LeetCode 3379 | Transformed Array | Java | Easy | Circular Array & Modular Arithmetic | CodeByTushu
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In today’s video, we solve LeetCode 3379 — Transformed Array using Array Simulation and Modular Arithmetic in Java.
We are given a circular array where each element indicates the number of steps to move: positive steps move right, negative steps move left, and zero keeps the position unchanged. Instead of executing multiple branching conditions or manually stepping through the array with a loop, all three movement rules can be mathematically unified into a single elegant circular transition formula.
By using the standardized modular formula targetIndex = ((i + nums[i]) % n + n) % n, we correctly handle both positive index overflows and negative dividends across all programming languages. This allows each transformed element to be constructed in constant O(1) time per element, achieving optimal linear O(n) runtime and minimal auxiliary space.
🔥 Problem Highlights
✅ Unified rightward, leftward, and stationary moves into one algebraic expression
✅ Handled negative remainders safely across languages with ((i + nums[i]) % n + n) % n
✅ Avoided costly step-by-step circular simulation loops
✅ Preserved original array immutability with clean independent index mapping
✅ Clean, idiomatic Java implementation
✅ Achieved optimal O(n) time and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:50 Problem Understanding & Constraints
02:00 Circular Movement & Modulo Insight
03:16 Universal Circular Index Strategy
04:09 Java Code Walkthrough
05:00 Execution & Test Cases
06:18 Time Complexity Analysis (O(n))
07:00 Space Complexity Analysis (O(1))
07:26 Outro & Wrap-up
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/5-February/3379-Transformed-Array
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/solution.js
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🔥 Perfect For
LeetCode Easy Problems
Java DSA Preparation
Circular Array Traversal
Modular Arithmetic Invariants
Array Transformation Problems
Time Complexity Optimization
Coding Interview Preparation
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💬 Question of the Day
Why do we compute ((i + nums[i]) % n + n) % n instead of simply (i + nums[i]) % n in languages like Java, C++, and JavaScript?
1️⃣ Because the % operator in Java/C++/JS returns a negative remainder when the dividend is negative
2️⃣ Because circular arrays always require two passes
3️⃣ Because array size n is not guaranteed to be positive
4️⃣ Because it prevents integer overflow beyond 64 bits
#LeetCode3379 #LeetCode #TransformedArray #Java #JavaDSA #Array #Simulation #ModularArithmetic #Math #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 3379 — Transformed Array using Array Simulation and Modular Arithmetic in Java.
We are given a circular array where each element indicates the number of steps to move: positive steps move right, negative steps move left, and zero keeps the position unchanged. Instead of executing multiple branching conditions or manually stepping through the array with a loop, all three movement rules can be mathematically unified into a single elegant circular transition formula.
By using the standardized modular formula targetIndex = ((i + nums[i]) % n + n) % n, we correctly handle both positive index overflows and negative dividends across all programming languages. This allows each transformed element to be constructed in constant O(1) time per element, achieving optimal linear O(n) runtime and minimal auxiliary space.
🔥 Problem Highlights
✅ Unified rightward, leftward, and stationary moves into one algebraic expression
✅ Handled negative remainders safely across languages with ((i + nums[i]) % n + n) % n
✅ Avoided costly step-by-step circular simulation loops
✅ Preserved original array immutability with clean independent index mapping
✅ Clean, idiomatic Java implementation
✅ Achieved optimal O(n) time and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:50 Problem Understanding & Constraints
02:00 Circular Movement & Modulo Insight
03:16 Universal Circular Index Strategy
04:09 Java Code Walkthrough
05:00 Execution & Test Cases
06:18 Time Complexity Analysis (O(n))
07:00 Space Complexity Analysis (O(1))
07:26 Outro & Wrap-up
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/5-February/3379-Transformed-Array
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/5-February/3379-Transformed-Array/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
Circular Array Traversal
Modular Arithmetic Invariants
Array Transformation Problems
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
Why do we compute ((i + nums[i]) % n + n) % n instead of simply (i + nums[i]) % n in languages like Java, C++, and JavaScript?
1️⃣ Because the % operator in Java/C++/JS returns a negative remainder when the dividend is negative
2️⃣ Because circular arrays always require two passes
3️⃣ Because array size n is not guaranteed to be positive
4️⃣ Because it prevents integer overflow beyond 64 bits
#LeetCode3379 #LeetCode #TransformedArray #Java #JavaDSA #Array #Simulation #ModularArithmetic #Math #DSA #CodingInterview #ProblemSolving #LeetCodeEasy #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description