In this tutorial, we will look at how to check if a list is sorted or not in Python with the help of some examples.

## How to check if a list is sorted in Python?

To check if a list is sorted, you can use the following methods –

- Compare the list with its corresponding sorted list.
- Iterate through the list and see if the elements are in order.

Let’s now take a look at the above methods in detail with the help of some examples.

### Compare the list with corresponding sorted list

To check if a list is sorted in Python, compare the list to the list resulting from a sort operation on the original list. If the list resulting after a sort operation is the same as the original list we can say that the original list is sorted.

You can use the python built-in `sorted()`

function to get the sorted list. Let’s take a look at an example.

# create a list ls = [1, 2, 3, 4, 5] # check if list is sorted print(ls == sorted(ls))

Output:

True

Here we get `True`

as the output because the original list `ls`

is already sorted.

Let’s look at another example.

# create a list ls = [1, 2, 4, 3, 5] # check if list is sorted print(ls == sorted(ls))

Output:

False

Here we get `False`

as the output because the list `ls`

does not contain elements in the sorted order.

The worst-case time complexity of this method is O(N Log N) where N is the size of the list.

### By Iterating through the list

You can also check if the list is sorted or not by iterating through the list elements. If any element is out of order (for example, it’s smaller than the previous element) we can say that the list is not sorted in ascending order.

Let’s look at an example.

# function to check if list is sorted in ascending order def is_list_sorted(ls): for i in range(1, len(ls)): # return False if the element is smaller than the previous element if ls[i] < ls[i-1]: return False return True # create a list ls = [1, 2, 3, 4, 5] # check if list is sorted print(is_list_sorted(ls))

Output:

True

We get the same result as above. The list `ls`

is in sorted order and thus we get `True`

as the output.

Let’s see what happens if we use this method on a list that is not sorted.

# create a list ls = [1, 2, 4, 3, 5] # check if list is sorted print(is_list_sorted(ls))

Output:

False

We get `False`

as the output.

The worst-case time complexity of this method is O(N) where N is the size of the list.

You can also use the above methods to check if a list is sorted in descending order or not.

# function to check if list is sorted in descending order def is_list_sorted_desc(ls): for i in range(1, len(ls)): # return False if the element is larger than the previous element if ls[i] > ls[i-1]: return False return True # create a list ls = [5, 4, 3, 2, 1] # check if list is sorted print(ls == sorted(ls, reverse=True)) print(is_list_sorted_desc(ls))

Output:

True True

Here we use both the methods to check if the list `ls`

is sorted in descending order or not.

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