LeetCode 3719 | Longest Balanced Subarray I | Java | Medium | Parity & D C | CodeByTushu
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In today’s video, we solve LeetCode 3719 — Longest Balanced Subarray I using Parity Tracking and High-Performance Distinct Counting in Java.
A contiguous subarray is balanced if the count of unique even numbers equals the count of unique odd numbers. Since the array length is at most 1500, we can explore all potential subarray boundaries in quadratic O(n^2) time. However, repeatedly creating and clearing standard hash sets inside nested loops can introduce garbage collection and memory overhead.
To improve performance, we use a global timestamp array that tracks previously seen elements in O(1) time without clearing or reallocating memory. For each left boundary, we increment the iteration timestamp and expand the right endpoint, updating the counts of distinct even and odd numbers along the way. We also use early pruning whenever the remaining elements cannot produce a subarray longer than our current maximum.
🔥 Problem Highlights
✅ Tracked the parity of distinct elements using the bitwise AND operation with val and 1
✅ Eliminated inner loop set allocations using a single global timestamp array
✅ Applied early search pruning when the remaining suffix length is at most the current maximum length
✅ Maintained separate counters for distinct even and odd numbers in O(1) time
✅ Clean and memory-efficient Java implementation
✅ Achieved O(n^2) time complexity and O(U) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding and Constraints
02:00 Timestamp Array and Early Pruning Insight
03:30 Subarray Scanning Strategy
04:48 Java Code Walkthrough
06:15 Execution and Test Cases
08:06 Time Complexity Analysis O(n^2)
08:55 Space Complexity Analysis O(maxVal)
09:33 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.js
🔗 LeetCode Source Code
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💬 Question of the Day
Why does using a timestamp array instead of clearing a HashSet or boolean array inside the outer loop improve performance?
1️⃣ Because checking whether lastSeen[val] differs from the current iteration avoids costly memory reallocations and repeated O(U) array initialization
2️⃣ Because HashSets cannot store negative numbers
3️⃣ Because timestamp arrays sort elements automatically
4️⃣ Because it reduces time complexity from O(n^2) to O(n)
#LeetCode3719 #LeetCode #LongestBalancedSubarrayI #Java #JavaDSA #Array #HashTable #Counting #Parity #Enumeration #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 3719 — Longest Balanced Subarray I using Parity Tracking and High-Performance Distinct Counting in Java.
A contiguous subarray is balanced if the count of unique even numbers equals the count of unique odd numbers. Since the array length is at most 1500, we can explore all potential subarray boundaries in quadratic O(n^2) time. However, repeatedly creating and clearing standard hash sets inside nested loops can introduce garbage collection and memory overhead.
To improve performance, we use a global timestamp array that tracks previously seen elements in O(1) time without clearing or reallocating memory. For each left boundary, we increment the iteration timestamp and expand the right endpoint, updating the counts of distinct even and odd numbers along the way. We also use early pruning whenever the remaining elements cannot produce a subarray longer than our current maximum.
🔥 Problem Highlights
✅ Tracked the parity of distinct elements using the bitwise AND operation with val and 1
✅ Eliminated inner loop set allocations using a single global timestamp array
✅ Applied early search pruning when the remaining suffix length is at most the current maximum length
✅ Maintained separate counters for distinct even and odd numbers in O(1) time
✅ Clean and memory-efficient Java implementation
✅ Achieved O(n^2) time complexity and O(U) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding and Constraints
02:00 Timestamp Array and Early Pruning Insight
03:30 Subarray Scanning Strategy
04:48 Java Code Walkthrough
06:15 Execution and Test Cases
08:06 Time Complexity Analysis O(n^2)
08:55 Space Complexity Analysis O(maxVal)
09:33 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/10-February/3719-Longest-Balanced-Subarray-I/solution.js
🔗 LeetCode Source Code
📌 Add the link here
🌐 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 Medium Problems
Java DSA Preparation
Parity and Modulo Patterns
Distinct Element Frequency Tracking
Timestamp Array Optimization
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 does using a timestamp array instead of clearing a HashSet or boolean array inside the outer loop improve performance?
1️⃣ Because checking whether lastSeen[val] differs from the current iteration avoids costly memory reallocations and repeated O(U) array initialization
2️⃣ Because HashSets cannot store negative numbers
3️⃣ Because timestamp arrays sort elements automatically
4️⃣ Because it reduces time complexity from O(n^2) to O(n)
#LeetCode3719 #LeetCode #LongestBalancedSubarrayI #Java #JavaDSA #Array #HashTable #Counting #Parity #Enumeration #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description