Merging $k$ Sorted Lists with a Thin Heap: A $\mathcal{O}(n \lg k)$ Algorithm

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SUMMARY

The discussion focuses on developing an $\mathcal{O}(n \lg k)$ algorithm for merging $k$ sorted lists into a single sorted list using a thin heap. A thin heap is utilized to efficiently manage the merging process by maintaining a min heap of size $k$, where $k$ represents the number of input lists. The algorithm involves inserting the first elements of each list into the heap, repeatedly extracting the minimum element, and replacing it with the next element from the corresponding list until all elements are merged.

PREREQUISITES
  • Understanding of heap data structures, specifically min heaps.
  • Familiarity with the concept of time complexity, particularly $\mathcal{O}(n \lg k)$.
  • Basic knowledge of sorting algorithms and their principles.
  • Experience with algorithm design and analysis.
NEXT STEPS
  • Study the implementation of min heaps in programming languages such as Python or Java.
  • Learn about the heapify operation and its role in maintaining heap properties.
  • Explore the concept of k-way merging and its applications in algorithms.
  • Investigate alternative merging techniques, such as using balanced binary search trees.
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Computer scientists, algorithm developers, and software engineers interested in optimizing merging processes and understanding advanced data structures.

evinda
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Hello! (Wave)

I am asked to write a $Ο (n \lg k)$ - time algorithm that merges $k$ sorted lists into one sorted list, where $n$ is the the total number of elements in all the input lists.
Hint: Use a thin heap for a $k$ -way merging.

Do you have an idea what could be meant with [m] thin heap [/m] ? (Worried)

Also, how could we merge $k$ sorted lists into one using a heap? :confused:
 
Last edited:
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Finally, a min heap is meant...
So do we have to have a heap with $k$ positions, put the elements of the first positions of the $k$ lists in the heap, heapify and delete the root, which will be the smallest element, and put it into the new list, then place at the root the second element from the list from which the minimum was, then heapify and continue the same procedure? (Thinking)
 

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