Python – Sort by Factor count

Last Updated : 28 Apr, 2023

Given element list, sort by factor count of each element.

Input : test_list = [12, 100, 22]
Output : [22, 12, 100]
Explanation : 3, 5, 8 factors respectively of elements.

Input : test_list = [6, 11]
Output : [11, 6]
Explanation : 1, 4 factors respectively of elements.

Method #1 : Using sort() + len() + list comprehension

In this, we perform task of sorting using sort(), and len() and list comprehension is used for task of getting the count of factors.

Python3

   # Python3 code to demonstrate working of
# Sort by Factor count
# Using sort() + len() + list comprehension
 
def factor_count(ele):
 
    # getting factors count
    return len([ele for idx in range(1, ele) if ele % idx == 0])
 
# initializing list
test_list = [12, 100, 360, 22, 200]
 
# printing original list
print("The original list is : " + str(test_list))
 
# performing sort
test_list.sort(key=factor_count)
 
# printing result
print("Sorted List : " + str(test_list))

Output:

The original list is : [12, 100, 360, 22, 200] Sorted List : [22, 12, 100, 200, 360]

Time Complexity: O(nlogn) where n is the number of elements in the list “test_list”.
Auxiliary Space: O(1) additional space is not needed

Method #2 : Using lambda + sorted() + len()

In this, task of sorting is done using sorted(), and lambda function is used to feed to sorted to get factors.

Python3

   # Python3 code to demonstrate working of
# Sort by Factor count
# Using lambda + sorted() + len()
 
# initializing list
test_list = [12, 100, 360, 22, 200]
 
# printing original list
print("The original list is : " + str(test_list))
 
# performing sort using sorted(), lambda getting factors
res = sorted(test_list, key=lambda ele: len(
    [ele for idx in range(1, ele) if ele % idx == 0]))
 
# printing result
print("Sorted List : " + str(res))
 
  

Output:

The original list is : [12, 100, 360, 22, 200] Sorted List : [22, 12, 100, 200, 360]

Method #3: Using a dictionary to store factor counts and sort by value

Algorithm:

Define an empty dictionary factor_counts.
Iterate over the elements of the input list test_list and for each element ele:
a. Define a variable count to store the number of factors of ele.
b. Iterate over the range from 1 to ele, inclusive, and increment count for each factor of ele.
c. Add a key-value pair to factor_counts with key ele and value count.
Sort test_list by factor count using the factor_counts dictionary. To do this, pass a lambda function to the sorted() function as the key parameter, which returns the value of the corresponding key-value pair in factor_counts for each element in test_list.
Return the sorted list.

Python3

   # Python3 code to demonstrate working of
# Sort by Factor count
# Using a dictionary to store factor counts and sort by value
 
# initializing list
test_list = [12, 100, 360, 22, 200]
 
# printing original list
print("The original list is : " + str(test_list))
 
# defining an empty dictionary to store factor counts
factor_counts = {}
 
# iterating over the elements of the input list and storing factor counts in the dictionary
for ele in test_list:
    count = 0
    for idx in range(1, ele + 1):
        if ele % idx == 0:
            count += 1
    factor_counts[ele] = count
 
# sorting the list by factor count using the dictionary
res = sorted(test_list, key=lambda ele: factor_counts[ele])
 
# printing result
print("Sorted List : " + str(res))
 
  

Output

The original list is : [12, 100, 360, 22, 200] Sorted List : [22, 12, 100, 200, 360]

Time complexity: O(n^2), where n is the length of the input list.
Auxiliary space: O(n), where n is the length of the input list.

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