题目截图

题目分析

排序后，限定了x和y的相对位置
假设y > x，随着y的移动，必须要保证2x >= y
所以可以使用滑动窗口维护一堆满足条件的x
这些x的异或值记录在Trie树中即可

ac code

class Node:__slots__ = 'children', 'cnt'def __init__(self):self.children = [None, None]self.cnt = 0  # 子树大小class Trie:HIGH_BIT = 19def __init__(self):self.root = Node()# 添加 valdef insert(self, val: int) -> None:cur = self.rootfor i in range(Trie.HIGH_BIT, -1, -1):bit = (val >> i) & 1if cur.children[bit] is None:cur.children[bit] = Node()cur = cur.children[bit]cur.cnt += 1  # 维护子树大小return cur# 删除 val，但不删除节点# 要求 val 必须在 trie 中def remove(self, val: int) -> None:cur = self.rootfor i in range(Trie.HIGH_BIT, -1, -1):cur = cur.children[(val >> i) & 1]cur.cnt -= 1  # 维护子树大小return cur# 返回 val 与 trie 中一个元素的最大异或和# 要求 trie 中至少有一个元素def max_xor(self, val: int) -> int:cur = self.rootans = 0for i in range(Trie.HIGH_BIT, -1, -1):bit = (val >> i) & 1# 如果 cur.children[bit^1].cnt == 0，视作空节点if cur.children[bit ^ 1] and cur.children[bit ^ 1].cnt:ans |= 1 << ibit ^= 1cur = cur.children[bit]return ansclass Solution:def maximumStrongPairXor(self, nums: List[int]) -> int:nums.sort()t = Trie()ans = left = 0for y in nums:t.insert(y)# 只考虑nums[left] * 2 >= y，否则滑走while nums[left] * 2 < y:t.remove(nums[left])left += 1ans = max(ans, t.max_xor(y))return ans

01Trie树模版

class Node:__slots__ = 'children', 'cnt'def __init__(self):self.children = [None, None]self.cnt = 0  # 子树大小class Trie:HIGH_BIT = 19def __init__(self):self.root = Node()# 添加 valdef insert(self, val: int) -> None:cur = self.rootfor i in range(Trie.HIGH_BIT, -1, -1):bit = (val >> i) & 1if cur.children[bit] is None:cur.children[bit] = Node()cur = cur.children[bit]cur.cnt += 1  # 维护子树大小return cur# 删除 val，但不删除节点# 要求 val 必须在 trie 中def remove(self, val: int) -> None:cur = self.rootfor i in range(Trie.HIGH_BIT, -1, -1):cur = cur.children[(val >> i) & 1]cur.cnt -= 1  # 维护子树大小return cur# 返回 val 与 trie 中一个元素的最大异或和# 要求 trie 中至少有一个元素def max_xor(self, val: int) -> int:cur = self.rootans = 0for i in range(Trie.HIGH_BIT, -1, -1):bit = (val >> i) & 1# 如果 cur.children[bit^1].cnt == 0，视作空节点if cur.children[bit ^ 1] and cur.children[bit ^ 1].cnt:ans |= 1 << ibit ^= 1cur = cur.children[bit]return ans

细节

__slot__加速