LeetCode 1622 | Fancy Sequence | Java | Hard | Modular Arithmetic | CodeByTushu
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In today’s video, we solve LeetCode 1622 — Fancy Sequence using Modular Arithmetic and Affine Transformation in Java.
The problem requires designing an API that supports dynamically appending integers, adding an increment to all existing numbers, multiplying all numbers by a multiplier, and retrieving the element at any index modulo 10^9 + 7. A naive approach of updating every element on each operation leads to O(N) updates per query and causes Time Limit Exceeded. Instead, every collective update on the sequence is mathematically represented as an affine transformation of the form f(x) = (a * x + b) % MOD.
By maintaining global scaling and offset variables (a, b), both addAll and multAll operations execute in strict O(1) time. When appending a new value, we store its base normalized value calculated as (val - b) * inv(a) % MOD. Because MOD is prime and the multiplier is non-zero modulo MOD, the modular multiplicative inverse of a is computed using Fermat's Little Theorem via binary exponentiation. This allows any subsequent index query to evaluate in O(1) time with optimal O(N) auxiliary space.
🔥 Problem Highlights
✅ Affine transformation invariant maintaining global scaling and additive offset (a, b)
✅ Strict O(1) constant time updates for both addAll and multAll operations
✅ Modular multiplicative inverse using Fermat's Little Theorem (a^(MOD - 2) % MOD)
✅ Base normalized value storage to decouple individual appends from future transformations
✅ Instant O(1) index retrieval without iterating over array elements
✅ Complete avoidance of O(N^2) TLE bottlenecks under 10^5 total API calls
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:24 Problem Understanding & API Operations
03:27 Key Insights: Global Affine Transformation & Modular Inverse
06:16 Mathematical Formulation & 4-Step Strategy
08:05 Clean Java Solution Walkthrough Line by Line
09:53 Test Cases Walkthrough & Step-by-Step Simulation
13:02 Time Complexity Analysis (O(1) Updates / O(log MOD) Append)
13:53 Space Complexity Analysis (O(N) Space - Zero Tree Overhead)
14:34 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/15-March/1622-Fancy-Sequence
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/solution.js
🔗 LeetCode Source Code
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🔥 Perfect For
LeetCode Hard Problems
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Modular Arithmetic and Inverses
Data Structure Design
Affine Transformation Patterns
Time Complexity Optimization
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💬 Question of the Day
According to Fermat's Little Theorem, if p is a prime number and a is not divisible by p, what is the modular multiplicative inverse of a modulo p?
1️⃣ a^(p - 1) % p
2️⃣ a^(p - 2) % p
3️⃣ a^p % p
4️⃣ (p - a) % p
#LeetCode1622 #LeetCode #FancySequence #Java #JavaDSA #Math #Design #DSA #CodingInterview #ProblemSolving #LeetCodeHard #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 1622 — Fancy Sequence using Modular Arithmetic and Affine Transformation in Java.
The problem requires designing an API that supports dynamically appending integers, adding an increment to all existing numbers, multiplying all numbers by a multiplier, and retrieving the element at any index modulo 10^9 + 7. A naive approach of updating every element on each operation leads to O(N) updates per query and causes Time Limit Exceeded. Instead, every collective update on the sequence is mathematically represented as an affine transformation of the form f(x) = (a * x + b) % MOD.
By maintaining global scaling and offset variables (a, b), both addAll and multAll operations execute in strict O(1) time. When appending a new value, we store its base normalized value calculated as (val - b) * inv(a) % MOD. Because MOD is prime and the multiplier is non-zero modulo MOD, the modular multiplicative inverse of a is computed using Fermat's Little Theorem via binary exponentiation. This allows any subsequent index query to evaluate in O(1) time with optimal O(N) auxiliary space.
🔥 Problem Highlights
✅ Affine transformation invariant maintaining global scaling and additive offset (a, b)
✅ Strict O(1) constant time updates for both addAll and multAll operations
✅ Modular multiplicative inverse using Fermat's Little Theorem (a^(MOD - 2) % MOD)
✅ Base normalized value storage to decouple individual appends from future transformations
✅ Instant O(1) index retrieval without iterating over array elements
✅ Complete avoidance of O(N^2) TLE bottlenecks under 10^5 total API calls
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:24 Problem Understanding & API Operations
03:27 Key Insights: Global Affine Transformation & Modular Inverse
06:16 Mathematical Formulation & 4-Step Strategy
08:05 Clean Java Solution Walkthrough Line by Line
09:53 Test Cases Walkthrough & Step-by-Step Simulation
13:02 Time Complexity Analysis (O(1) Updates / O(log MOD) Append)
13:53 Space Complexity Analysis (O(N) Space - Zero Tree Overhead)
14:34 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/15-March/1622-Fancy-Sequence
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/15-March/1622-Fancy-Sequence/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 Hard Problems
Java DSA Preparation
Modular Arithmetic and Inverses
Data Structure Design
Affine Transformation Patterns
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
According to Fermat's Little Theorem, if p is a prime number and a is not divisible by p, what is the modular multiplicative inverse of a modulo p?
1️⃣ a^(p - 1) % p
2️⃣ a^(p - 2) % p
3️⃣ a^p % p
4️⃣ (p - a) % p
#LeetCode1622 #LeetCode #FancySequence #Java #JavaDSA #Math #Design #DSA #CodingInterview #ProblemSolving #LeetCodeHard #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
