliding Window
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
Input: nums = [1,2,3,4,5]
Output: 7
Explanation: There are 11 strong pairs in the array nums: (1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (3, 5), (4, 4), (4, 5) and (5, 5).
The maximum XOR possible from these pairs is 3 XOR 4 = 7.
Python Solution
class Trie:
__slots__ = ("children", "cnt")
def __init__(self):
self.children: List[Trie | None] = [None, None]
self.cnt = 0
def insert(self, x: int):
node = self
for i in range(20, -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: int) -> int:
node = self
ans = 0
for i in range(20, -1, -1):
v = x >> i & 1
if node.children[v ^ 1] and node.children[v ^ 1].cnt:
ans |= 1 << i
node = node.children[v ^ 1]
else:
node = node.children[v]
return ans
def remove(self, x: int):
node = self
for i in range(20, -1, -1):
v = x >> i & 1
node = node.children[v]
node.cnt -= 1
class Solution:
def maximumStrongPairXor(self, nums: List[int]) -> int:
nums.sort()
tree = Trie()
ans = i = 0
for y in nums:
tree.insert(y)
while y > nums[i] * 2:
tree.remove(nums[i])
i += 1
ans = max(ans, tree.search(y))
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.