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From C# to Python, the functions and logic of the sorted arrays differ in all programming languages.

For instance in case of C# we input the “System.Array.Sort(array)” logic, while in the case of Python we input the “array.sort()” function.

The main function of a sorted array is to optimize the search operation by making use of the binary search function. This can be potentially tracked by **merging two sorted arrays**.

Along with merging two sorted arrays, the programmers also have to keep track of time complexity which generally arises because of the recursive calls.

The problem with time complexity is made null and void by an efficient concept in programming i.e the Binary Tree data structure.

This blog will unfold the process of finding solutions for the time complexity issue while also discussing **successor or predecessor** within the Binary Tree.

**How to merge two sorted arrays by managing the time complexity?**

The function used for merging two sorted arrays is the “SortedMerge()” logic. Take for instance that you have been given two sorted arrays a and b.

Now, in order to merge the two you will have to use the SortedMerge() function by the end of each array. This logic will ultimately merge the two sorted arrays and generate your desired output by the end of the program.

The only default or loophole in this concept is managing the factor of time complexity. This is what we will be elaborating on further.

Merging two sorted arrays by maintaining the time complexity can only be explained by the use of examples.

**Example**

You have been provided two sorted arrays and you are being asked to merge the two in the similar sorted order.

**Input**

arr1 [] = {1,2,3,4} and arr2 [] = {5,6,7,8}

**Output**

arr3 [] = {1,2,3,4,5,6,7,8}

**Input**

arr1 [] = {5,6,7,8} and arr2 [] = {1,2,3,4}

**Output**

arr3 [] = {8,7,6,5,4,3,2,1}

In the above example, we have been given two different sorted arrays to solve in order to merge keeping track of the time complexity.

Now, in order to solve the problem we can employ three different methods and figure out which one of them has a direct impact in reducing the time complexity.

**Method 1:**

**The Native Approach**

The native approach for solving the time constraint while merging two given sorted arrays is to apply the brute force method.

This method is known for searching and optimising all the possible outputs for a given problem in order to figure out the best possible solution.

If you solve the time constraint problem by merging the two given sorted arrays i.e a and b respectively using the C++ language, you will observe the following time constraint:

**Time complexity:** O ((a+b) log(a+b)), the size of the entire arr3 is a+b

**Method 2: **

**Using Extra Space**

The function employed for this method can be written as follows:

(O(a1×a2) Time and O(a1+a2) Extra Space)

Take a look at the steps you need to imply in order to apply the extra space logic:

- Firstly start with creating an arr3[] of the sizes a1 and a2
- Next up, copy all the a1 elements from the arr1[] to arr3[]
- Traverse the arr2[] array and start inserting elements from the arr1[] and the arr3[] list.

**Time complexity:** O(a1 × a2)

**Method 3:**

**Using Merge Sort Algorithm**

In the third method we will employ the Merge Sort Algorithm which is depicted by the Merge () function.

The idea here is to follow the sequence mentioned below:

- Design an array of arr3[] of size a1+a2
- Next up, traverse the two formerly given arrays from the example question i.e arr1[] and arr2[]
- If you figure out that there are a few remaining elements in the two input arrays, copy them in the third array i.e arr3[]

**Time complexity:** O(a1 + a2)

**Method 4:**

**Using Maps, Time and Extra Space**

The functions used for depicting the logic for maps is (O(alog(a) + blog(b)).

In order to apply this method, you will have to mention the elements from both the arrays within the maps in the form of keys.

This will help you derive the required output by printing the keys of the map.

**Time complexity:** (O(alog(a) + blog(b))

The key takeaway from all the approaches is that the best approach for merging two sorted arrays while defying the time complexity is using the native approach.

This employs the brute force method which is closely based on the recursion method in programming. In this approach you can solve the problem with all the possible procedures and decide the easiest method for yourself.

As a programmer, the main motive of designing data structures is to combat the time complexity. Within this context, the Binary Tree is definitely worth mentioning.

Due to its hierarchical structure, the time complexity for figuring out the location of elements and data is efficiently reduced. This is initiated by the successor and the predecessor of the Binary Tree.

Take a quick look at the definition.

**What are successor and predecessor in context of the Binary Trees?**

The successor of a Binary Tree is referred to as the smallest value of node present in the left part of the subtree.

On the other hand, the predecessor of a Binary Tree is referred to as the highest value present in the left subtree of a Binary Tree.

**Wrapping Up**

The whole concept of data structures and algorithms is based on effectively managing the time constraint while dealing with huge quantities of data.

By using the brute force approach for **merging two sorted arrays** can certainly help reduce the time complexity. When it comes to hierarchically structuring data for enhancing productivity, implementing the Binary Trees is definitely the first choice for the programmers. This is owed to the **successor and predecessor** of the Binary Tree structure.