LeetCode 1536 | Minimum Swaps to Arrange a Binary Grid | Java | Medium | CodeByTushu
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In today’s video, we solve LeetCode 1536 — Minimum Swaps to Arrange a Binary Grid using Greedy and Adjacent Row Swapping in Java.
We are given an n by n binary grid where we can swap any two adjacent rows in one step. A valid grid requires all cells above the main diagonal to be zero. Therefore, each row must have a minimum number of trailing zeros based on its position. Our goal is to find the minimum adjacent row swaps needed to satisfy this condition or return minus one if it is impossible.
We first calculate the trailing zero count for every row. Then, moving from top to bottom, we greedily find the nearest available row that has enough trailing zeros for the current position. We count the required adjacent swaps and move that row into its correct position by shifting the intermediate rows.
This greedy approach guarantees the minimum number of swaps. It achieves O(n^2) time complexity and O(n) auxiliary space without modifying the original grid.
🔥 Problem Highlights
✅ Converted the diagonal condition into trailing zero counts
✅ Used greedy selection to find the nearest qualifying row
✅ Simulated adjacent swaps using array shifting
✅ Detected impossible cases and returned minus one
✅ Used an efficient two-phase approach without modifying the input grid
✅ Achieved O(n^2) time and O(n) auxiliary space
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:21 Problem Understanding & Diagonal Invariant
03:08 Key Observation: 1D Trailing Zeros & Greedy Choice
04:47 Greedy First-Match Strategy (4 Steps)
06:19 Clean Java Solution Walkthrough Line by Line
08:00 Test Cases Walkthrough & Step-by-Step Simulation
10:02 Time Complexity Analysis (O(N^2) Time - 1 ms)
11:15 Space Complexity Analysis (O(N) Space)
12:06 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/solution.js
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🔥 Perfect For
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Adjacent Swaps and Inversions
Time Complexity Optimization
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💬 Question of the Day
Why is choosing the first qualifying row at or below the current position always optimal?
1️⃣ It minimizes the number of adjacent swaps while preserving valid candidates for later rows
2️⃣ It avoids checking whether the grid is square
3️⃣ Rows can only be swapped downwards, not upwards
4️⃣ It guarantees that all diagonal elements become one
#LeetCode1536 #LeetCode #MinimumSwapsToArrangeABinaryGrid #Java #JavaDSA #Greedy #Matrix #Array #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 1536 — Minimum Swaps to Arrange a Binary Grid using Greedy and Adjacent Row Swapping in Java.
We are given an n by n binary grid where we can swap any two adjacent rows in one step. A valid grid requires all cells above the main diagonal to be zero. Therefore, each row must have a minimum number of trailing zeros based on its position. Our goal is to find the minimum adjacent row swaps needed to satisfy this condition or return minus one if it is impossible.
We first calculate the trailing zero count for every row. Then, moving from top to bottom, we greedily find the nearest available row that has enough trailing zeros for the current position. We count the required adjacent swaps and move that row into its correct position by shifting the intermediate rows.
This greedy approach guarantees the minimum number of swaps. It achieves O(n^2) time complexity and O(n) auxiliary space without modifying the original grid.
🔥 Problem Highlights
✅ Converted the diagonal condition into trailing zero counts
✅ Used greedy selection to find the nearest qualifying row
✅ Simulated adjacent swaps using array shifting
✅ Detected impossible cases and returned minus one
✅ Used an efficient two-phase approach without modifying the input grid
✅ Achieved O(n^2) time and O(n) auxiliary space
⏱️ TIMESTAMPS
00:00 Intro & Daily Challenge Overview
01:21 Problem Understanding & Diagonal Invariant
03:08 Key Observation: 1D Trailing Zeros & Greedy Choice
04:47 Greedy First-Match Strategy (4 Steps)
06:19 Clean Java Solution Walkthrough Line by Line
08:00 Test Cases Walkthrough & Step-by-Step Simulation
10:02 Time Complexity Analysis (O(N^2) Time - 1 ms)
11:15 Space Complexity Analysis (O(N) Space)
12:06 Outro & Resources
📂 Problem Folder & Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/March/2-March/1536-Minimum-Swaps-to-Arrange-a-Binary-Grid/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
Greedy Algorithms
Matrix Manipulation
Adjacent Swaps and Inversions
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 is choosing the first qualifying row at or below the current position always optimal?
1️⃣ It minimizes the number of adjacent swaps while preserving valid candidates for later rows
2️⃣ It avoids checking whether the grid is square
3️⃣ Rows can only be swapped downwards, not upwards
4️⃣ It guarantees that all diagonal elements become one
#LeetCode1536 #LeetCode #MinimumSwapsToArrangeABinaryGrid #Java #JavaDSA #Greedy #Matrix #Array #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
