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

The optimal solution has a time complexity of O(n^2).

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

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

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

Insight 1: The strong pair condition |x - y| ≤ min(x, y) implies that the two numbers in the pair are relatively close to each other. This doesn't significantly reduce the search space but helps in understanding the nature of the pairs. Insight 2: The brute-force approach of checking all possible pairs is sufficient given the constraints (array size ≤ 50). More efficient algorithms (like those using tries or specialized data structures) aren't necessary due to the small input size.

Which companies ask LeetCode 2932. Maximum Strong Pair XOR I in interviews?

This problem is frequently asked at ZScaler.

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Maximum Strong Pair XOR I - LeetCode 2932 Solution

Q: How to solve Maximum Strong Pair XOR I?

The provided code uses a brute-force approach. It iterates through all possible pairs of numbers in the input array `nums`. For each pair, it checks if the pair is strong according to the given condition. If it is, it calculates the bitwise XOR of the pair and updates the `max_xor` variable if the current XOR is greater. Finally, it returns the `max_xor` value.

Maximum Strong Pair XOR I - Complete Solution Guide

Maximum Strong Pair XOR I is LeetCode problem 2932, a Easy 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 asks you to find the maximum bitwise XOR result from pairs of numbers in an input array that satisfy a specific condition. A pair (x, y) is considered "strong" if the absolute difference between x and y is less than or equal to the minimum of x and y (|x - y| ≤ min(x, y)). The goal is to find the maximum XOR value among all such strong pairs. The input is a 0-indexed integer array `nums`, and the output is a single integer representing the maximum XOR value.

Solution Approach

The provided code uses a brute-force approach. It iterates through all possible pairs of numbers in the input array `nums`. For each pair, it checks if the pair is strong according to the given condition. If it is, it calculates the bitwise XOR of the pair and updates the `max_xor` variable if the current XOR is greater. Finally, it returns the `max_xor` value.

Step-by-Step Algorithm

Step 1: Initialize `max_xor` to 0. This variable will store the maximum XOR value found so far.
Step 2: Use nested loops to iterate through all possible pairs of numbers in the `nums` array (including pairs with the same number).
Step 3: For each pair (x, y), check if it is a strong pair using the condition `abs(x - y) <= min(x, y)`.
Step 4: If the pair is strong, calculate the bitwise XOR (x ^ y) and update `max_xor` if this XOR value is greater than the current `max_xor`.
Step 5: After checking all pairs, return the final `max_xor` value.

Key Insights

Insight 1: The strong pair condition |x - y| ≤ min(x, y) implies that the two numbers in the pair are relatively close to each other. This doesn't significantly reduce the search space but helps in understanding the nature of the pairs.
Insight 2: The brute-force approach of checking all possible pairs is sufficient given the constraints (array size ≤ 50). More efficient algorithms (like those using tries or specialized data structures) aren't necessary due to the small input size.
Insight 3: The XOR operation is commutative (x ^ y = y ^ x), so the order of elements in a pair doesn't matter. This doesn't change the algorithm but is worth noting for optimization considerations in larger datasets.

Complexity Analysis

Time Complexity: O(n^2)

Space Complexity: O(1)

Topics

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

Companies

Asked at: ZScaler.

Maximum Strong Pair XOR I | LeetCode 2932

Easy

Array Hash Table Bit Manipulation Trie Sliding Window

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

Solution Code

Available in:

class Solution {
    public int maximumStrongPairXor(int[] nums) {
        int maxXor = 0;
        for (int i = 0; i < nums.length; i++) {
            for (int j = 0; j < nums.length; j++) {
                int x = nums[i];
                int y = nums[j];
                if (Math.abs(x - y) <= Math.min(x, y)) {
                    maxXor = Math.max(maxXor, x ^ y);
                }
            }
        }
        return maxXor;
    }
}

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Problem Info

Description

Problem 2932: Maximum Strong Pair XOR I - Detailed explanation and solution approaches.

Topics

Array Hash Table Bit Manipulation Trie Sliding Window

Companies

ZScaler

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:2932

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

5 steps

1
Step 1: Initialize `max_xor` to 0. This variable will store the maximum XOR value found so far.
2
Step 2: Use nested loops to iterate through all possible pairs of numbers in the `nums` array (including pairs with the same number).
3
Step 3: For each pair (x, y), check if it is a strong pair using the condition `abs(x - y) <= min(x, y)`.
4
Step 4: If the pair is strong, calculate the bitwise XOR (x ^ y) and update `max_xor` if this XOR value is greater than the current `max_xor`.
5
Step 5: After checking all pairs, return the final `max_xor` value.

💡

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^2)

🐢 Quadratic - Gets slow with large inputs. Consider optimization!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The nested loops iterate through all possible pairs of numbers in the array, resulting in O(n^2) time complexity. The algorithm uses only a constant amount of extra space to store variables like `max_xor`, resulting in O(1) space complexity.

📖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

Optimized Brute Force (with early exit)

Instead of checking all pairs, we can optimize by sorting the array and then iterating through pairs in a way that limits the number of comparisons (e.g. only compare numbers within a certain range based on the min(x,y) condition).

Time: Potentially slightly better than O(n^2) in some cases, but still O(n^2) in the worst case.Space: O(1) or O(n) depending on sorting algorithm used.

Trade-offs: Offers marginal improvement for larger input sizes but adds complexity. For the given constraints, the original solution is simpler and sufficient.

✨

Key Insights

1Insight 1: The strong pair condition |x - y| ≤ min(x, y) implies that the two numbers in the pair are relatively close to each other. This doesn't significantly reduce the search space but helps in understanding the nature of the pairs.
2Insight 2: The brute-force approach of checking all possible pairs is sufficient given the constraints (array size ≤ 50). More efficient algorithms (like those using tries or specialized data structures) aren't necessary due to the small input size.
3Insight 3: The XOR operation is commutative (x ^ y = y ^ x), so the order of elements in a pair doesn't matter. This doesn't change the algorithm but is worth noting for optimization considerations in larger datasets.

📚

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: An array with only one element. The solution correctly handles this by forming a strong pair with the element itself (x, x), resulting in an XOR of 0.
Edge case 2: An array with all identical numbers. The solution correctly identifies that the maximum XOR will be 0 (x ^ x = 0 for any x).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly implementing the strong pair condition. Pay close attention to the absolute value and the minimum function.

✗Common mistake 2: Forgetting to handle the case where the array is empty or contains only one element.

✅

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: This type of problem might be encountered in network security or cryptography, where analyzing pairs of data with certain relationships (like proximity) and their XOR values might be important.

🏦Real-world scenario 2: In data analysis, finding pairs with specific properties (like being close in value) and then performing bitwise operations could be useful for feature engineering or pattern recognition.

💡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

Use this problem to demonstrate clean code practices. Variable naming and code organization matter.

⚠️ 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: July 23, 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

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

@InterviewProJul 14

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

@CodeNinjaJul 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 should I approach easy-level algorithm problems?

Easy problems build foundational skills. Focus on: 1) Understanding the problem completely before coding, 2) Writing clean, readable code, 3) Handling all edge cases (empty input, single element, etc.), 4) Analyzing time and space complexity. Don't skip complexity analysis even for simple solutions.

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 Strong Pair XOR I"?

For this easy-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.

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