LeetCode 3634 | Minimum Removals to Balance Array | Java | Medium | Two Pointers | CodeByTushu
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In today’s video, we solve LeetCode 3634 - Minimum Removals to Balance Array using Sorting and Two Pointers in Java.
The problem requires removing the minimum number of elements from an array such that the maximum value of the remaining array is at most k times its minimum value. Minimizing removals is mathematically equivalent to maximizing the number of kept elements. When elements are sorted in non-decreasing order, any valid balanced subset with minimum element nums at the left boundary and maximum element nums at the right boundary can include all intermediate numbers without violating the balance condition.
By sorting the array and maintaining a sliding window with two pointers, we expand the right boundary as long as the value at nums[right] is less than or equal to the value at nums[left] multiplied by k. We track the maximum window size across all candidate left boundaries and calculate the minimum removals as n minus maxKept. We also safeguard against 32-bit integer multiplication overflow using 64-bit long casting, achieving an optimal O(n log n) runtime dominated by sorting.
🔥 Problem Highlights
✅ Inverted the problem from minimizing removals to maximizing kept elements
✅ Identified that any optimal kept subset forms a contiguous range in the sorted array
✅ Implemented an efficient two-pointer sliding window running in linear O(n) scan time
✅ Prevented 32-bit integer overflow by using 64-bit long casting for the multiplication of nums[left] and k
✅ Handled single-element and already-balanced array edge cases seamlessly
✅ Achieved optimal O(n log n) time and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:56 Problem Understanding and Constraints
01:55 Key Mathematical Insight and Monotonicity
03:21 Sorting and Two-Pointer Sliding Window Strategy
04:26 Java Code Walkthrough
05:24 Execution and Test Cases
06:52 Time Complexity Analysis - O(n log n)
07:42 Space Complexity Analysis - O(1)
08:17 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/solution.js
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💬 Question of the Day
Why is it always optimal to keep a contiguous subarray in the sorted array when finding the maximum balanced subset?
1️⃣ Because including any number between the chosen minimum and maximum never changes the minimum or maximum values
2️⃣ Because elements in an array cannot be removed out of order
3️⃣ Because the problem requires the remaining array to be strictly increasing
4️⃣ Because sliding window only works on contiguous subarrays in unsorted arrays
#LeetCode3634 #LeetCode #MinimumRemovalsToBalanceArray #Java #JavaDSA #Array #Sorting #TwoPointers #SlidingWindow #BinarySearch #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 3634 - Minimum Removals to Balance Array using Sorting and Two Pointers in Java.
The problem requires removing the minimum number of elements from an array such that the maximum value of the remaining array is at most k times its minimum value. Minimizing removals is mathematically equivalent to maximizing the number of kept elements. When elements are sorted in non-decreasing order, any valid balanced subset with minimum element nums at the left boundary and maximum element nums at the right boundary can include all intermediate numbers without violating the balance condition.
By sorting the array and maintaining a sliding window with two pointers, we expand the right boundary as long as the value at nums[right] is less than or equal to the value at nums[left] multiplied by k. We track the maximum window size across all candidate left boundaries and calculate the minimum removals as n minus maxKept. We also safeguard against 32-bit integer multiplication overflow using 64-bit long casting, achieving an optimal O(n log n) runtime dominated by sorting.
🔥 Problem Highlights
✅ Inverted the problem from minimizing removals to maximizing kept elements
✅ Identified that any optimal kept subset forms a contiguous range in the sorted array
✅ Implemented an efficient two-pointer sliding window running in linear O(n) scan time
✅ Prevented 32-bit integer overflow by using 64-bit long casting for the multiplication of nums[left] and k
✅ Handled single-element and already-balanced array edge cases seamlessly
✅ Achieved optimal O(n log n) time and O(1) auxiliary space complexity
⏱️ TIMESTAMPS
00:00 Intro
00:56 Problem Understanding and Constraints
01:55 Key Mathematical Insight and Monotonicity
03:21 Sorting and Two-Pointer Sliding Window Strategy
04:26 Java Code Walkthrough
05:24 Execution and Test Cases
06:52 Time Complexity Analysis - O(n log n)
07:42 Space Complexity Analysis - O(1)
08:17 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/6-February/3634-Minimum-Removals-to-Balance-Array/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
Two Pointers Technique
Sliding Window Pattern
Sorting and Greedy Invariants
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 it always optimal to keep a contiguous subarray in the sorted array when finding the maximum balanced subset?
1️⃣ Because including any number between the chosen minimum and maximum never changes the minimum or maximum values
2️⃣ Because elements in an array cannot be removed out of order
3️⃣ Because the problem requires the remaining array to be strictly increasing
4️⃣ Because sliding window only works on contiguous subarrays in unsorted arrays
#LeetCode3634 #LeetCode #MinimumRemovalsToBalanceArray #Java #JavaDSA #Array #Sorting #TwoPointers #SlidingWindow #BinarySearch #DSA #CodingInterview #ProblemSolving #LeetCodeMedium #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description