Smallest Range Covering Elements from K Lists | Leetcode 632 | 3 Approaches | Java |Developer Coder

In this video, we'll tackle the LeetCode problem Smallest Range Covering Elements from K Lists. I will walk you through 3 different approaches to solve the problem efficiently:
1️⃣ Brute Force Approach
2️⃣ Priority Queue Approach
3️⃣ 2 Pointers (Sliding Window) Approach
Whether you're preparing for coding interviews or simply expanding your knowledge, this video will provide you with the skills you need to tackle similar problems. I’ll be using Java for implementing these solutions, providing a clear and in-depth explanation of each approach.
Stay tuned and improve your coding skills with me on Developer Coder! 🚀
👨‍💻 Topics Covered:
LeetCode 632 Solution
Brute Force Approach
Priority Queue Approach
Sliding Window Approach
2 Pointers Technique
Java Implementation
📌 Check out more videos on LeetCode solutions and coding challenges on Developer Coder!
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[@DeveloperCoder]

2 pointers: (Sliding Window) (Time Complexity)
Let n be the total number of elements across all lists and k be the number of lists.
Time complexity: O(nlogn)
The first nested loop iterates over k lists, and for each list, it iterates through its elements. In the worst case, this requires O(n) time since we are processing all elements once.
After merging, we sort the merged array which contains n elements. Sorting has a time complexity of O(nlogn).
The two-pointer approach iterates through the merged list once (with the right pointer) and may also move the left pointer forward multiple times. In total, each pointer will traverse the merged list at most n times.
Combining these steps, the overall time complexity is: O(nlogn)

[@DeveloperCoder]

Priority Queue: (Time Complexity)
Let n be the total number of elements across all lists and k be the number of lists.
Time complexity: O(nlogk)
The initial loop that inserts the first element from each list into the priority queue runs in O(k). The while loop continues until we have exhausted one of the lists in the priority queue. Each iteration of the loop involves:
Extracting the minimum element from the priority queue, which takes O(logk).
Inserting a new element from the same list into the priority queue, which also takes O(logk).
In the worst case, we will process all n elements, leading to a total complexity of O(nlogk).

[@DeveloperCoder]

Brute Force: (Time Complexity )
Let n be the total number of elements across all lists and k be the number of lists.
Time complexity: O(n⋅k)
In each iteration of the while (true) loop, we traverse all k lists to find the current minimum and maximum. This takes O(k) time.
The loop continues until at least one of the lists is fully traversed. In the worst case, every element from every list is visited, and the total number of elements across all lists is n. Therefore, the loop runs O(n) times.
Overall, the time complexity becomes O(n⋅k).