💡 Problem Formulation: This article tackles the challenge of generating all possible substrings of length n that consist of exactly k distinct characters from a given string. For example, given the string "aabac" and the parameters n=3 and k=2, a desirable output would be ["aab", "aba", "bac"].
Method 1: Brute-Force Approach
Using a brute-force approach, we create all possible substrings of length n and then filter those with exactly k distinct characters. We achieve this by iterating through the string with nested loops, checking the distinct character count using a set.
Here’s an example:
def n_sized_substrings_with_k_distinct(s, n, k):
result = []
for i in range(len(s) - n + 1):
substr = s[i:i+n]
if len(set(substr)) == k:
result.append(substr)
return result
print(n_sized_substrings_with_k_distinct("aabac", 3, 2))
Output:
["aab", "aba", "bac"]
This code snippet defines a function n_sized_substrings_with_k_distinct() that takes a string s, and integers n and k as parameters. It iterates over all possible substrings and collects those matching the criterion in a list result.
Method 2: Using Sliding Window Technique
The sliding window technique involves moving a window of length n across the string and checking for k unique characters within this window. It’s an optimized approach for substrings when compared to the brute-force method, saving valuable computation time.
Here’s an example:
from collections import defaultdict
def n_sized_substring_with_k_distinct(s, n, k):
if n > len(s):
return []
char_count = defaultdict(int)
result, distinct_count = [], 0
for i in range(len(s)):
if char_count[s[i]] == 0:
distinct_count += 1
char_count[s[i]] += 1
if i >= n:
char_count[s[i-n]] -= 1
if char_count[s[i-n]] == 0:
distinct_count -= 1
if i >= n - 1 and distinct_count == k:
result.append(s[i-n+1:i+1])
return result
print(n_sized_substring_with_k_distinct("aabac", 3, 2))
Output:
["aab", "aba", "bac"]
This code uses a dictionary char_count to keep track of the count of each character in the current window and a counter distinct_count for the number of distinct characters. The sliding window is moved by incrementing the start and end positions while updating the counter and dictionary.
Method 3: Using List Comprehension and Set Operations
This method is a more pythonic approach using a combination of list comprehension and set operations to filter substrings of length n that contain exactly k distinct characters.
Here’s an example:
n, k = 3, 2 s = "aabac" substrings = [s[i:i+n] for i in range(len(s) - n + 1) if len(set(s[i:i+n])) == k] print(substrings)
Output:
["aab", "aba", "bac"]
This concise snippet generates a list of substrings using list comprehension with a condition that selects only those substrings where the count of unique characters, determined by converting the substring into a set, equals k.
Method 4: Utilizing itertools and filter
By leveraging the itertools library and the filter function, this method generates all possible substrings and filters out those that do not meet the distinct character requirement. This approach introduces functional programming concepts into our solution.
Here’s an example:
from itertools import combinations
def valid_substring(t):
return len(set(t)) == k
s = "aabac"
n, k = 3, 2
substrings = filter(valid_substring, (s[i:j] for i, j in combinations(range(len(s) + 1), 2) if j - i == n))
print(list(substrings))
Output:
["aab", "aba", "bac"]
This method uses a generator expression to create all possible substrings and the filter function with a custom predicate to keep only the substrings with k distinct characters.
Bonus One-Liner Method 5: Using Generator Expression
As a bonus, here’s a compact one-liner using a generator expression that does the job succinctly. This is for those who enjoy minimalistic and elegant solutions.
Here’s an example:
n, k = 3, 2 s = "aabac" print([s[i:i+n] for i in range(len(s) - n + 1) if len(set(s[i:i+n])) == k])
Output:
["aab", "aba", "bac"]
This one-liner is essentially the inline version of Method 3. It’s a compact and efficient way to write the solution, although it might be less readable for beginners.
Summary/Discussion
- Method 1: Brute-Force Approach. Simple and straightforward. It might not be suitable for long strings due to its O(n^2) complexity.
- Method 2: Sliding Window Technique. More efficient than brute force with a complexity of O(n). However, it requires managing the sliding window complexities.
- Method 3: List Comprehension and Set Operations. Pythonic and concise. Readability may suffer for users not familiar with list comprehensions.
- Method 4: Utilizing itertools and filter. Showcases functional programming in Python. Can be less intuitive for those not used to such paradigms.
- Bonus Method 5: Generator Expression One-Liner. Extremely concise. Not the best in terms of readability or when debugging is required.