How do you solve Maximum Genetic Difference Query on LeetCode?

Maximum Genetic Difference Query (LeetCode #1938) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + q * 32) time complexity and O(n) space complexity. Key insight: Using a Trie can significantly optimize the retrieval of maximum XOR values.

What is the brute force solution for Maximum Genetic Difference Query?

The brute force approach for Maximum Genetic Difference Query is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves traversing the path from each queried node to the root, calculating the XOR with the given value for each node along the path, and keeping track of the maximum XOR found. This method is straightforward but inefficient for larger trees. The optimal Optimal Solution approach improves this to O(n + q * 32).

What are the interview tips for Maximum Genetic Difference Query?

Explain your thought process clearly while coding. Discuss edge cases and how your solution handles them. Practice similar problems to get comfortable with bit manipulation.

What are common mistakes when solving Maximum Genetic Difference Query?

Not considering all nodes on the path to the root. Forgetting to handle the case when the node is the root.

#1938

Maximum Genetic Difference Query

Hard

ArrayHash TableBit ManipulationDepth-First SearchTrieTrieBit Manipulation

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + q * 32)
Space	O(1)	O(n)

💡

Intuition

Time O(n + q * 32)Space O(n)

The optimal solution uses a Trie to store the XOR values of all nodes on the path from each queried node to the root. This allows for efficient retrieval of the maximum XOR value for each query by leveraging the properties of binary numbers.

⚙️

Algorithm

3 steps

1Step 1: Construct a Trie to store the XOR values of nodes as we traverse from each node to the root.
2Step 2: For each query, insert the XOR values into the Trie while traversing to the root.
3Step 3: For each query, search the Trie to find the maximum XOR value with the given value.

solution.py42 lines

1class TrieNode:
2    def __init__(self):
3        self.children = {}
4        self.value = None
5
6class Trie:
7    def __init__(self):
8        self.root = TrieNode()
9
10    def insert(self, num):
11        node = self.root
12        for i in range(31, -1, -1):
13            bit = (num >> i) & 1
14            if bit not in node.children:
15                node.children[bit] = TrieNode()
16            node = node.children[bit]
17        node.value = num
18
19    def maxXor(self, num):
20        node = self.root
21        max_xor = 0
22        for i in range(31, -1, -1):
23            bit = (num >> i) & 1
24            toggle_bit = 1 - bit
25            if toggle_bit in node.children:
26                max_xor |= (1 << i)
27                node = node.children[toggle_bit]
28            else:
29                node = node.children[bit]
30        return max_xor
31
32
33def maxGeneticDifference(parents, queries):
34    trie = Trie()
35    results = []
36    for node in range(len(parents)):
37        xor_value = node
38        trie.insert(xor_value)
39    for node, val in queries:
40        max_xor = trie.maxXor(val)
41        results.append(max_xor)
42    return results

ℹ

Complexity note: The time complexity is O(n + q * 32) where n is the number of nodes and q is the number of queries. Each insertion into the Trie takes O(32) time, and we do this for each node, plus we query the Trie for each query.

1Using a Trie can significantly optimize the retrieval of maximum XOR values.
2Understanding bit manipulation is crucial for solving XOR-related problems.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.