What is the time complexity of Minimize Manhattan Distances?

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

What data structures are used to solve Minimize Manhattan Distances?

This problem uses Array, Math, Geometry, Sorting, Ordered Set. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Minimize Manhattan Distances?

Insight 1: The maximum Manhattan distance is determined by extreme points. Specifically, the maximum range of x+y and x-y values. Insight 2: Instead of calculating distances between all pairs of points after removing each point, it's more efficient to identify potential candidates for the point to remove. These candidates are the points that contribute to the extreme values of x+y and x-y.

Which companies ask LeetCode 3102. Minimize Manhattan Distances in interviews?

This problem is frequently asked at Deutsche Bank.

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Minimize Manhattan Distances - LeetCode 3102 Solution

Minimize Manhattan Distances - Complete Solution Guide

Minimize Manhattan Distances is LeetCode problem 3102, 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 an array points representing integer coordinates of some points on a 2D plane, where points[i] = [x i , y i ] . The distance between two points is defined as their Manhattan distance . Return the minimum possible value for maximum distance between any two points by removing exactly one point . Example 1: Input: points = [[3,10],[5,15],[10,2],[4,4]] Output: 12 Explanation: The maximum distance after removing each point is the following: After removing the 0 th point the maximum dist

Detailed Explanation

The problem requires finding the minimum possible value for the maximum Manhattan distance between any two points in a given set of 2D points after removing exactly one point. Manhattan distance between two points (x1, y1) and (x2, y2) is defined as |x1 - x2| + |y1 - y2|.

Solution Approach

The solution leverages the properties of Manhattan distance to optimize the search for the minimum possible maximum distance after removing one point. It focuses on finding the points that define the extreme ranges of `x + y` and `x - y` coordinates. By sorting the points based on these sums and differences, the algorithm identifies the points that are most likely to contribute to the maximum Manhattan distance. The algorithm then iterates through these candidate points, removes each one in turn, and calculates the new maximum Manhattan distance based on the remaining points. Finally, it returns the minimum of these maximum distances.

Step-by-Step Algorithm

Step 1: Calculate `x + y` and `x - y` for each point and store them along with their original index.
Step 2: Sort the points based on `x + y` and `x - y` values, creating two sorted lists.
Step 3: Identify the indices of the points with the minimum and maximum `x + y` values and the minimum and maximum `x - y` values. These indices are potential candidates to be removed.
Step 4: Iterate through the set of candidate indices.
Step 5: For each candidate index, simulate removing the corresponding point.
Step 6: After removing the point, determine the new minimum and maximum `x + y` and `x - y` values based on the remaining points.
Step 7: Calculate the maximum of the ranges of `x + y` and `x - y`. This represents the maximum Manhattan distance after removing the current candidate point.
Step 8: Update the minimum maximum distance found so far.
Step 9: Return the minimum maximum distance found after considering all candidate points.

Key Insights

Insight 1: The maximum Manhattan distance is determined by extreme points. Specifically, the maximum range of x+y and x-y values.
Insight 2: Instead of calculating distances between all pairs of points after removing each point, it's more efficient to identify potential candidates for the point to remove. These candidates are the points that contribute to the extreme values of x+y and x-y.
Insight 3: Sorting the points based on x+y and x-y allows for easy identification of the minimum and maximum values, and consequently, the candidate points to remove.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Math, Geometry, Sorting, Ordered Set.

Companies

Asked at: Deutsche Bank.

Minimize Manhattan Distances | LeetCode 3102

Hard

Array Math Geometry Sorting Ordered Set

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

Solution Code

Available in:

import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;

class Solution {
    public int minimumDistance(int[][] points) {
        int n = points.length;

        List<Pair> sums = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            sums.add(new Pair(points[i][0] + points[i][1], i));
        }
        sums.sort((a, b) -> a.val - b.val);

        List<Pair> diffs = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            diffs.add(new Pair(points[i][0] - points[i][1], i));
        }
        diffs.sort((a, b) -> a.val - b.val);

        Set<Integer> candidateIndices = new HashSet<>();
        candidateIndices.add(sums.get(0).index);
        candidateIndices.add(sums.get(n - 1).index);
        candidateIndices.add(diffs.get(0).index);
        candidateIndices.add(diffs.get(n - 1).index);

        int minMaxDist = Integer.MAX_VALUE;

        for (int removedIdx : candidateIndices) {
            int sMinVal = sums.get(0).val;
            int sMaxVal = sums.get(n - 1).val;

            if (sums.get(0).index == removedIdx) {
                sMinVal = sums.get(1).val;
            }

            if (sums.get(n - 1).index == removedIdx) {
                sMaxVal = sums.get(n - 2).val;
            }

            int maxURange = sMaxVal - sMinVal;

            int dMinVal = diffs.get(0).val;
            int dMaxVal = diffs.get(n - 1).val;

            if (diffs.get(0).index == removedIdx) {
                dMinVal = diffs.get(1).val;
            }

            if (diffs.get(n - 1).index == removedIdx) {
                dMaxVal = diffs.get(n - 2).val;
            }

            int maxVRange = dMaxVal - dMinVal;

            int currentMaxDist = Math.max(maxURange, maxVRange);
            minMaxDist = Math.min(minMaxDist, currentMaxDist);
        }

        return minMaxDist;
    }

    private static class Pair {
        int val;
        int index;

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

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Problem Info

Description

Problem 3102: Minimize Manhattan Distances - Detailed explanation and solution approaches.

Topics

Array Math Geometry Sorting Ordered Set

Companies

Deutsche Bank

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:3102

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

9 steps

1
Step 1: Calculate `x + y` and `x - y` for each point and store them along with their original index.
2
Step 2: Sort the points based on `x + y` and `x - y` values, creating two sorted lists.
3
Step 3: Identify the indices of the points with the minimum and maximum `x + y` values and the minimum and maximum `x - y` values. These indices are potential candidates to be removed.
4
Step 4: Iterate through the set of candidate indices.
5
Step 5: For each candidate index, simulate removing the corresponding point.
6
Step 6: After removing the point, determine the new minimum and maximum `x + y` and `x - y` values based on the remaining points.
7
Step 7: Calculate the maximum of the ranges of `x + y` and `x - y`. This represents the maximum Manhattan distance after removing the current candidate point.
8
Step 8: Update the minimum maximum distance found so far.
9
Step 9: Return the minimum maximum distance found after considering all candidate points.

💡

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(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The dominant operation is sorting the points by the sums and differences. Sorting is typically O(n log n). Other operations, like calculating the max distance for each removed point, take O(1) once the sorted lists have been created. Therefore the overall complexity remains O(n log n).

📖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 each point, remove it, and then calculate the maximum Manhattan distance between all remaining pairs of points. Keep track of the minimum of these maximum distances.

Time: O(n^3), where n is the number of points. Removing a point takes O(1), calculating maximum manhattan distance between all remaining points takes O(n^2), and you do this for n points.Space: O(1). Constant extra space is used.

Trade-offs: The brute force approach is simple to implement but inefficient for large datasets. It is suitable only for small numbers of points due to its high time complexity. The presented optimized approach is better for larger sets of data points.

✨

Key Insights

1Insight 1: The maximum Manhattan distance is determined by extreme points. Specifically, the maximum range of x+y and x-y values.
2Insight 2: Instead of calculating distances between all pairs of points after removing each point, it's more efficient to identify potential candidates for the point to remove. These candidates are the points that contribute to the extreme values of x+y and x-y.
3Insight 3: Sorting the points based on x+y and x-y allows for easy identification of the minimum and maximum values, and consequently, the candidate points to remove.

📚

Key Terms Explained

For Students

Array

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

Sorting

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

Edge Cases

Edge case 1: All points are the same. Removing any point will result in a maximum distance of 0. The algorithm correctly identifies this and returns 0.
Edge case 2: Three points are collinear on the same line parallel to the x or y axis. The algorithm will handle this case correctly as the difference between min and max sum/difference will be the min possible maximum distance.
Edge case 3: Points with same x+y or x-y value. Code handles these edge cases correctly because sorting algorithms preserve original order of elements with same value. Also, the min/max range logic selects the next smallest/largest if the point to be removed equals to the min/max point found during sorting. Therefore correct maximum manhattan distance is calculated.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the Manhattan distance, forgetting the absolute value or adding instead of taking the difference.

✗Common mistake 2: Failing to consider all potential candidate points for removal. Only considering the extreme x and y coordinates is insufficient; the sums and differences must be considered.

✗Common mistake 3: Not handling edge cases where all points have the same coordinates, or where points are collinear.

✗Common mistake 4: Getting the logic wrong on how to calculate the maximum distance with a point removed. This is especially true if the minimum or maximum `x + y` or `x - y` values are associated with the removed point. In these situations, it's necessary to recalculate the minimum and maximum values from the remaining points.

✅

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: Optimizing facility placement. Imagine a city planning to remove one of its existing emergency service locations (fire station, hospital). This algorithm could help find the best location to remove so that the maximum response time (approximated by Manhattan distance in a grid-like city) to any two points in the city is minimized.

🏦Real-world scenario 2: Data outlier removal. In data analysis, identifying and removing outliers is crucial. This algorithm can be adapted to find the best data point to remove such that the maximum 'distance' (defined by some metric) between any two remaining data points is minimized, essentially reducing the overall data spread.

💡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: July 13, 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 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 "Minimize Manhattan Distances"?

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 math techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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