What is the time complexity of Magnetic Force Between Two Balls?

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

What data structures are used to solve Magnetic Force Between Two Balls?

This problem uses Array, Binary Search, Sorting. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Magnetic Force Between Two Balls?

Insight 1: The problem is an optimization problem, aiming to maximize a minimum value, suggesting a binary search approach. Insight 2: The feasibility of a given minimum distance can be checked greedily. Given a minimum distance 'dist', we can place the balls one by one, always placing the next ball in the closest available basket that is at least 'dist' away from the previously placed ball.

Which companies ask LeetCode 1552. Magnetic Force Between Two Balls in interviews?

This problem is frequently asked at PhonePe, Roblox.

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Magnetic Force Between Two Balls - LeetCode 1552 Solution

Magnetic Force Between Two Balls - Complete Solution Guide

Magnetic Force Between Two Balls is LeetCode problem 1552, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

In the universe Earth C-137, Rick discovered a special form of magnetic force between two balls if they are put in his new invented basket. Rick has n empty baskets, the i th basket is at position[i] , Morty has m balls and needs to distribute the balls into the baskets such that the minimum magnetic force between any two balls is maximum . Rick stated that magnetic force between two different balls at positions x and y is |x - y| . Given the integer array position and the integer m . Return the

Detailed Explanation

The problem asks us to maximize the minimum magnetic force between 'm' balls placed in 'n' baskets located at different positions. The magnetic force between any two balls is the absolute difference of their positions. We are given an array 'position' representing the locations of the baskets and an integer 'm' representing the number of balls. The goal is to find the largest possible minimum distance between any two balls when we place all 'm' balls into the baskets.

Solution Approach

The solution uses a binary search algorithm to find the maximum possible minimum magnetic force. First, the 'position' array is sorted. Then, binary search is performed on the range of possible distances, from 1 (minimum possible distance) to 'position[n-1] - position[0]' (maximum possible distance). For each candidate distance 'mid', a helper function 'is_feasible' greedily checks if it's possible to place all 'm' balls such that the minimum distance between any two balls is at least 'mid'. If 'is_feasible' returns true, it means that 'mid' is a valid minimum distance, and we try to increase the distance by searching in the upper half of the range. Otherwise, 'mid' is too large, and we need to reduce the distance by searching in the lower half of the range. The binary search continues until the lower and upper bounds converge, and the last feasible distance found is the result.

Step-by-Step Algorithm

Step 1: Sort the 'position' array in ascending order.
Step 2: Initialize 'low' to 1 and 'high' to 'position[n-1] - position[0]', where n is the length of 'position'.
Step 3: Initialize 'result' to 0. This will store the maximum feasible distance found so far.
Step 4: While 'low' is less than or equal to 'high', calculate the middle value 'mid' as 'low + (high - low) // 2'.
Step 5: Call the 'is_feasible' function with 'mid' as the minimum distance.
Step 6: If 'is_feasible(mid)' returns true, update 'result' to 'mid' and move 'low' to 'mid + 1' to search for a larger feasible distance.
Step 7: If 'is_feasible(mid)' returns false, move 'high' to 'mid - 1' to search for a smaller feasible distance.
Step 8: After the while loop finishes, return 'result'.

Key Insights

Insight 1: The problem is an optimization problem, aiming to maximize a minimum value, suggesting a binary search approach.
Insight 2: The feasibility of a given minimum distance can be checked greedily. Given a minimum distance 'dist', we can place the balls one by one, always placing the next ball in the closest available basket that is at least 'dist' away from the previously placed ball.
Insight 3: Sorting the positions allows for efficient binary search and greedy feasibility checks.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(1)

Topics

This problem involves: Array, Binary Search, Sorting.

Companies

Asked at: PhonePe, Roblox.

Magnetic Force Between Two Balls | LeetCode 1552

Medium

Array Binary Search Sorting

Time: O(n log n)Space: O(1)

Solution Code

Available in:

import java.util.Arrays;

class Solution {
    public int maxDistance(int[] position, int m) {
        Arrays.sort(position);
        int n = position.length;

        // Define a helper function to check if a given distance is feasible
        class Helper {
            public boolean isFeasible(int dist) {
                // Greedily check if m balls can be placed with at least dist distance apart.
                int count = 1;
                int last_pos = position[0];
                for (int i = 1; i < n; i++) {
                    if (position[i] - last_pos >= dist) {
                        count += 1;
                        last_pos = position[i];
                    }
                }
                return count >= m;
            }
        }

        Helper helper = new Helper();

        int low = 1;
        int high = position[n - 1] - position[0];
        int result = 0;

        while (low <= high) {
            int mid = low + (high - low) / 2;
            if (helper.isFeasible(mid)) {
                // This distance is feasible, try for a larger one
                result = mid;
                low = mid + 1;
            } else {
                // This distance is not feasible, need to reduce it
                high = mid - 1;
            }
        }

        return result;
    }
}

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Problem Info

Description

Problem 1552: Magnetic Force Between Two Balls - Detailed explanation and solution approaches.

Topics

Array Binary Search Sorting

Companies

PhonePeRoblox

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:1552

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 'position' array in ascending order.
2
Step 2: Initialize 'low' to 1 and 'high' to 'position[n-1] - position[0]', where n is the length of 'position'.
3
Step 3: Initialize 'result' to 0. This will store the maximum feasible distance found so far.
4
Step 4: While 'low' is less than or equal to 'high', calculate the middle value 'mid' as 'low + (high - low) // 2'.
5
Step 5: Call the 'is_feasible' function with 'mid' as the minimum distance.
6
Step 6: If 'is_feasible(mid)' returns true, update 'result' to 'mid' and move 'low' to 'mid + 1' to search for a larger feasible distance.
7
Step 7: If 'is_feasible(mid)' returns false, move 'high' to 'mid - 1' to search for a smaller feasible distance.
8
Step 8: After the while loop finishes, return 'result'.

💡

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)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is dominated by sorting the 'position' array, which takes O(n log n) time. The binary search loop runs in O(log(max(position) - min(position))) time, which can be simplified to O(log n) since the constraint on position values is at most 10^9 and there are n elements. The is_feasible function iterates through the sorted array in O(n) time. Hence the binary search contributes O(n log n). Since sorting also contributes O(n log n), the overall time complexity is O(n log n). The space complexity is O(1) because the solution sorts the array in-place and uses only a few constant extra variables.

📖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

Trying all possible combinations of 'm' baskets and calculating the minimum distance for each combination. Then take the maximum of the minimum distances.

Time: O(C(n, m) * m) which is O(n^m) in the worst caseSpace: O(m)

Trade-offs: This approach is not efficient and will time out for larger values of 'n' and 'm'. The main approach is better because of its logarithmic time complexity.

✨

Key Insights

1Insight 1: The problem is an optimization problem, aiming to maximize a minimum value, suggesting a binary search approach.
2Insight 2: The feasibility of a given minimum distance can be checked greedily. Given a minimum distance 'dist', we can place the balls one by one, always placing the next ball in the closest available basket that is at least 'dist' away from the previously placed ball.
3Insight 3: Sorting the positions allows for efficient binary search and greedy feasibility checks.

📚

Key Terms Explained

For Students

Array

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

Binary Search

Halving the search space each step. Like finding a word in a dictionary!

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: `m = 2`. The algorithm correctly handles the case where there are only two balls to place. It finds the two baskets that maximize the distance between them.
Edge case 2: All positions are close together. The binary search correctly narrows down the maximum minimum distance by adjusting the high pointer.
Edge case 3: Positions are widely spread. The algorithm searches for the largest minimum distance and appropriately adjusts the low pointer during the binary search process.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the 'position' array. Sorting is crucial for the greedy feasibility check to work correctly.

✗Common mistake 2: Incorrectly implementing the binary search. Off-by-one errors in calculating 'mid', updating 'low' and 'high', and the termination condition can lead to incorrect results.

✗Common mistake 3: Not handling integer overflow when calculating the 'mid' value. Using 'low + (high - low) // 2' prevents overflow.

✅

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: Placing cell towers to maximize coverage while minimizing interference between towers. The positions of the towers are given, and you want to place a certain number of towers such that the minimum distance between any two towers is maximized.

🏦Real-world scenario 2: Distributing servers in a data center to minimize network latency. You have a set of available locations for servers, and you need to place a certain number of servers such that the maximum distance between any two servers is minimized, or conversely, the minimum distance is maximized for redundancy reasons.

💡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

If stuck, try working through a small example by hand. The pattern often reveals itself.

⚠️ 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: September 21, 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.

How do I avoid off-by-one errors in binary search?

Use the pattern: left = 0, right = n-1, while (left <= right), mid = left + (right - left) / 2. Decide if you need to find exact match, first occurrence, or last occurrence, and adjust your conditions accordingly. Test with arrays of size 0, 1, and 2.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

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 "Magnetic Force Between Two Balls"?

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

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