LeetCode 3453 | Separate Squares I | Java | Medium | Binary Search on Answer | CodeByTushu
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In today’s video, we solve LeetCode 3453 - Separate Squares I using Array, Binary Search, and Geometry in Java.
We are given a collection of axis-aligned squares on a 2D plane and need to identify the minimum y-coordinate of a horizontal line that partitions the total square area into two equal halves. Notice that as our test horizontal line moves upward, the cumulative area of squares lying below this line is strictly monotonic non-decreasing. This monotonicity makes the problem an ideal candidate for Binary Search on the Answer.
We first compute the target half-area by summing up the areas of all squares. Then, we establish a search range from y = 0 to the maximum upper boundary of any square. In each iteration, we calculate the area beneath the midpoint line across all squares by summing the clipped heights. When the area beneath meets or exceeds our target, we adjust our upper bound downward to ensure we converge to the minimum qualifying y-coordinate. Running 80 iterations delivers extreme numerical precision far beyond the required 10^-5 threshold in optimal O(N) time per iteration.
🔥 Problem Highlights
✅ Binary Search on the Answer leveraging monotonic area accumulation
✅ Target half-area threshold determination with 64-bit integer overflow protection
✅ Precise horizontal slice height calculation per square in O(1)
✅ Convergence to the lowest valid y-coordinate when multiple horizontal lines qualify
✅ Full high-precision code across Java, Python, C++, and JavaScript
✅ Optimal O(N * log((maxY - minY) / eps)) time complexity and O(1) auxiliary space
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding
02:12 Observation & Monotonic Area Growth
03:02 Continuous Binary Search Approach
04:05 Java Code Walkthrough
05:06 Test Cases & Precision Walkthrough
06:26 Time Complexity Analysis
07:03 Space Complexity Analysis
07:33 Summary & Outro
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/January/13-January/3453-Separate-Squares-I
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/solution.js
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LeetCode Medium Problems
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💬 Question of the Day
Why does setting high equal to mid when areaBelow of mid is greater than or equal to targetArea guarantee finding the MINIMUM valid y-coordinate?
1️⃣ It biases the binary search toward the lower half whenever a valid equal-area line is found
2️⃣ It stops the binary search immediately upon finding the first valid line
3️⃣ It ensures that areas above the line are always calculated first
4️⃣ It prevents floating-point inaccuracies from accumulating
#LeetCode3453 #LeetCode #SeparateSquaresI #Java #JavaDSA #BinarySearch #Geometry #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 3453 - Separate Squares I using Array, Binary Search, and Geometry in Java.
We are given a collection of axis-aligned squares on a 2D plane and need to identify the minimum y-coordinate of a horizontal line that partitions the total square area into two equal halves. Notice that as our test horizontal line moves upward, the cumulative area of squares lying below this line is strictly monotonic non-decreasing. This monotonicity makes the problem an ideal candidate for Binary Search on the Answer.
We first compute the target half-area by summing up the areas of all squares. Then, we establish a search range from y = 0 to the maximum upper boundary of any square. In each iteration, we calculate the area beneath the midpoint line across all squares by summing the clipped heights. When the area beneath meets or exceeds our target, we adjust our upper bound downward to ensure we converge to the minimum qualifying y-coordinate. Running 80 iterations delivers extreme numerical precision far beyond the required 10^-5 threshold in optimal O(N) time per iteration.
🔥 Problem Highlights
✅ Binary Search on the Answer leveraging monotonic area accumulation
✅ Target half-area threshold determination with 64-bit integer overflow protection
✅ Precise horizontal slice height calculation per square in O(1)
✅ Convergence to the lowest valid y-coordinate when multiple horizontal lines qualify
✅ Full high-precision code across Java, Python, C++, and JavaScript
✅ Optimal O(N * log((maxY - minY) / eps)) time complexity and O(1) auxiliary space
⏱️ TIMESTAMPS
00:00 Intro
00:58 Problem Understanding
02:12 Observation & Monotonic Area Growth
03:02 Continuous Binary Search Approach
04:05 Java Code Walkthrough
05:06 Test Cases & Precision Walkthrough
06:26 Time Complexity Analysis
07:03 Space Complexity Analysis
07:33 Summary & Outro
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/January/13-January/3453-Separate-Squares-I
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/January/13-January/3453-Separate-Squares-I/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 Medium Problems
Java DSA Preparation
Binary Search on Answer
Geometry Algorithms
Floating Point Precision
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 setting high equal to mid when areaBelow of mid is greater than or equal to targetArea guarantee finding the MINIMUM valid y-coordinate?
1️⃣ It biases the binary search toward the lower half whenever a valid equal-area line is found
2️⃣ It stops the binary search immediately upon finding the first valid line
3️⃣ It ensures that areas above the line are always calculated first
4️⃣ It prevents floating-point inaccuracies from accumulating
#LeetCode3453 #LeetCode #SeparateSquaresI #Java #JavaDSA #BinarySearch #Geometry #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description