Python - Reorder for consecutive elements

Given a List perform reordering to get similar elements in consecution.

Input : test_list = [4, 7, 5, 4, 1, 4, 1, 6, 7, 5] 
Output : [4, 4, 4, 7, 7, 5, 5, 1, 1, 6] 
Explanation : All similar elements are assigned to be consecutive.

Input : test_list = [4, 7, 5, 1, 4, 1, 6, 7, 5] 
Output : [4, 4, 7, 7, 5, 5, 1, 1, 6] 
Explanation : All similar elements are assigned to be consecutive. 

Method #1 : Using Counter() + loop + items()

In this, we perform the task of computing frequency using Counter(), and loop and items() are used to reorder elements according to count, and access frequencies respectively.

# Python3 code to demonstrate working of 
# Reorder for consecutive elements
# Using Counter() + loop + items()
from collections import Counter

# initializing list
test_list = [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]

# printing original lists
print("The original list is : " + str(test_list))

# getting frequency
freqs = Counter(test_list)
res = []

# reordering basis of frequency
for val, cnt in freqs.items():
    res.extend([val]*cnt)

# printing result 
print("Reordered List : " + str(res))

Output:

The original list is : [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]
Reordered List : [4, 4, 4, 7, 7, 5, 5, 1, 1, 6]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #2 : Using Counter() + elements()

In this, we perform the task of reordering the counted frequencies using elements(), providing a concise solution.

# Python3 code to demonstrate working of 
# Reorder for consecutive elements
# Using Counter() + elements()
from collections import Counter

# initializing list
test_list = [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]

# printing original lists
print("The original list is : " + str(test_list))

# reordering using elements()
res = list(Counter(test_list).elements())

# printing result 
print("Reordered List : " + str(res))

Output:

The original list is : [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]
Reordered List : [4, 4, 4, 7, 7, 5, 5, 1, 1, 6]

Time Complexity: O(n)
Auxiliary Space: O(n)

Method #3: Using dictionary comprehension + sorted()

• Convert the list into a dictionary with keys as elements and values as their frequency using dictionary comprehension.
• Use sorted() function to sort the dictionary based on frequency in descending order.
• Create an empty list 'res' to store the reordered list.
• Iterate through the sorted dictionary and append each element to the 'res' list as many times as its frequency.
• Return the reordered list.

# Python3 code to demonstrate working of 
# Reorder for consecutive elements
# Using dictionary comprehension + sorted()

# initializing list
test_list = [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]

# printing original lists
print("The original list is : " + str(test_list))

# convert list to dictionary with frequency
freq_dict = {val: test_list.count(val) for val in test_list}

# sort dictionary based on frequency in descending order
sorted_dict = dict(sorted(freq_dict.items(), key=lambda item: item[1], reverse=True))

# create empty list to store reordered list
res = []

# iterate through sorted dictionary and append element to res as many times as its frequency
for key, value in sorted_dict.items():
    res.extend([key] * value)

# print result
print("Reordered List : " + str(res))

The original list is : [4, 7, 5, 4, 1, 4, 1, 6, 7, 5]
Reordered List : [4, 4, 4, 7, 7, 5, 5, 1, 1, 6]

Time complexity: O(n log n), where n is the length of the list. This is because of the sorting operation performed on the dictionary.
Auxiliary space: O(n), where n is the length of the list. This is because of the dictionary and list created to store the frequency and reordered list, respectively.