2932. Maximum Strong Pair XOR I

# Time:  O(nlogn + nlogr) = O(nlogr), r = max(nums)
# Space: O(t)

# bit manipulation, greedy, trie, sort, two pointers
class Solution(object):
    def maximumStrongPairXor(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        class Trie(object):
            def __init__(self, bit_length):
                self.__nodes = []
                self.__cnts = []
                self.__new_node()
                self.__bit_length = bit_length
            
            def __new_node(self):
                self.__nodes.append([-1]*2)
                self.__cnts.append(0)
                return len(self.__nodes)-1

            def update(self, num, d):
                curr = 0
                for i in reversed(xrange(self.__bit_length)):
                    x = num>>i
                    if self.__nodes[curr][x&1] == -1:
                        self.__nodes[curr][x&1] = self.__new_node()
                    curr = self.__nodes[curr][x&1]
                    self.__cnts[curr] += d
                        
            def query(self, num):
                result = curr = 0
                for i in reversed(xrange(self.__bit_length)):
                    result <<= 1
                    x = num>>i
                    if self.__nodes[curr][1^(x&1)] != -1 and self.__cnts[self.__nodes[curr][1^(x&1)]]:
                        curr = self.__nodes[curr][1^(x&1)]
                        result |= 1
                    else:
                        curr = self.__nodes[curr][x&1]
                return result
    
        nums.sort()
        trie = Trie(nums[-1].bit_length())
        result = j = 0
        for i, num in enumerate(nums):
            trie.update(num, +1)
            while not (nums[i] <= 2*nums[j]) :
                trie.update(nums[j], -1)
                j += 1
            result = max(result, trie.query(num))
        return result


# Time:  O(nlogr), r = max(nums)
# Space: O(t)
# bit manipulation, greedy, trie, dp
class Solution2(object):
    def maximumStrongPairXor(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        class Trie(object):
            def __init__(self, bit_length):
                self.__nodes = []
                self.__mins = []
                self.__maxs = []
                self.__new_node()
                self.__bit_length = bit_length
            
            def __new_node(self):
                self.__nodes.append([-1]*2)
                self.__mins.append(float("inf"))
                self.__maxs.append(float("-inf"))
                return len(self.__nodes)-1

            def insert(self, num):
                curr = 0
                for i in reversed(xrange(self.__bit_length)):
                    x = num>>i
                    if self.__nodes[curr][x&1] == -1:
                        self.__nodes[curr][x&1] = self.__new_node()
                    curr = self.__nodes[curr][x&1]
                    self.__mins[curr] = min(self.__mins[curr], num)
                    self.__maxs[curr] = max(self.__maxs[curr], num)
                        
            def query(self, num):
                result = curr = 0
                for i in reversed(xrange(self.__bit_length)):
                    result <<= 1
                    x = num>>i
                    y = (result|1)^x
                    assert(x != y) 
                    if (self.__nodes[curr][y&1] != -1 and
                        ((x > y and num <= 2*self.__maxs[self.__nodes[curr][y&1]]) or
                         (x < y and self.__mins[self.__nodes[curr][y&1]] <= 2*num))):
                        result |= 1
                        curr = self.__nodes[curr][y&1]
                    else:
                        curr = self.__nodes[curr][1^(y&1)]
                return result
    
        trie = Trie(max(nums).bit_length())
        result = 0
        for num in nums:
            trie.insert(num)
            result = max(result, trie.query(num))
        return result


# Time:  O(nlogr), r = max(nums)
# Space: O(n)
# bit manipulation, greedy, dp
class Solution3(object):
    def maximumStrongPairXor(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        result = 0
        for i in reversed(xrange(max(nums).bit_length())):
            prefix_min, prefix_max = {}, {}
            for x in nums:
                y = x>>i
                if y not in prefix_min:
                    prefix_min[y] = prefix_max[y] = x
                prefix_min[y] = min(prefix_min[y], x)
                prefix_max[y] = max(prefix_max[y], x)
            result <<= 1
            for x in prefix_min.iterkeys():
                y = (result|1)^x
                assert(x != y)
                if y in prefix_max and prefix_min[max(x, y)] <= 2*prefix_max[min(x, y)]:
                    result |= 1
                    break
        return result


# Time:  O(n^2)
# Space: O(1)
# bit manipulation, brute force
class Solution4(object):
    def maximumStrongPairXor(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        return max(nums[i]^nums[j] for i in xrange(len(nums)) for j in xrange(i, len(nums)) if abs(nums[i]-nums[j]) <= min(nums[i], nums[j]))

Solution from kamyu104/LeetCode-Solutions · MIT