What is the time complexity of Minimum Time Visiting All Points?

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

What data structures are used to solve Minimum Time Visiting All Points?

This problem uses Array, Math, Geometry. Understanding these concepts is essential for solving this Easy problem efficiently.

What are the key insights for solving Minimum Time Visiting All Points?

The diagonal movement rule is key: moving one unit in X and one unit in Y simultaneously still costs only 1 second. This isn't a Euclidean `sqrt(2)` movement in terms of time, but a combined Manhattan-like step. Optimal path between two points: To minimize time between `(x1, y1)` and `(x2, y2)`, one should always maximize diagonal moves. You effectively make `min(abs(x2-x1), abs(y2-y1))` diagonal steps, and then `abs(abs(x2-x1) - abs(y2-y1))` straight (horizontal or vertical) steps. The sum of these two components `min(dx, dy) + abs(dx - dy)` directly equals `max(dx, dy)`.

Which companies ask LeetCode 1266. Minimum Time Visiting All Points in interviews?

This problem is frequently asked at Media.net.

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Minimum Time Visiting All Points - LeetCode 1266 Solution

Minimum Time Visiting All Points - Complete Solution Guide

Minimum Time Visiting All Points is LeetCode problem 1266, 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

On a 2D plane, there are n points with integer coordinates points[i] = [x i , y i ] . Return the minimum time in seconds to visit all the points in the order given by points . You can move according to these rules: In 1 second, you can either: move vertically by one unit, move horizontally by one unit, or move diagonally sqrt(2) units (in other words, move one unit vertically then one unit horizontally in 1 second). You have to visit the points in the same order as they appear in the array. You

Detailed Explanation

Imagine navigating a city grid where, in addition to moving straight along streets, you can also cut diagonally across blocks at the same speed. This problem asks for the minimum total time to visit a predefined sequence of points. You're given `n` points, and you *must* visit them in the exact order they appear in the input array. The crucial part is understanding the movement rules: moving one unit horizontally, one unit vertically, or one unit diagonally (meaning one unit horizontally *and* one unit vertically simultaneously) all take exactly 1 second. This simultaneous diagonal movement is what makes this problem distinct from a simple Manhattan distance calculation.

Solution Approach

The elegance of the solution lies in a crucial observation about the diagonal move. When traveling from `(x1, y1)` to `(x2, y2)`, let `dx = abs(x2 - x1)` and `dy = abs(y2 - y1)`. Each second, you can reduce `dx` by 1, or `dy` by 1, or both `dx` and `dy` by 1. To reach the target `(x2, y2)` in minimum time, you'll always want to leverage the diagonal move as much as possible, as it effectively reduces two 'distances' (X and Y) for the cost of one second. You can perform `min(dx, dy)` diagonal moves. Each of these moves reduces both `dx` and `dy` by one. After `min(dx, dy)` seconds, one of the differences (either `dx` or `dy`) will be zero, and the other will be `abs(dx - dy)`. The remaining `abs(dx - dy)` units of movement must then be covered purely horizontally or vertically, each taking 1 second per unit. Therefore, the total time for a single segment is `min(dx, dy) + abs(dx - dy)`, which mathematically simplifies directly to `max(dx, dy)`. The overall solution simply iterates through the given points, calculating this `max(dx, dy)` time for each consecutive pair and summing them up.

Step-by-Step Algorithm

Step 1: Initialize a variable `totalTime` to 0. This variable will accumulate the total time taken to visit all points.
Step 2: Iterate through the `points` array from the first point to the second to last point. For each point `i`, consider it along with the next point `i+1`.
Step 3: Calculate the absolute difference in x-coordinates (`dx`) and the absolute difference in y-coordinates (`dy`) between points `i` and `i+1`.
Step 4: Find the maximum of `dx` and `dy`. This represents the minimum time needed to travel from point `i` to point `i+1`.
Step 5: Add this maximum value to `totalTime`.
Step 6: After iterating through all consecutive pairs of points, return the final `totalTime`.

Key Insights

The diagonal movement rule is key: moving one unit in X and one unit in Y simultaneously still costs only 1 second. This isn't a Euclidean `sqrt(2)` movement in terms of time, but a combined Manhattan-like step.
Optimal path between two points: To minimize time between `(x1, y1)` and `(x2, y2)`, one should always maximize diagonal moves. You effectively make `min(abs(x2-x1), abs(y2-y1))` diagonal steps, and then `abs(abs(x2-x1) - abs(y2-y1))` straight (horizontal or vertical) steps. The sum of these two components `min(dx, dy) + abs(dx - dy)` directly equals `max(dx, dy)`.
The problem constraint to visit points 'in order' simplifies it significantly. We don't need to consider permutations or complex graph traversal algorithms like Traveling Salesperson. We just calculate the minimum time for each consecutive segment and add them up, leveraging the previously described optimal strategy for a single segment.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Geometry.

Companies

Asked at: Media.net.

Minimum Time Visiting All Points | LeetCode 1266

Easy

Array Math Geometry

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

Solution Code

Available in:

class Solution {
    public int minTimeToVisitAllPoints(int[][] points) {
        int totalTime = 0;
        for (int i = 0; i < points.length - 1; i++) {
            int x1 = points[i][0];
            int y1 = points[i][1];
            int x2 = points[i + 1][0];
            int y2 = points[i + 1][1];
            int dx = Math.abs(x2 - x1);
            int dy = Math.abs(y2 - y1);
            totalTime += Math.max(dx, dy);
        }
        return totalTime;
    }
}

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Problem Info

Description

Problem 1266: Minimum Time Visiting All Points - Detailed explanation and solution approaches.

Topics

Array Math Geometry

Companies

Media.net

Stats

Difficulty:Easy

Languages:4

Companies:1

Problem #:1266

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

6 steps

1
Step 1: Initialize a variable `totalTime` to 0. This variable will accumulate the total time taken to visit all points.
2
Step 2: Iterate through the `points` array from the first point to the second to last point. For each point `i`, consider it along with the next point `i+1`.
3
Step 3: Calculate the absolute difference in x-coordinates (`dx`) and the absolute difference in y-coordinates (`dy`) between points `i` and `i+1`.
4
Step 4: Find the maximum of `dx` and `dy`. This represents the minimum time needed to travel from point `i` to point `i+1`.
5
Step 5: Add this maximum value to `totalTime`.
6
Step 6: After iterating through all consecutive pairs of points, return the final `totalTime`.

💡

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)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is linear because the number of operations is directly proportional to the number of points. The space complexity is constant because the memory used does not depend on the input size; it remains the same regardless of the number of points.

📖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

✨

Key Insights

1The diagonal movement rule is key: moving one unit in X and one unit in Y simultaneously still costs only 1 second. This isn't a Euclidean `sqrt(2)` movement in terms of time, but a combined Manhattan-like step.
2Optimal path between two points: To minimize time between `(x1, y1)` and `(x2, y2)`, one should always maximize diagonal moves. You effectively make `min(abs(x2-x1), abs(y2-y1))` diagonal steps, and then `abs(abs(x2-x1) - abs(y2-y1))` straight (horizontal or vertical) steps. The sum of these two components `min(dx, dy) + abs(dx - dy)` directly equals `max(dx, dy)`.
3The problem constraint to visit points 'in order' simplifies it significantly. We don't need to consider permutations or complex graph traversal algorithms like Traveling Salesperson. We just calculate the minimum time for each consecutive segment and add them up, leveraging the previously described optimal strategy for a single segment.

📚

Key Terms Explained

For Students

Array

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

Edge Cases

Edge case 1: If the `points` array contains only one point, the function correctly returns 0 because there's no travel time needed.
Edge case 2: If the points are collinear, the algorithm still correctly calculates the minimum time as the distance between the farthest points along the line (either x or y axis).

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly calculating the distance between two points using the Euclidean distance formula instead of focusing on the maximum of the absolute differences in x and y coordinates.

✗Common mistake 2: Forgetting to take the absolute difference between the coordinates. Negative differences can lead to incorrect calculations of the minimum time.

✅

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 the route of a robot that needs to visit several locations in a warehouse. The algorithm helps minimize the robot's travel time.

🏦Real-world scenario 2: Planning the shortest path for a drone delivering packages to multiple addresses. The algorithm can assist in optimizing the delivery route to reduce travel time and fuel consumption.

💡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

Even for easy problems, discuss time and space complexity. Interviewers want to see you think analytically.

⚠️ 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: June 7, 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

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

@AlgoMasterJul 13

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

@InterviewProJul 6

❓

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 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 "Minimum Time Visiting All Points"?

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

🤖 Need help with this problem?