#Mergex sort time complexity mod
#Mergex sort time complexity code

This can be derived as follows:( Here 2 is base) Advantages:Д«est and worst-case efficiency is O(nlog2n). Hence the time complexity of Merge Sort is O(n log2 n). In each iteration, n elements are merged. The total number of iterations in Merge sort is log2n. Step 3: Merge (a,0,4) i.e Merge a and a into a which is the sorted list Time Complexity of Merge Sort The same steps are carried out for the second half. Merge a and a into a i.e ДЄt this stage, the first half is sorted. Let us consider an example of sorting 5 elementsДЄ=. In such a case, one sublist will have an additional element. However, the list may contain an odd number of elements. In the example shown, each sublist has an even number of elements. The sublists are then merged to create sorted lists until all elements are in a single list. The elements are successively divided into two sublists till we cannot divide any further i.e, we reach a sublist of size 1. Let us consider the example of sorting 20,40,50,15,30,35,10 and 5.Initially they are all in one list. Merge (a,mid,high) //Merge sorted sub lists MergeSort (a,mid+1,high) //sort second sub list MergeSort (a,low,mid) //Sort first sub list Mid=(low + high) / 2 //Divide a into two sublists It's obvious that you have to store the elements somewhere, but it's always better to mention 'additional memory' to keep purists at bay.Void MergeSort (int a, int low, int high) That requires constant additional memory.Д) That's why in merge-sort complexity analysis people mention 'additional space requirement' or things like that.

In case of lists it's O(log n) additional memory.Д¬) Lists only need some pointers changed during the merge process. If you count stack frames, then it's O(n)+ O(log n), so still O(n) in case of arrays. In practice implementations using lists need additional space for list pointers, so unless you already have the list in memory it shouldn't matter. Additional space complexity can be O(n) in an implementation using arrays and O(1) in linked list implementations. In not-so-perfect-world you don't have infinite number of processors and context-switching and synchronization offsets any potential gains.Д«) Space complexity is always О©(n) as you have to store the elements somewhere.

n/4, n/2, n - they can't be parallelized), which gives O(n). After returning from previous mergesort only do you create the next array.ДЄ) Yes - in a perfect world you'd have to do log n merges of size n, n/2, n/4.

#Mergex sort time complexity mod

note: doing and operator is much faster in hardware than calculating the mod (%) Void mergesort(X a, int n) // X is a type using templates Notice how number of space used is 32 = 2n = 2*16

#Mergex sort time complexity code

For example, in the case where N = 16, this is how the code for mergesort executes. However, this is not the case in the actual code as it does not execute in parallel. Where height of tree is O(logn) => Space complexity is O(nlogn + n) = O(nlogn). However, as the code is a Depth First code, you will always only be expanding along one branch of the tree, therefore, the total space usage required will always be bounded by O(3n) = O(n).ДЇor example, if you draw the space tree out, it seems like it is O(nlgn) 16 | 16 If you draw the space tree out, it will seem as though the space complexity is O(nlgn). Merge Sort space complexity will always be O(n) including with arrays. MergeSort time Complexity is O(nlgn) which is a fundamental knowledge.