LeetCode 1266 | Minimum Time Visiting All Points | Java | Easy | Chebyshev Distance | CodeByTushu
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In today’s video, we solve LeetCode 1266 — Minimum Time Visiting All Points using Array, Math, and Geometry in Java.
Given an ordered sequence of 2D points with integer coordinates, we need to find the minimum time in seconds required to visit all points sequentially. Within one second, we can move horizontally, vertically, or diagonally. Moving diagonally is optimal because it reduces both the horizontal and vertical distances simultaneously by one unit, making the shortest time between any two points equal to their Chebyshev distance, which is max(|x2 - x1|, |y2 - y1|).
By iterating through consecutive adjacent coordinate pairs in the array, we calculate the absolute horizontal difference and absolute vertical difference for each segment. The maximum of these two values represents the minimum seconds needed for that transition. Summing these values across the entire array yields the total optimal time in a single linear pass with O(1) auxiliary space.
🔥 Problem Highlights
✅ Direct application of the Chebyshev distance (L-infinity metric)
✅ Maximizing diagonal moves to reduce both x and y distances concurrently
✅ Clean single-pass iteration across consecutive point pairs
✅ Optimal handling of negative coordinates and arbitrary point sequences
✅ Production-quality implementations across Java, Python, C++, and JavaScript
✅ Optimal O(N) time complexity and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:52 Problem Understanding
01:57 Observation & Chebyshev Distance Intuition
02:54 Linear Scan Approach
03:45 Java Code Walkthrough
04:49 Test Cases & Edge Cases Walkthrough
06:01 Time Complexity Analysis
06:41 Space Complexity Analysis
07:10 Summary & Outro
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.js
🔗 LeetCode Source Code
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🔥 Perfect For
LeetCode Easy Problems
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Chebyshev Distance
Grid Traversal
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💬 Question of the Day
What mathematical distance metric defines the minimum steps needed between (x1, y1) and (x2, y2) when diagonal moves cost 1 unit of time?
1️⃣ Euclidean Distance (L2 norm)
2️⃣ Manhattan Distance (L1 norm)
3️⃣ Chebyshev Distance (L-infinity norm)
4️⃣ Hamming Distance
#LeetCode1266 #LeetCode #MinimumTimeVisitingAllPoints #Java #JavaDSA #Geometry #Math #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 1266 — Minimum Time Visiting All Points using Array, Math, and Geometry in Java.
Given an ordered sequence of 2D points with integer coordinates, we need to find the minimum time in seconds required to visit all points sequentially. Within one second, we can move horizontally, vertically, or diagonally. Moving diagonally is optimal because it reduces both the horizontal and vertical distances simultaneously by one unit, making the shortest time between any two points equal to their Chebyshev distance, which is max(|x2 - x1|, |y2 - y1|).
By iterating through consecutive adjacent coordinate pairs in the array, we calculate the absolute horizontal difference and absolute vertical difference for each segment. The maximum of these two values represents the minimum seconds needed for that transition. Summing these values across the entire array yields the total optimal time in a single linear pass with O(1) auxiliary space.
🔥 Problem Highlights
✅ Direct application of the Chebyshev distance (L-infinity metric)
✅ Maximizing diagonal moves to reduce both x and y distances concurrently
✅ Clean single-pass iteration across consecutive point pairs
✅ Optimal handling of negative coordinates and arbitrary point sequences
✅ Production-quality implementations across Java, Python, C++, and JavaScript
✅ Optimal O(N) time complexity and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:52 Problem Understanding
01:57 Observation & Chebyshev Distance Intuition
02:54 Linear Scan Approach
03:45 Java Code Walkthrough
04:49 Test Cases & Edge Cases Walkthrough
06:01 Time Complexity Analysis
06:41 Space Complexity Analysis
07:10 Summary & Outro
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/12-January/1266-Minimum-Time-Visiting-All-Points/solution.js
🔗 LeetCode Source Code
📌 Admin type the link
🌐 All LeetCode Solutions
📌 https://codebytushu.com/Leetcode/
📱 Join CodeByTushu on WhatsApp
Get coding updates, LeetCode solutions, daily questions and more.
https://whatsapp.com/channel/0029Vb8k5mCDeON2Xwlkxe40
❤️ Donate
📌 https://codebytushu.com/Leetcode/donate.php
🔥 Perfect For
LeetCode Easy Problems
Java DSA Preparation
Geometry Algorithms
Chebyshev Distance
Grid Traversal
Time Complexity Optimization
Coding Interview Preparation
Problem Solving
🌐 Follow CodeByTushu
▶️ YouTube: https://www.youtube.com/@codebytushu
📸 Instagram: https://instagram.com/codebytushu/
💼 LinkedIn: https://www.linkedin.com/company/codebytushu/
🌍 Website: https://codebytushu.com
🐙 GitHub: https://github.com/CodeByTushu
❤️ Don’t Forget To
👍 Like | 💬 Comment | 🔔 Subscribe
💬 Question of the Day
What mathematical distance metric defines the minimum steps needed between (x1, y1) and (x2, y2) when diagonal moves cost 1 unit of time?
1️⃣ Euclidean Distance (L2 norm)
2️⃣ Manhattan Distance (L1 norm)
3️⃣ Chebyshev Distance (L-infinity norm)
4️⃣ Hamming Distance
#LeetCode1266 #LeetCode #MinimumTimeVisitingAllPoints #Java #JavaDSA #Geometry #Math #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
