Merge sort is one of the most efficient sorting algorithms. It works on the principle of **Divide and Conquer** based on the idea of breaking down a list into several sub-lists until each sublist consists of a single element and merging those sublists in a manner that results into a sorted list.

## Merge Sort Working Rule

The concept of Divide and Conquer involves three steps:

- Divide the problem into multiple subproblems.
- Solve the Sub Problems. The idea is to break down the problem into atomic subproblems, where they are actually solved.
- Combine the solutions of the subproblems to find the solution of the actual problem.

So, the merge sort working rule involves the following steps:

- Divide the unsorted array into subarray, each containing a single element.
- Take adjacent pairs of two single-element array and merge them to form an array of 2 elements.
- Repeat the process till a single sorted array is obtained.

An array of Size ‘N’ is divided into two parts ‘N/2’ size of each. then those arrays are further divided till we reach a single element.

The base case here is reaching one single element. When the base case is hit, we start merging the left part and the right part and we get a sorted array at the end.

Merge sort repeatedly breaks down an array into several subarrays until each subarray consists of a single element and merging those subarrays in a manner that results in a sorted array.

### Merge Sort Algorithm Flow

Array = *{70,50,30,10,20,40,60}*

We repeatedly break down the array in two parts, the left part, and the right part. the division takes place from the mid element.

We divide until we reach a single element and then we start combining them to form a Sorted Array.

## Merge Sort Implementations

We will implement the merge sort algorithm in Java, C, and Python.

### 1. Java Implementation

```
package com.journaldev.ds;
public class MergeSort {
public static void main(String[] args) {
int[] arr = { 70, 50, 30, 10, 20, 40, 60 };
int[] merged = mergeSort(arr, 0, arr.length - 1);
for (int val : merged) {
System.out.print(val + " ");
}
}
public static int[] mergeTwoSortedArrays(int[] one, int[] two) {
int[] sorted = new int[one.length + two.length];
int i = 0;
int j = 0;
int k = 0;
while (i < one.length && j < two.length) {
if (one[i] < two[j]) {
sorted[k] = one[i];
k++;
i++;
} else {
sorted[k] = two[j];
k++;
j++;
}
}
if (i == one.length) {
while (j < two.length) {
sorted[k] = two[j];
k++;
j++;
}
}
if (j == two.length) {
while (i < one.length) {
sorted[k] = one[i];
k++;
i++;
}
}
return sorted;
}
public static int[] mergeSort(int[] arr, int lo, int hi) {
if (lo == hi) {
int[] br = new int[1];
br[0] = arr[lo];
return br;
}
int mid = (lo + hi) / 2;
int[] fh = mergeSort(arr, lo, mid);
int[] sh = mergeSort(arr, mid + 1, hi);
int[] merged = mergeTwoSortedArrays(fh, sh);
return merged;
}
}
```

**OUTPUT**

### 2. C Implementation

```
#include <stdio.h>
void merge(int arr[], int l, int m, int r)
{
int i, j, k;
int n1 = m - l + 1;
int n2 = r - m;
/* create temp arrays */
int L[n1], R[n2];
/* Copy data to temp arrays L[] and R[] */
for (i = 0; i < n1; i++)
L[i] = arr[l + i];
for (j = 0; j < n2; j++)
R[j] = arr[m + 1+ j];
/* Merge the temp arrays back into arr[l..r]*/
i = 0; // Initial index of first subarray
j = 0; // Initial index of second subarray
k = l; // Initial index of merged subarray
while (i < n1 && j < n2)
{
if (L[i] <= R[j])
{
arr[k] = L[i];
i++;
}
else
{
arr[k] = R[j];
j++;
}
k++;
}
/* Copy the remaining elements of L[], if there
are any */
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
/* Copy the remaining elements of R[], if there
are any */
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
/* l is for left index and r is the right index of the
sub-array of arr to be sorted */
void mergeSort(int arr[], int l, int r)
{
if (l < r)
{
// Same as (l+r)/2, but avoids overflow for
// large l and h
int m = l+(r-l)/2;
// Sort first and second halves
mergeSort(arr, l, m);
mergeSort(arr, m+1, r);
merge(arr, l, m, r);
}
}
void printArray(int A[], int size)
{
int i;
for (i=0; i < size; i++)
printf("%d ", A[i]);
printf("\n");
}
/* Driver program to test above functions */
int main()
{
int arr[] = {70, 50, 30, 10, 20, 40,60};
int arr_size = sizeof(arr)/sizeof(arr[0]);
printf("Given array is \n");
printArray(arr, arr_size);
mergeSort(arr, 0, arr_size - 1);
printf("\nSorted array is \n");
printArray(arr, arr_size);
return 0;
}
```

**OUTPUT**

### 3. Python Implementation

```
def merge_sort(unsorted_array):
if len(unsorted_array) > 1:
mid = len(unsorted_array) // 2 # Finding the mid of the array
left = unsorted_array[:mid] # Dividing the array elements
right = unsorted_array[mid:] # into 2 halves
merge_sort(left)
merge_sort(right)
i = j = k = 0
# data to temp arrays L[] and R[]
while i < len(left) and j < len(right):
if left[i] < right[j]:
unsorted_array[k] = left[i]
i += 1
else:
unsorted_array[k] = right[j]
j += 1
k += 1
# Checking if any element was left
while i < len(left):
unsorted_array[k] = left[i]
i += 1
k += 1
while j < len(right):
unsorted_array[k] = right[j]
j += 1
k += 1
# Code to print the list
def print_list(array1):
for i in range(len(array1)):
print(array1[i], end=" ")
print()
# driver code to test the above code
if __name__ == '__main__':
array = [12, 11, 13, 5, 6, 7]
print("Given array is", end="\n")
print_list(array)
merge_sort(array)
print("Sorted array is: ", end="\n")
print_list(array)
```

**OUTPUT**

## Merge Sort Time and Space Complexity

### 1. Space Complexity

Auxiliary Space: **O(n)**

Sorting In Place: No

Algorithm :** Divide and Conquer**

### 2. Time Complexity

Merge Sort is a recursive algorithm and time complexity can be expressed as following recurrence relation.

**T(n) = 2T(n/2) + O(n)**

The solution of the above recurrence is ** O(nLogn)**. The list of size N is divided into a max of Logn parts, and the merging of all sublists into a single list takes O(N) time, the worst-case run time of this algorithm is O(nLogn)

Best Case Time Complexity: **O(n*log n)**

Worst Case Time Complexity: **O(n*log n)**

Average Time Complexity: **O(n*log n)**

The time complexity of MergeSort is O(n*Log n) in all the 3 cases (worst, average and best) as the mergesort always divides the array into two halves and takes linear time to merge two halves.

**Further Readings:**