In this program, we need to accept a list and sort it based on the length of the elements present within. Examples:
Input : list = ["rohan", "amy", "sapna", "muhammad",
"aakash", "raunak", "chinmoy"]
Output : ['amy', 'rohan', 'sapna', 'aakash', 'raunak',
'chinmoy', 'muhammad']
Input : list = [["ram", "mohan", "aman"], ["gaurav"],
["amy", "sima", "ankita", "rinku"]]
Output : [['gaurav'], ['ram', 'mohan', 'aman'],
['amy', 'sima', 'ankita', 'rinku']]
Note: The first example comprises of Strings whose
length can be calculated. The second example comprises
of sublists, which is also arranged according to their
length.
There are many ways of performing this. Anyone can use their own algorithmic technique, but Python provides us with various built-in functions to perform these. The built-in functions include sort() and sorted() along with the key parameter. We can perform these in two ways. One way is to sort the list by creating a new list and another way is to sort within the given list, saving space. The syntax for sorting by creating a new list is:
sorted_list = sorted(unsorted_list, key=len)
Python3
# Python code to sort a list by creating# another list Use of sorted()def Sorting(lst): lst2 = sorted(lst, key=len) return lst2 # Driver codelst = ["rohan", "amy", "sapna", "muhammad", "aakash", "raunak", "chinmoy"]print(Sorting(lst)) |
Time complexity: O(n*logn), as sorted() function is used.
Auxiliary Space: O(n), where n is length of lst list.
The syntax for sorting without creating a new list is:
unsorted_list.sort(key=len)
Python3
# Python code to sort a list without# creating another list Use of sort()def Sorting(lst): lst.sort(key=len) return lst # Driver codelst = ["rohan", "amy", "sapna", "muhammad", "aakash", "raunak", "chinmoy"]print(Sorting(lst)) |
Output:
['amy', 'rohan', 'sapna', 'aakash', 'raunak', 'chinmoy', 'muhammad']
Working: These key function of Python’s while sorting implemented is known as the decorate-sort-undecorate design pattern. It follows the following steps:
- Each element of the list is temporarily replaced with a “decorated” version that includes the result of the key function applied to the element.
- The list is sorted based upon the natural order of the keys.
- The decorated elements are replaced by the original elements.
Time Complexity: O(n log n)
Auxiliary Space: O(1)
The code for sorting by creating a new dummy list is:
Python3
import numpydef Sorting(lst): # list for storing the length of each string in list lenlist=[] for x in lst: lenlist.append(len(x)) # return a list with the index of the sorted # items in the list sortedindex = numpy.argsort(lenlist) # creating a dummy list where we will place the # word according to the sortedindex list lst2 = ['dummy']*len(lst) # print(sortedindex,lenlist) for i in range(len(lst)): # placing element in the lst2 list by taking the # value from original list lst where it should belong # in the sorted list by taking its index from sortedindex lst2[i] = lst[sortedindex[i]] return lst2 # Driver codelst = ["rohan", "amy", "sapna", "muhammad", "aakash", "raunak", "chinmoy"]print(Sorting(lst)) |
Output:
['amy', 'rohan', 'sapna', 'aakash', 'raunak', 'chinmoy', 'muhammad']
Reference: stackoverflow
Method: Using lambda function and sorted() function
This approach uses sorted() function along with a lambda function as the key parameter to sort the list based on the length of the elements present within. Below is the algorithm
Algorithm:
Create a list with the elements to.
UsING the sorted() function with a lambda function as the key parameter sort the list based on the length of its elements.
The lambda function takes an element x and returns the length of that element len(x).
The sorted() function returns a new sorted list.
Print the sorted list.
Python3
myList = ["rohan", "amy", "sapna", "muhammad", "aakash", "raunak", "chinmoy"]sortedList = sorted(myList, key=lambda x: len(x))print(sortedList) |
['amy', 'rohan', 'sapna', 'aakash', 'raunak', 'chinmoy', 'muhammad']
Time Complexity: O(n log n)
The sorted() function has a time complexity of O(n log n) for the average and worst-case scenarios, where n is the number of elements in the list.
Space Complexity: O(n)
The auxiliary space of this method is O(n) because a new sorted list is created with n elements.
