What is the time complexity of Maximum Strong Pair XOR II?

The optimal solution has a time complexity of O(n*log(n)).

What data structures are used to solve Maximum Strong Pair XOR II?

This problem uses Array, Hash Table, Bit Manipulation, Trie, Sliding Window. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Maximum Strong Pair XOR II?

Insight 1: Sorting the array allows the use of a sliding window approach to efficiently identify strong pairs. Insight 2: A Trie data structure is highly effective for finding the maximum XOR value with a given number within a set of numbers.

Which companies ask LeetCode 2935. Maximum Strong Pair XOR II in interviews?

This problem is frequently asked at ZScaler.

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Maximum Strong Pair XOR II - LeetCode 2935 Solution

Q: How to solve Maximum Strong Pair XOR II?

The provided solution utilizes a combination of sorting, a sliding window, and a Trie data structure. First, the input array `nums` is sorted. Then, a sliding window is maintained using two pointers, `l` (left) and `r` (right). The right pointer iterates through the sorted array, while the left pointer adjusts to ensure that the numbers within the window can form strong pairs with the number at the right pointer. A Trie is used to store the numbers within the sliding window. For each number `y` at the right pointer, the algorithm finds the maximum XOR value between `y` and any number currently in the Trie using Trie's query_max_xor function. Numbers outside of the condition |x - y| <= min(x, y) i.e. nums[l] * 2 < y are removed from trie, shrinking the sliding window. Finally, the algorithm returns the maximum XOR value found during the iteration.

Maximum Strong Pair XOR II - Complete Solution Guide

Maximum Strong Pair XOR II is LeetCode problem 2935, 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

You are given a 0-indexed integer array nums . A pair of integers x and y is called a strong pair if it satisfies the condition: |x - y| <= min(x, y) You need to select two integers from nums such that they form a strong pair and their bitwise XOR is the maximum among all strong pairs in the array. Return the maximum XOR value out of all possible strong pairs in the array nums . Note that you can pick the same integer twice to form a pair. Example 1: Input: nums = [1,2,3,4,5] Output: 7 Explanati

Detailed Explanation

The problem requires finding the maximum XOR value between any two numbers in a given array `nums` that form a 'strong pair'. A strong pair (x, y) satisfies the condition |x - y| <= min(x, y). The goal is to iterate through the array, identify strong pairs, calculate their XOR values, and return the maximum XOR value found. The same number can be picked twice to form a pair, which simplifies some edge cases but doesn't change the fundamental complexity.

Solution Approach

The provided solution utilizes a combination of sorting, a sliding window, and a Trie data structure. First, the input array `nums` is sorted. Then, a sliding window is maintained using two pointers, `l` (left) and `r` (right). The right pointer iterates through the sorted array, while the left pointer adjusts to ensure that the numbers within the window can form strong pairs with the number at the right pointer. A Trie is used to store the numbers within the sliding window. For each number `y` at the right pointer, the algorithm finds the maximum XOR value between `y` and any number currently in the Trie using Trie's query_max_xor function. Numbers outside of the condition |x - y| <= min(x, y) i.e. nums[l] * 2 < y are removed from trie, shrinking the sliding window. Finally, the algorithm returns the maximum XOR value found during the iteration.

Step-by-Step Algorithm

Step 1: Sort the input array `nums` in ascending order.
Step 2: Initialize a Trie data structure and a variable `max_xor_val` to 0.
Step 3: Initialize two pointers, `l` and `r`, to 0. `l` represents the left boundary of the sliding window, and `r` the right.
Step 4: Iterate through the sorted array with the right pointer `r` from 0 to the end of `nums`.
Step 5: For each `nums[r]` (denoted as `y`), insert it into the Trie.
Step 6: While `nums[l] * 2 < y`, remove `nums[l]` from the Trie and increment `l`. This ensures the condition |x - y| <= min(x,y) is met which simplifies to x*2 >= y for strong pair property check given x <= y (because nums is sorted)
Step 7: Query the Trie for the maximum XOR value with `y` and update `max_xor_val` if a larger value is found.
Step 8: After iterating through all numbers in `nums`, return `max_xor_val`.

Key Insights

Insight 1: Sorting the array allows the use of a sliding window approach to efficiently identify strong pairs.
Insight 2: A Trie data structure is highly effective for finding the maximum XOR value with a given number within a set of numbers.
Insight 3: The crucial optimization is efficiently maintaining the Trie with only the numbers that can form a strong pair with the current number under consideration. This limits the size of the Trie and improves the time complexity.

Complexity Analysis

Time Complexity: O(n*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Hash Table, Bit Manipulation, Trie, Sliding Window.

Companies

Asked at: ZScaler.

Maximum Strong Pair XOR II | LeetCode 2935

Hard

Array Hash Table Bit Manipulation Trie Sliding Window

Time: O(n*log(n))Space: O(n)

Solution Code

Available in:

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;

class Solution {
    public int maximumStrongPairXor(int[] nums) {
        final int MAX_BITS = 20;

        class TrieNode {
            Map<Integer, TrieNode> children;
            int count;

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

        class Trie {
            TrieNode root;

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

            public void insert(int num) {
                TrieNode node = this.root;
                node.count++;
                for (int i = MAX_BITS - 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;
                node.count--;
                for (int i = MAX_BITS - 1; i >= 0; i--) {
                    int bit = (num >> i) & 1;
                    TrieNode child = node.children.get(bit);
                    child.count--;
                    if (child.count == 0) {
                        node.children.remove(bit);
                        return;
                    }
                    node = child;
                }
            }

            public int query_max_xor(int num) {
                if (this.root.count == 0) {
                    return 0;
                }

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

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

        Arrays.sort(nums);
        Trie trie = new Trie();
        int max_xor_val = 0;
        int l = 0;
        for (int r = 0; r < nums.length; r++) {
            int y = nums[r];
            trie.insert(y);

            while (nums[l] * 2 < y) {
                trie.remove(nums[l]);
                l++;
            }

            max_xor_val = Math.max(max_xor_val, trie.query_max_xor(y));

        }
        return max_xor_val;
    }
}

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Problem Info

Description

Problem 2935: Maximum Strong Pair XOR II - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Trie Sliding Window

Companies

ZScaler

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2935

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: Sort the input array `nums` in ascending order.
2
Step 2: Initialize a Trie data structure and a variable `max_xor_val` to 0.
3
Step 3: Initialize two pointers, `l` and `r`, to 0. `l` represents the left boundary of the sliding window, and `r` the right.
4
Step 4: Iterate through the sorted array with the right pointer `r` from 0 to the end of `nums`.
5
Step 5: For each `nums[r]` (denoted as `y`), insert it into the Trie.
6
Step 6: While `nums[l] * 2 < y`, remove `nums[l]` from the Trie and increment `l`. This ensures the condition |x - y| <= min(x,y) is met which simplifies to x*2 >= y for strong pair property check given x <= y (because nums is sorted)
7
Step 7: Query the Trie for the maximum XOR value with `y` and update `max_xor_val` if a larger value is found.
8
Step 8: After iterating through all numbers in `nums`, return `max_xor_val`.

💡

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*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting the array takes O(n*log(n)) time. The outer loop iterates `n` times. Inside the loop, inserting and removing from the Trie and querying the Trie each take O(log(MAX_NUM)) time, where MAX_NUM is the maximum number in the array. Since MAX_NUM is limited to 2^20 - 1, log(MAX_NUM) is effectively a constant (20). The sliding window's while loop in the outer loop takes O(n) time in total across all iterations because l can only increment. Therefore, the overall time complexity is dominated by the sorting step O(n*log(n)). Space complexity is determined by the Trie, which can, in the worst case, hold all `n` numbers in `nums`, resulting in O(n) space complexity. The sorting algorithm used in-place could take O(log n) extra space

📖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

Iterate through all possible pairs of numbers in the array, check if they form a strong pair, and calculate their XOR value. Keep track of the maximum XOR value found.

Time: O(n^2)Space: O(1)

Trade-offs: This approach is simple to implement but very inefficient for larger arrays. It's only suitable for small input sizes. It avoids the overhead of sorting and the Trie data structure but has a much higher time complexity.

✨

Key Insights

1Insight 1: Sorting the array allows the use of a sliding window approach to efficiently identify strong pairs.
2Insight 2: A Trie data structure is highly effective for finding the maximum XOR value with a given number within a set of numbers.
3Insight 3: The crucial optimization is efficiently maintaining the Trie with only the numbers that can form a strong pair with the current number under consideration. This limits the size of the Trie and improves the time complexity.

📚

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!

Sliding Window

A moving "window" over data to process subsets efficiently.

Edge Cases

Edge case 1: Empty array. The provided code will handle this case gracefully as the loop won't execute, and the initial `max_xor_val` (0) will be returned.
Edge case 2: Array with only one element. The loop will execute once, the number will be inserted into the trie, and since the trie can only pair with itself it will result in value xor itself and return 0 which is correct.
Edge case 3: Array with duplicate elements. The solution works correctly because the Trie handles duplicates by incrementing the `count` attribute of the nodes.
Edge case 4: Large range of numbers (close to the constraint limit). The code considers the `MAX_BITS` constraint (20) correctly, ensuring that the Trie implementation will handle numbers up to 2^20 - 1.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the array. Without sorting, it's much harder to efficiently find strong pairs, and you'd likely resort to a brute-force approach.

✗Common mistake 2: Incorrectly implementing the Trie data structure. A faulty Trie implementation can lead to incorrect XOR calculations and affect the correctness of the overall solution.

✗Common mistake 3: Not handling the sliding window correctly, leading to invalid pairs being considered.

✗Common mistake 4: Memory leaks in C implementation, which can happen when allocating TrieNodes but not freeing them after use.

✅

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: Data compression. Finding similar data patterns (which can be represented by low XOR values) can be used to compress data more efficiently.

🏦Real-world scenario 2: Anomaly detection. Identifying data points that have significantly different bit patterns (high XOR values) from the rest of the dataset can indicate anomalies or outliers in network traffic analysis or fraud detection systems.

🎮Real-world scenario 3: Bioinformatics. Comparing genetic sequences to identify differences. XOR can highlight the differences which can be used in phylogenetic analysis or disease research.

💡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

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

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

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: July 26, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 14

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 7

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

How do I choose between a hash map and a hash set?

Use a hash set when you only need to track presence/absence of elements (like detecting duplicates). Use a hash map when you need to associate values with keys (like counting frequencies or storing indices). Hash sets use less memory since they only store keys.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Maximum Strong Pair XOR II"?

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?