Python - Factors Frequency Dictionary

Given a list with elements, construct a dictionary with frequency of factors.

Input : test_list = [2, 4, 6, 8] Output : {1: 4, 2: 4, 3: 1, 4: 2, 5: 0, 6: 1, 7: 0, 8: 1} Explanation : All factors count mapped, e.g 2 is divisible by all 4 values, hence mapped with 4. Input : test_list = [1, 2] Output : {1: 2, 2 : 1} Explanation : Similar as above, 1 is factor of all.

Method #1 : Using loop 

 This is brute way in which this task can be performed. In this, the elements are iterated and required number is checked for being a factor, if yes, its frequency is increased in dictionary corresponding to its key.

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}

Time Complexity: O(n*n), where n is the elements of dictionary
Auxiliary Space: O(n), where n is the size of dictionary

Method #2 : Using sum() + loop

This is almost similar approach to above problem. The difference being sum() is used for summation rather than a manual loop for solving problem. 

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0}

Time Complexity: O(n*n) where n is the number of elements in the list “test_list”. 
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”. 

Method #3: Using collections.Counter() and itertools.chain()

Import the collections module.
Initialize a dictionary res with all the keys as integers from 1 to the maximum value in the test_list.
Convert the test_list into a list of factors using a nested list comprehension.
Flatten the list of factors into a single list using the itertools.chain() method.
Count the frequency of each factor using the collections.Counter() method and store the result in res.
Print the resulting dictionary res.

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 7: 0, 8: 2, 9: 2, 10: 0, 11: 0, 12: 1, 13: 0, 14: 0, 15: 1, 16: 1, 17: 0, 18: 1}

The time complexity O(n^2), where n is the length of the test_list.
 The auxiliary space  O(n^2), since we are creating a list of factors for each element in test_list, and then flattening it into a single list.

Method #4 : Using list(),set() and count() methods

Approach

1. Find the factors of each number of test_list using nested for loops and append them to an empty list x
2. Remove the duplicates from x using list(),set() and store it in y, create an empty dictionary res
3. Initiate a for loop over list y and initialise the dictionary with elements of y as keys and count of these elements in x as values
4. Display res

The original list : [2, 4, 6, 8, 3, 9, 12, 15, 16, 18]
The constructed dictionary : {1: 10, 2: 7, 3: 6, 4: 4, 5: 1, 6: 3, 8: 2, 9: 2, 12: 1, 15: 1, 16: 1}

Time Complexity : O(M*N) M - length of range 1 to max(test_list) N - length of test_list

Auxiliary Space : O(N) N - length of res dictionary