LeetCode 3013 | Divide an Array Into Subarrays With Minimum Cost II | Java | Hard | CodeByTushu
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In today’s video, we solve LeetCode 3013 - Divide an Array Into Subarrays With Minimum Cost II using Sliding Window and Two Balanced Sets in Java.
We are given an array of integers and need to partition it into k contiguous disjoint subarrays such that the difference between the starting indices of the second and kth subarrays does not exceed dist. Since the first subarray must start at index 0, its cost is unconditionally fixed to nums[0].
The problem naturally reduces to finding a sliding window of size at most dist + 1 across the array elements from index 1 to the end that minimizes the sum of its k - 1 smallest elements. We dynamically maintain the k - 1 smallest elements using two balanced TreeSets in O(N log(dist)) time and O(dist) auxiliary space, rebalancing dynamically as the window advances.
🔥 Problem Highlights
✅ Subarray 1 cost is unconditionally fixed to nums[0]
✅ The distance constraint between the starting indices of the kth and second subarrays cannot exceed dist, restricting all k - 1 selected starting indices to a sliding window of size dist + 1
✅ Reduction to finding the sum of the k - 1 smallest elements in a sliding window
✅ Dynamic Top-M maintenance using two balanced TreeSets called Left and Right
✅ O(log(dist)) insertion, deletion, and rebalancing per window shift
✅ Time Complexity: O(N log(dist)) | Space Complexity: O(dist) auxiliary memory
⏱️ TIMESTAMPS
00:00 Intro
00:46 Problem Understanding
01:53 Key Observation
02:54 Two Balanced TreeSets and Sliding Window
04:01 Java Code Walkthrough
05:17 Execution and Test Cases
06:14 Time Complexity Analysis
07:02 Space Complexity Analysis
07:40 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/solution.js
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💬 Question of the Day
Why does the distance constraint between the starting indices of the second and kth subarrays imply that all k - 1 selected starting indices must lie inside a window of size at most dist + 1?
1️⃣ Because the array elements are sorted in non-decreasing order
2️⃣ Because the difference between the maximum and minimum index among the k - 1 choices cannot exceed dist, bounding their entire index span to at most dist + 1
3️⃣ Because the first subarray must have a length of exactly dist
4️⃣ Because all k subarrays must have equal lengths
#LeetCode3013 #LeetCode #DivideAnArrayIntoSubarraysWithMinimumCostII #Java #JavaDSA #SlidingWindow #Heap #DataStructure #TreeSet #DSA #CodingInterview #ProblemSolving #LeetCodeHard #DailyLeetCode #JavaProgramming #CodeByTushu
👇 Comment the answer below! 🔥 Close description
In today’s video, we solve LeetCode 3013 - Divide an Array Into Subarrays With Minimum Cost II using Sliding Window and Two Balanced Sets in Java.
We are given an array of integers and need to partition it into k contiguous disjoint subarrays such that the difference between the starting indices of the second and kth subarrays does not exceed dist. Since the first subarray must start at index 0, its cost is unconditionally fixed to nums[0].
The problem naturally reduces to finding a sliding window of size at most dist + 1 across the array elements from index 1 to the end that minimizes the sum of its k - 1 smallest elements. We dynamically maintain the k - 1 smallest elements using two balanced TreeSets in O(N log(dist)) time and O(dist) auxiliary space, rebalancing dynamically as the window advances.
🔥 Problem Highlights
✅ Subarray 1 cost is unconditionally fixed to nums[0]
✅ The distance constraint between the starting indices of the kth and second subarrays cannot exceed dist, restricting all k - 1 selected starting indices to a sliding window of size dist + 1
✅ Reduction to finding the sum of the k - 1 smallest elements in a sliding window
✅ Dynamic Top-M maintenance using two balanced TreeSets called Left and Right
✅ O(log(dist)) insertion, deletion, and rebalancing per window shift
✅ Time Complexity: O(N log(dist)) | Space Complexity: O(dist) auxiliary memory
⏱️ TIMESTAMPS
00:00 Intro
00:46 Problem Understanding
01:53 Key Observation
02:54 Two Balanced TreeSets and Sliding Window
04:01 Java Code Walkthrough
05:17 Execution and Test Cases
06:14 Time Complexity Analysis
07:02 Space Complexity Analysis
07:40 Outro and Wrap-up
📂 Problem Folder and Detailed Documentation:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/tree/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II
☕ Java Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/Solution.java
🐍 Python Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/solution.py
⚡ C++ Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/solution.cpp
🌐 JavaScript Solution:
https://github.com/CodeByTushu/CodeByTushu-LeetCode-Solutions/blob/main/solutions/February/2-February/3013-Divide-an-Array-Into-Subarrays-With-Minimum-Cost-II/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 Hard Problems
Java DSA Preparation
Sliding Window Top-K
Two Balanced Multisets and TreeSets
Priority Queue 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 the distance constraint between the starting indices of the second and kth subarrays imply that all k - 1 selected starting indices must lie inside a window of size at most dist + 1?
1️⃣ Because the array elements are sorted in non-decreasing order
2️⃣ Because the difference between the maximum and minimum index among the k - 1 choices cannot exceed dist, bounding their entire index span to at most dist + 1
3️⃣ Because the first subarray must have a length of exactly dist
4️⃣ Because all k subarrays must have equal lengths
#LeetCode3013 #LeetCode #DivideAnArrayIntoSubarraysWithMinimumCostII #Java #JavaDSA #SlidingWindow #Heap #DataStructure #TreeSet #DSA #CodingInterview #ProblemSolving #LeetCodeHard #DailyLeetCode #JavaProgramming #CodeByTushu
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