LeetCode 1727 | Largest Submatrix With Rearrangements | Java | Medium | Matrix | CodeByTushu
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In today’s video, we solve LeetCode 1727 — Largest Submatrix With Rearrangements using Matrix Processing and Greedy Sorting in Java.
The problem asks us to find the area of the largest submatrix consisting entirely of 1s in a binary matrix, where we are allowed to arbitrarily rearrange entire columns in any order. For every cell at row i and column j, we first determine how many consecutive 1s extend vertically upward ending at that cell. This continuous height accumulation transforms each row into a histogram problem where the horizontal order of bars is completely customizable.
Because columns can be permuted freely, the optimal strategy for any given row is to greedily sort its heights in non-decreasing order. Once sorted, a rectangle of height equal to the current column's height can span across all remaining taller columns to its right. We evaluate this candidate rectangular area for every position across all rows, maintaining the global maximum area in optimal O(m * n log n) time and O(n) auxiliary space.
🔥 Problem Highlights
✅ In-place column height accumulation capturing vertical consecutive 1s in linear time
✅ Greedy sorting invariant leveraging unconstrained column permutation
✅ Transformation of 2D matrix search into independent optimal 1D histogram evaluations
✅ Direct area calculation using height multiplied by available horizontal width
✅ Optimal O(m * n log n) overall runtime executing in under 15 milliseconds
✅ Minimal O(n) auxiliary memory overhead without modifying original input structures
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:16 Problem Understanding & Column Rearrangements
02:45 Key Insights: Consecutive 1s & Greedy Sorting
04:49 Algorithmic Strategy & 3-Step Implementation
06:18 Clean Java Solution Walkthrough Line by Line
07:44 Test Cases Walkthrough & Step-by-Step Simulation
09:59 Time Complexity Analysis (O(M·N log N) - 10 ms)
11:09 Space Complexity Analysis (O(N) Auxiliary Space)
11:52 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/solution.js
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🔥 Perfect For
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Histogram Area Optimization
Time Complexity Optimization
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💬 Question of the Day
If a row in an m by n matrix has sorted heights [0, 2, 2, 4, 5], what is the maximum area of a submatrix with all 1s ending at this row?
1️⃣ 5
2️⃣ 8
3️⃣ 9
4️⃣ 10
#LeetCode1727 #LeetCode #LargestSubmatrixWithRearrangements #Java #JavaDSA #Matrix #Greedy #Sorting #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 1727 — Largest Submatrix With Rearrangements using Matrix Processing and Greedy Sorting in Java.
The problem asks us to find the area of the largest submatrix consisting entirely of 1s in a binary matrix, where we are allowed to arbitrarily rearrange entire columns in any order. For every cell at row i and column j, we first determine how many consecutive 1s extend vertically upward ending at that cell. This continuous height accumulation transforms each row into a histogram problem where the horizontal order of bars is completely customizable.
Because columns can be permuted freely, the optimal strategy for any given row is to greedily sort its heights in non-decreasing order. Once sorted, a rectangle of height equal to the current column's height can span across all remaining taller columns to its right. We evaluate this candidate rectangular area for every position across all rows, maintaining the global maximum area in optimal O(m * n log n) time and O(n) auxiliary space.
🔥 Problem Highlights
✅ In-place column height accumulation capturing vertical consecutive 1s in linear time
✅ Greedy sorting invariant leveraging unconstrained column permutation
✅ Transformation of 2D matrix search into independent optimal 1D histogram evaluations
✅ Direct area calculation using height multiplied by available horizontal width
✅ Optimal O(m * n log n) overall runtime executing in under 15 milliseconds
✅ Minimal O(n) auxiliary memory overhead without modifying original input structures
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:16 Problem Understanding & Column Rearrangements
02:45 Key Insights: Consecutive 1s & Greedy Sorting
04:49 Algorithmic Strategy & 3-Step Implementation
06:18 Clean Java Solution Walkthrough Line by Line
07:44 Test Cases Walkthrough & Step-by-Step Simulation
09:59 Time Complexity Analysis (O(M·N log N) - 10 ms)
11:09 Space Complexity Analysis (O(N) Auxiliary Space)
11:52 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/17-March/1727-Largest-Submatrix-With-Rearrangements/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
Matrix Manipulation Algorithms
Greedy and Sorting Paradigms
Histogram Area 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
If a row in an m by n matrix has sorted heights [0, 2, 2, 4, 5], what is the maximum area of a submatrix with all 1s ending at this row?
1️⃣ 5
2️⃣ 8
3️⃣ 9
4️⃣ 10
#LeetCode1727 #LeetCode #LargestSubmatrixWithRearrangements #Java #JavaDSA #Matrix #Greedy #Sorting #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
