What is the time complexity of Maximum Genetic Difference Query?

The optimal solution has a time complexity of O(N*L + Q*L).

What is the space complexity of Maximum Genetic Difference Query?

The space complexity is O(N*L).

What data structures are used to solve Maximum Genetic Difference Query?

This problem uses Array, Hash Table, Bit Manipulation, Depth-First Search, Trie. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Genetic Difference Query?

Insight 1: The core operation is maximizing the XOR value. A Trie data structure is highly suitable for this because it allows efficient searching for numbers with the most different bits. Insight 2: We need to consider only the nodes on the path from a node to the root. A Depth-First Search (DFS) traversal is perfect for exploring the tree and maintaining the path from the root to the current node.

Which companies ask LeetCode 1938. Maximum Genetic Difference Query in interviews?

This problem is frequently asked at Media.net.

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Maximum Genetic Difference Query - LeetCode 1938 Solution

Maximum Genetic Difference Query - Complete Solution Guide

Maximum Genetic Difference Query is LeetCode problem 1938, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

There is a rooted tree consisting of n nodes numbered 0 to n - 1 . Each node's number denotes its unique genetic value (i.e. the genetic value of node x is x ). The genetic difference between two genetic values is defined as the bitwise- XOR of their values. You are given the integer array parents , where parents[i] is the parent for node i . If node x is the root of the tree, then parents[x] == -1 . You are also given the array queries where queries[i] = [node i , val i ] . For each query i , f

Detailed Explanation

The problem asks us to find the maximum genetic difference between a given value `val` and any node's genetic value on the path from a given node to the root of a tree. The genetic difference is defined as the bitwise XOR of the two values. We are given the tree structure through the `parents` array, where `parents[i]` is the parent of node `i`, and the queries in the `queries` array, where each query is a pair `[node, val]`. The output should be an array of the maximum genetic differences for each query.

Solution Approach

The solution uses a Trie to store the genetic values of the nodes on the path from the root to the current node during a DFS traversal of the tree. For each query `[node, val]`, the DFS reaches the specified `node`. At this point, the Trie contains the genetic values of all nodes on the path from the root to `node` (inclusive). The solution then queries the Trie to find the value that maximizes the XOR with `val`. After processing the queries for a node, the node's genetic value is removed from the Trie before exploring its children. This ensures that when the DFS backtracks, only the values on the correct path are present in the Trie.

Step-by-Step Algorithm

Step 1: Build the adjacency list representation of the tree from the `parents` array. Also, identify the root node.
Step 2: Create a list of queries for each node to efficiently process all queries when a node is visited during the DFS.
Step 3: Initialize a Trie data structure to store genetic values.
Step 4: Perform a DFS traversal starting from the root node.
Step 5: In the DFS function, insert the current node's genetic value into the Trie.
Step 6: Process the queries associated with the current node by querying the Trie for the maximum XOR value for each query and storing the result in the `ans` array.
Step 7: Recursively call the DFS function for each child of the current node.
Step 8: After processing the children, remove the current node's genetic value from the Trie before returning from the DFS function.

Key Insights

Insight 1: The core operation is maximizing the XOR value. A Trie data structure is highly suitable for this because it allows efficient searching for numbers with the most different bits.
Insight 2: We need to consider only the nodes on the path from a node to the root. A Depth-First Search (DFS) traversal is perfect for exploring the tree and maintaining the path from the root to the current node.
Insight 3: Inserting and removing nodes from the Trie during the DFS is crucial. This ensures that for each query, only the genetic values on the path from the query node to the root are considered when finding the maximum XOR.

Complexity Analysis

Time Complexity: O(N*L + Q*L)

Space Complexity: O(N*L)

Topics

This problem involves: Array, Hash Table, Bit Manipulation, Depth-First Search, Trie.

Companies

Asked at: Media.net.

Maximum Genetic Difference Query | LeetCode 1938

Hard

Array Hash Table Bit Manipulation Depth-First Search Trie

Time: O(N*L + Q*L)Space: O(N*L)

Solution Code

Available in:

import java.util.*;

class Solution {
    public int[] maxGeneticDifference(int[] parents, int[][] queries) {
        class TrieNode {
            Map<Integer, TrieNode> children;
            int count;

            public TrieNode() {
                this.children = new HashMap<>();
                this.count = 0;
            }
        }

        class Trie {
            TrieNode root;
            int L;

            public Trie(int L) {
                this.root = new TrieNode();
                this.L = L;
            }

            public void insert(int num) {
                TrieNode node = this.root;
                for (int i = this.L - 1; i >= 0; i--) {
                    int bit = (num >> i) & 1;
                    if (!node.children.containsKey(bit)) {
                        node.children.put(bit, new TrieNode());
                    }
                    node = node.children.get(bit);
                    node.count++;
                }
            }

            public void remove(int num) {
                TrieNode node = this.root;
                for (int i = this.L - 1; i >= 0; i--) {
                    int bit = (num >> i) & 1;
                    node = node.children.get(bit);
                    node.count--;
                }
            }

            public int find_max_xor(int num) {
                TrieNode node = this.root;
                int max_xor = 0;
                for (int i = this.L - 1; i >= 0; i--) {
                    int bit = (num >> i) & 1;
                    int opposite_bit = 1 - bit;

                    if (node.children.containsKey(opposite_bit) && node.children.get(opposite_bit).count > 0) {
                        max_xor |= (1 << i);
                        node = node.children.get(opposite_bit);
                    } else {
                        node = node.children.get(bit);
                    }
                }
                return max_xor;
            }
        }

        int n = parents.length;
        List<List<Integer>> adj = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            adj.add(new ArrayList<>());
        }
        int root = -1;
        for (int i = 0; i < n; i++) {
            int p = parents[i];
            if (p == -1) {
                root = i;
            } else {
                adj.get(p).add(i);
            }
        }

        List<List<Pair>> queries_by_node = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            queries_by_node.add(new ArrayList<>());
        }

        for (int i = 0; i < queries.length; i++) {
            int node = queries[i][0];
            int val = queries[i][1];
            queries_by_node.get(node).add(new Pair(val, i));
        }

        int[] ans = new int[queries.length];

        int L = 18;
        Trie trie = new Trie(L);

        class Pair {
            int val;
            int idx;

            public Pair(int val, int idx) {
                this.val = val;
                this.idx = idx;
            }
        }

        function dfs = new function() {
            public void apply(int u) {
                trie.insert(u);

                for (Pair p : queries_by_node.get(u)) {
                    ans[p.idx] = trie.find_max_xor(p.val);
                }

                for (int v : adj.get(u)) {
                    apply(v);
                }

                trie.remove(u);
            }
        };

        dfs.apply(root);

        return ans;
    }
    
    interface function {
        void apply(int u);
    }
}

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Problem Info

Description

Problem 1938: Maximum Genetic Difference Query - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Depth-First Search Trie

Companies

Media.net

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:1938

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Build the adjacency list representation of the tree from the `parents` array. Also, identify the root node.
2
Step 2: Create a list of queries for each node to efficiently process all queries when a node is visited during the DFS.
3
Step 3: Initialize a Trie data structure to store genetic values.
4
Step 4: Perform a DFS traversal starting from the root node.
5
Step 5: In the DFS function, insert the current node's genetic value into the Trie.
6
Step 6: Process the queries associated with the current node by querying the Trie for the maximum XOR value for each query and storing the result in the `ans` array.
7
Step 7: Recursively call the DFS function for each child of the current node.
8
Step 8: After processing the children, remove the current node's genetic value from the Trie before returning from the DFS function.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(N*L + Q*L)

💾

Space Complexity

O(N*L)

📝

Why?

The time complexity is determined by the DFS traversal and the Trie operations. The space complexity is mainly determined by the size of the Trie, which depends on the number of nodes and the length of the binary representation of the genetic values.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

For each query, traverse the path from the given node to the root, calculate the XOR with each node on the path, and find the maximum XOR value. This approach doesn't utilize any special data structures.

Time: O(N*Q), where N is the number of nodes and Q is the number of queries. In the worst case, we might have to traverse the entire tree for each query.Space: O(1), if path is traced back during execution, O(N) to store the full path to the root.

Trade-offs: This approach is much simpler to implement but has a significantly higher time complexity, making it unsuitable for large inputs. It is only viable if the constraints are very small.

✨

Key Insights

1Insight 1: The core operation is maximizing the XOR value. A Trie data structure is highly suitable for this because it allows efficient searching for numbers with the most different bits.
2Insight 2: We need to consider only the nodes on the path from a node to the root. A Depth-First Search (DFS) traversal is perfect for exploring the tree and maintaining the path from the root to the current node.
3Insight 3: Inserting and removing nodes from the Trie during the DFS is crucial. This ensures that for each query, only the genetic values on the path from the query node to the root are considered when finding the maximum XOR.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Hash Table

A data structure that maps keys to values for quick lookup. Like a dictionary!

Edge Cases

Edge case 1: A single-node tree (only the root node). The algorithm handles this correctly because the DFS will only visit the root node and process any queries associated with it.
Edge case 2: A skewed tree (all nodes are in a single path). The algorithm will correctly insert and remove nodes from the Trie as it traverses the path, ensuring that only the nodes on the path from the current node to the root are considered.
Edge case 3: Queries on the root node itself. The code correctly handles this because when the DFS reaches the root node, its value is inserted into the trie, and the queries associated to the root will be correctly computed.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not removing nodes from the Trie after processing a subtree. This would result in incorrect XOR calculations because the Trie would contain values from paths that are not relevant to the current query.

✗Common mistake 2: Incorrectly constructing the tree from the `parents` array. The code must correctly identify the root node and create the appropriate adjacency list to represent the parent-child relationships.

✗Common mistake 3: Integer overflow. Although the values in queries can reach 2 * 10^5, during the xor operation, it should not overflow, considering that the maximum number of bits is fixed to 18.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Network routing. Finding the path with the maximum difference in network addresses (represented as bit strings) can be useful for optimizing routing protocols.

🏦Real-world scenario 2: Data compression. XOR operations are used in various data compression algorithms to reduce redundancy and improve storage efficiency. Finding values that maximize the XOR difference can be useful in this context.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: October 17, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

What is the most efficient way to solve array problems?

Most array problems can be solved efficiently using techniques like two-pointer approach, sliding window, or hash maps. The key is identifying whether the array is sorted (enabling binary search) or if tracking seen elements (using hash maps) can help achieve O(n) time complexity.

What are the trade-offs of using a hash table?

Hash tables provide O(1) average-case lookup, insertion, and deletion, making them ideal for frequency counting and duplicate detection. However, they use O(n) extra space and may degrade to O(n) time in worst-case scenarios with hash collisions. Always consider if the space trade-off is acceptable.

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Maximum Genetic Difference Query"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether hash table techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?