Python Program to Group Strings by K length Using Suffix

Given Strings List, the task is to write a Python program to group them into K-length suffixes.

Input : test_list = [“food”, “peak”, “geek”, “good”, “weak”, “sneek”], K = 3 
Output : {‘ood’: [‘food’, ‘good’], ‘eak’: [‘peak’, ‘weak’], ‘eek’: [‘geek’, ‘sneek’]} 
Explanation : words ending with ood are food and good, hence grouped.

Input : test_list = [“peak”, “geek”, “good”, “weak”], K = 3 
Output : {‘ood’: [‘good’], ‘eak’: [‘peak’, ‘weak’], ‘eek’: [‘geek’]} 
Explanation : word ending with ood is good, hence grouped. 

Method 1 : Using try/except + loop

In this, we extract the last K characters and form a string, and append it to the existing key’s list corresponding to it, if not found, it goes through catch flow and creates a new key with a list with the first word initialized.

Python3

# Python3 code to demonstrate working of
# Group Strings by K length Suffix
# Using try/except + loop

# initializing list
test_list = ["food", "peak", "geek",
             "good", "weak", "sneek"]

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

# initializing K
K = 3

res = {}
for ele in test_list:

    # extracting suffix
    suff = ele[-K : ]

    # appending if key found, else creating new one
    try:
        res[suff].append(ele)
    except:
        res[suff] = [ele]

# printing result
print("The grouped suffix Strings : " + str(res))

Output

The original list is : ['food', 'peak', 'geek', 'good', 'weak', 'sneek']
The grouped suffix Strings : {'ood': ['food', 'good'], 'eak': ['peak', 'weak'], 'eek': ['geek', 'sneek']}

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

Method 2 : Using defaultdict() + loop.

This method avoids the need of using try/except block as default list initialization is handled by defaultdict().

Python3

# Python3 code to demonstrate working of
# Group Strings by K length Suffix
# Using defaultdict() + loop
from collections import defaultdict

# initializing list
test_list = ["food", "peak", "geek",
             "good", "weak", "sneek"]

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

# initializing K
K = 3

res = defaultdict(list)
for ele in test_list:

    # extracting suffix
    suff = ele[-K : ]

    # appending into matched suffix key
    res[suff].append(ele)

# printing result
print("The grouped suffix Strings : " + str(dict(res)))

Output

The original list is : ['food', 'peak', 'geek', 'good', 'weak', 'sneek']
The grouped suffix Strings : {'ood': ['food', 'good'], 'eak': ['peak', 'weak'], 'eek': ['geek', 'sneek']}

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

Method 3 : Using for loops and endswith()

Python3

# Python3 code to demonstrate working of
# Group Strings by K length Suffix

# initializing list
test_list = ["food", "peak", "geek",
            "good", "weak", "sneek"]

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

# initializing K
K = 3
res=dict()
x=[]
for i in test_list:
    if i[-K:] not in x:
        x.append(i[-K:])
for i in x:
    a=[]
    for j in test_list:
        if(j.endswith(i)):
            a.append(j)
    res[i]=a
print(res)

Output

The original list is : ['food', 'peak', 'geek', 'good', 'weak', 'sneek']
{'ood': ['food', 'good'], 'eak': ['peak', 'weak'], 'eek': ['geek', 'sneek']}

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

Method 4 :  Using Dictionary and List Comprehensions

This approach uses a combination of dictionary and list comprehensions to group the strings based on their K-length suffix. We first create a list of all unique K-length suffixes from the strings in the input list. Then, using the list comprehension, we create a dictionary where each key is a suffix, and its value is a list of all the strings that end with that suffix.

Python3

# Python3 code to demonstrate working of
# Group Strings by K length Suffix

# initializing list
test_list = ["food", "peak", "geek",
            "good", "weak", "sneek"]

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

# initializing K
K = 3

result = {suffix: [word for word in test_list if word.endswith(suffix)] for suffix in set([word[-K:] for word in test_list])}

# printing result
print("The grouped suffix Strings : " + str(result))

Output

The original list is : ['food', 'peak', 'geek', 'good', 'weak', 'sneek']
The grouped suffix Strings : {'ood': ['food', 'good'], 'eek': ['geek', 'sneek'], 'eak': ['peak', 'weak']}

Time Complexity: O(n * m) where n is the length of the input list test_list and m is the length of the longest string in test_list.
Auxiliary Space: O(n)

Method 5 : Using itertools.groupby()

 step-by-step approach 

1. Import the itertools module for working with iterators.
2. Sort the list of words by their last K characters using the key parameter of the sorted() function.
3. Use the itertools.groupby() function to group the words by their last K characters.
4. Create an empty dictionary to store the grouped suffix strings.
5. Loop through the groups of words and add them to the dictionary.
6. Print the final dictionary.

Python3

import itertools

# initializing list
test_list = ["food", "peak", "geek", "good", "weak", "sneek"]

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

# initializing K
K = 3

# sort list by last K characters
sorted_list = sorted(test_list, key=lambda word: word[-K:])

# group words by suffix using itertools.groupby()
groups = itertools.groupby(sorted_list, key=lambda word: word[-K:])

# create empty dictionary to store grouped suffix strings
result = {}

# loop through groups and add them to dictionary
for suffix, words in groups:
    result[suffix] = list(words)

# printing result
print("The grouped suffix Strings : " + str(result))

Output

The original list is : ['food', 'peak', 'geek', 'good', 'weak', 'sneek']
The grouped suffix Strings : {'eak': ['peak', 'weak'], 'eek': ['geek', 'sneek'], 'ood': ['food', 'good']}

The time complexity of this method is O(n*log(n)), where n is the length of the original list, due to the sorting operation.
The auxiliary space used by this method is O(n), where n is the length of the original list, due to the creation of the sorted list.