Smallest Range Covering Elements from K Lists | Multiple Ways | Leetcode 632 | codestorywithMIK
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This is the 20th Video of our Playlist "Heap (Priority Queue): Popular Interview Problems" by codestorywithMIK
In this video we will try to solve a good Array Interval based Problem : Smallest Range Covering Elements from K Lists | Brute Force | Better | Optimal | Intuition | Leetcode 632 | codestorywithMIK
I will do the complete dry run as well which will help you visualize how the algorithm works.
I will explain the intuition so easily that you will never forget and start seeing this as cakewalk EASYYY.
We will do live coding after explanation and see if we are able to pass all the test cases.
Also, please note that my Github solution link below contains both C++ as well as JAVA code.
Problem Name : Smallest Range Covering Elements from K Lists | Brute Force | Better | Optimal | Intuition | Leetcode 632 | codestorywithMIK
Company Tags : will update
My solutions on Github(C++ & JAVA) - https://github.com/MAZHARMIK/Interview_DS_Algo/blob/master/Heap/Smallest Range Covering Elements from K Lists.cpp
Leetcode Link : https://leetcode.com/problems/smallest-range-covering-elements-from-k-lists
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Summary :
Approach 1 (Using Vector of Indexes)
Time Complexity: O(n×k) where n is the average size of each list and k is the number of lists.
Space Complexity: O(k) for the vector that tracks the indices of the elements being considered from each list.
Method:
This approach uses a vector (vec) to track the current index for each list. In each iteration, it scans all lists to find the current minimum and maximum elements and then updates the result range if the new range is smaller. After processing, the index of the list containing the minimum element is advanced. This process continues until one list is exhausted.
Pros: Simple and intuitive.
Cons: The approach scans all lists in every iteration, making it less efficient for large inputs.
Approach 2 (Using Heap/Priority Queue)
Time Complexity: O(n×logk) where n is the average size of each list and k is the number of lists.
Space Complexity: O(k) for the priority queue.
Method:
This approach uses a min-heap (priority queue) to always fetch the smallest element among the current elements from all lists. The heap stores triples containing the current element, the index of the list it belongs to, and its position within the list. In each iteration, the smallest element is popped, and if the current range is smaller than the previous, the result range is updated. The next element from the list containing the smallest element is pushed into the heap, and the maximum element is updated.
✨ Timelines✨
00:00 - Introduction
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