How to analyze the time complexity for Merge Sort?

- This video helps a lot: https://www.youtube.com/watch?v=0nlPxaC2lTw
- Prereq: how to understand O(n), O(logn) etc.
- say the input size is
*n*, then for a loop that starts from 0 and iterates until*n*or*n-1*, we say the time complexity for such a loop is O(n)- this is also why, we say the time complexity for binary search is O(logn)
- i.e. in binary search, we only need to search logn times before we find the target because each time, we cut the size of the input by half, e.g. with an input of size 16, we only need to cut 4 times or less before we could find the target, thus, it’s log16 = 4

- this is also why, we say the time complexity for insertion sort is O(n2)
- i.e. insertion sort has two for loops, each for loop has to iterate
*n*times, thus, in total, it will have to loop through n*n, i.e. O(n2) time complexity

- i.e. insertion sort has two for loops, each for loop has to iterate

- this is also why, we say the time complexity for binary search is O(logn)

- say the input size is
- Then based on the code below, merge sort time complexity could be analyzed as below:
- we use T(n) to denote the total running time for merge sort to work on an array of size n
- then sort(arr, l, m) takes T(n/2) time since it’s the half size of the original size
- similarly, sort(arr, m+1, r) also takes T(n/2) time
- but merge(arr, l, m, r) will take
*c**n time (*c*as a constant) since we’ll have to go through the entire array - then T(n) could be represented as
**T(n) = 2*T(n/2) +**(*c**n*function*1) - apparently, this is a recursive function, from it, we could get the following:
- T(n/2) = 2*T(n/4) +
*c**n - T(n/4) = 2*T(n/8) +
*c**n - T(n/8) = 2*T(n/16) +
*c**n - …..
- take those into the
*function 1*and simplify it, we could get:**T(n) = 2kT(n/2k)+ k***(*c**n*function*2) - we want to deduce this to T(1) because that is what we know: constant time.
- in that case, n/2k will be equal to 1 , then we could get n = 2k, then we use log on both sides of the equation, we could get
**logn = log2k = k**(*function*3) - we take
*function*3 into*function*2, we could get**T(n) = 2lognT(1)+ logn***, then using Big O notation (lower-order and constant operations could be neglected), this could be represented as T(n) = O(logn*n) = O(n*logn), i.e. the so-called O(nlogn) time complexity for MergeSort.*c**n

- T(n/2) = 2*T(n/4) +

Merge Sort is a stable sorting algorithm: the order of equal keys in the original input is respected.

Merge Sort is NOT an in-place algorithm: the amount of extra space needed for merge sort is proportional to the size of the input array, this is different from other sorting algorithms like insertion sort or bubble sort which requires only a temp variable to do swap. However, in merge sort, we divide the arrays into two sub-arrays and we make copy of the entire two sub-arrays. Thus, the extra memory required for doing merge sort is proportional to the size of the input array. We say the * space* complexity of merge sort is O(n) which means its extra memory required is proportional to the size of the input array.

Merge sort works like this:

It partitions the original array into the most atomic elements, then compare and combine them, thus the code looks like the following:

public static void main(String... args){ int[] a = new int[]{4,3,9,8,7}; MergeSort test = new MergeSort(); test.sort(a, 0, a.length-1); } public void sort(int[] arr, int l, int r){ if(l &amp;amp;amp;lt; r){ int m = (l+r)/2; sort(arr, l, m); sort(arr, m+1, r); merge(arr, l, m, r); } } private void merge(int[] arr, int l, int m, int r) { //find sizes of two subarrays that are to be merged int size1 = m-l+1; int size2 = r-m; //copy the two subarrays into two temp arrays int[] tempL = new int[size1]; int[] tempR = new int[size2]; for(int i = 0; i &amp;amp;amp;lt; size1; i++){ tempL[i] = arr[l+i]; } for(int i = 0; i &amp;amp;amp;lt; size2; i++){ tempR[i] = arr[m+i+1]; } //now we merge the two subarrays //initial indices of the two subarrays int i = 0, j = 0; //initial index of the merged subarray array int k = l; while(i &amp;amp;amp;lt; size1 &amp;amp;amp;amp;&amp;amp;amp;amp; j &amp;amp;amp;lt; size2){ if(tempL[i] &amp;amp;amp;lt;= tempR[j]){ arr[k] = tempL[i]; i++; } else { arr[k] = tempR[j]; j++; } k++; } //copy remaining list into arr if any while(i &amp;amp;amp;lt; size1){ arr[k++] = tempL[i++]; } while(j &amp;amp;amp;lt; size2){ arr[k++] = tempR[j++]; } }

A sample walkthrough of the code (omitted *merge()* function), recursion is so cool/powerful!