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Sorting by combining Insertion Sort and Merge Sort algorithms

  • Last Updated : 10 Nov, 2021

Insertion sort: The array is virtually split into a sorted and an unsorted part. Values from the unsorted part are picked and placed at the correct position in the sorted part.
Advantages: Following are the advantages of insertion sort:

  • If the size of the list to be sorted is small, insertion sort runs faster
  • Insertion sort takes O(N) time when elements are already sort
  • It is an in-place algorithm O(1), no auxiliary space is required

Merge sort: Merge Sort is a Divide and Conquer algorithm. It divides the input array into two halves, calls itself for the two halves, and then merges the two sorted halves.

Advantages: Following are the advantages of merge sort:

  • The division of the main problem into sub-problem has no major cost

From the above two comparisons, the advantages of both the sorting algorithms can be combined, and the resulting algorithm will have time complexity O(N[K+log(N/K)]). Below is the derivation of the time complexity of this combined algorithm:
Let, no. of elements in the list = N
Divide:

  • We first divide these N elements into (N/K) groups of size K

Sorting:

  • For each division of subarray of size K, perform the insertion sort operation to sort this subarray
  • The total cost of insertion sort for a single bock of K elements:
    • For best case: O(K)
    • For worst case: O(K^{2})
  • Since there are (N/K) such blocks each of size K, we get the total cost of applying insertion sort as:
    • For best case: (N/K) * K = O(N)    <– (1)
    • For worst case: (N/K) * K^{2} = O(NK)    <– (2)

Merging:

  • After applying Insertion sort on (N/K) groups each of K sorted elements
  • For merging these (N/K) groups:

  • Let’s say we take i iterations of merge sort. So, for the loop to stop we will need to equate as:
  • (2^i) * K = N
  • (2^i) = N/K
  • i*log(2) = log(N/K)        Taking log on both sides
  • i = log(N/K)
  • Cost of merging = O(N)
  • Total cost of merging = No. of iteration * Cost of iteration
    • = log(N/K)*N
    • = N*log(N/K)
    • = O(N*Log(N/K))      <– (3)

The total cost of the algorithm (insertion + merge) is:

  • Best case:  N+Nlog(N/K)             <– from (1) and (3)
  • Worst case: NK + Nlog(N/K)       <– from (2) and (3)

If K = 1, then it is complete merge sort which is better in terms of time complexity
If K = N, then it is complete Insertion sort which is better in terms of space complexity

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