Learning Algorithms  -  Merge Sort
Tags: computerscience, algorithms, python, mergesort
Published: 2020-10-08 08:53:53

Previously, we had a look of [Insertion Sort](https://medium.com/python-in-plain-english/learning-algorithms-insertion-sort-71e64a6388d) and implemented it. 

However, its worst-case scenario was ***O(n²)***, which is really slow isn't it? and I promised you to more sophisticated and faster algorithms to sort an array that is **Merge Sort**.

# Divide and Conquer
So how can we save time for sorting when it comes to a list of elements? The answer is to **divide and conquer**. I assume you must know **binary search** at this juncture, where you already started learning algorithms or you can go back to learn [binary search](https://medium.com/python-in-plain-english/learning-algorithms-binary-search-601289e455db). Let me give you a big picture of the whole plan. Firstly, we will have an array of size **n**, we will call it **A**. Secondly, We will then divide in a half (until it is not dividable), which will be two arrays Left and Right. Next, the two arrays will be sorted. Lastly, we will **merge** them in a **sorted order** back in array **A**.

![1_Y2FZaJLxvUDfcaaj3ckj9w](https://dev-to-uploads.s3.amazonaws.com/i/dnxt5ezu2uq5hw61xotc.png)
 

# Recursion
Unfortunately, to use Merge Sort, we need to know this notoriously confusing concept, so-called **recursion**. In a nutshell, recursion is **a function call that calls itself**. Let’s take a look of code right below.
```python
def call_me(arg):
    call_me(arg)
    print(arg)
call_me("I am calling me")
```
![1_z5aRQeU2pBiJMEKMTK4C9g](https://dev-to-uploads.s3.amazonaws.com/i/y9d0qh3zd5o8h2bu98zr.png)
The function **call_me(arg)** calls itself, then prints **arg**. What will happen in this case? It will call itself infinitely printing “I am calling me” like a black hole. It will cause, the notorious name of error, **stack overflow**, which is literally function calls being stacked up and program used up memory spaces to store those calls, so thrown an error. So all we need to do is let this recursion finishing at some point, to do that, we need a **condition to end** the recursion, we call it **base case**.
![1_XXDyk0U73T8KeFDZioshXg](https://dev-to-uploads.s3.amazonaws.com/i/8i4snoqb7i4uehz46h86.png)
Let me stop this recursion with a base case. the call_me function is taking one more argument, i, that increments by 1 every execution, and if i becomes 5 the function is going to be returned.
```python
def call_me(arg, i):
    if i == 5:  # if i == 5, return
        return
   
    i += 1
    call_me(arg, i)  # otherwise keep executing with incremented i
    print(arg)
call_me("I am calling me", 0)
```
![1_sKQApV0u2_ZYWdcAcXRuPA](https://dev-to-uploads.s3.amazonaws.com/i/t1evtrr31gju2y79kb2w.png)


# Merge Sort
### Copying halves of the array A
I assume we are now ready to get cracking. We have this unsorted array A as follows.
```python
A = [4, 7, 20, 3, 9, 13, 5, 11]
```
As I mentioned earlier in a planning phase, we are going to copy arrays of left half(L) and right half(R) respectively.
```python
def merge_sort(A):
    mid = len(A) // 2
    # copy slicing from the 0 to mid from array A
    L = A[:mid]
    # copy slicing from mid to the end from array A
    R = A[mid:] 
```
![1_Li9UK-FOoiTmD5T5v0w20w](https://dev-to-uploads.s3.amazonaws.com/i/r5xnu5hmko66csdksrpg.png)
We are going to recursively halve the arrays until it is not dividable. As I explained earlier we need a base case to stop recursion at some point, and it will be until the length of the array **A** becomes **1** since there is nothing to sort and the array is individable.
```python
 def merge_sort(A):
    if len(A) > 1:
        ... previous code
        # recursively halve arrays
        merge_sort(L)
        merge_sort(R)
        
        # indices for array A, L, R 
        # for short term, i = j = k = 0 works as well.
        i = 0
        j = 0
        k = 0
```
![1_cv-aRDjjDsTR0fLJMP1Tzw](https://dev-to-uploads.s3.amazonaws.com/i/5d7w1wnfegxf1ndfm5x9.png)
As a result of the code above, arrays will look like this. Note that, each level of halving process happens simultaneously, so in this case, we took 4 steps to halve an array length of 8.

### Merge and sort
Now, it is time to merge and sort back to an array A, to sort elements, we need to compare which one should come first. the first one would be the smaller ones, the larger one comes next. Implementation of the merge and sort will be like so, once an element is placed back in array A, index moves to a next element.
```python
# while indices are within a range of array L and R
# note that indices i for L, j for R, k for A
while i < len(L) and j < len(R):
    if L[i] < R[j]:
        A[k] = L[i]
        i += 1
    else:
        A[k] = R[j]
        j += 1
    k += 1
```
There is the last thing left, that is to insert an element that hasn’t been moved to array A from L or R. This can happen when the length of L and R is different, in other words when the length of A is an odd number like 7.
```python
... previous code
while i < len(L):
    A[k] = L[i]
    i += 1
    k += 1
while j < len(R):
    A[k] = R[j]
    j += 1
    k += 1
```
The process of the merge would look like this.
![1_AECCzpdvdQBk-7aJyiWxqw](https://dev-to-uploads.s3.amazonaws.com/i/bvfcl0p1zhqpom9pwg5j.png)


# Time and Space Complexity Analysis
### Time Complexity of Merge Sort
Seemingly, the time complexity of this algorithm **O(n log n)** because of divide and conquer phases (the halving arrays). But what about the merge and sort phases? _It loops through each array to sort and merge_, which is **O(n)**. When worst-case exists, the other case doesn’t matter, so the time complexity is **O(n)**.

### Space Complexity of Merge Sort
How many memory spaces merge sort uses? A lot! it copies every half of the array and that means extra spaces, depending on the number of elements. Therefore space complexity of Merge Sort is O(n). In comparison to Insertion sort, it would be Merge sort O(n) > Insertion sort O(1).
It has been a long journey to the end, I hope you now have a good grasp of the concept.