leetcode: 2935. Find the maximum XOR value of strong number pairs II [The maximum XOR value still depends on the 01Trie tree! 】

Question screenshot

Question analysis

After sorting, the relative positions of x and y are limited
Assuming y > x, as y moves, it must be guaranteed that 2x >= y< a i=2> So you can use the sliding window to maintain a bunch of x that meet the conditions The XOR values ​​of these x can be recorded in the Trie tree

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 = 19

    def __init__(self):
        self.root = Node()

    # 添加 val
    def insert(self, val: int) -> None:
        cur = self.root
        for i in range(Trie.HIGH_BIT, -1, -1):
            bit = (val >> i) & 1
            if 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.root
        for 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.root
        ans = 0
        for 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 << i
                bit ^= 1
            cur = cur.children[bit]
        return ans

class Solution:
    def maximumStrongPairXor(self, nums: List[int]) -> int:
        nums.sort()
        t = Trie()
        ans = left = 0
        for 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 += 1
            ans = max(ans, t.max_xor(y))
        return ans

01Trie tree template

class Node:
    __slots__ = 'children', 'cnt'

    def __init__(self):
        self.children = [None, None]
        self.cnt = 0  # 子树大小

class Trie:
    HIGH_BIT = 19

    def __init__(self):
        self.root = Node()

    # 添加 val
    def insert(self, val: int) -> None:
        cur = self.root
        for i in range(Trie.HIGH_BIT, -1, -1):
            bit = (val >> i) & 1
            if 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.root
        for 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.root
        ans = 0
        for 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 << i
                bit ^= 1
            cur = cur.children[bit]
        return ans

detail

__slot__ speed up