5 Best Ways to Count K-Length Substrings Occurring More Than Once in a Python String

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πŸ’‘ Problem Formulation: You are given a string and are tasked to find and count the substrings of length k that appear more than once within that string. For example, in the string “ababc”, for k=2, the substring “ab” occurs twice; thus, the desired output is 1, indicating there is one such substring.

Method 1: Brute Force with Substring Slicing

This traditional approach involves iterating through the string while slicing out substrings of length k and incrementing their count in a hash map. It’s a straightforward method but can be inefficient for larger strings due to its O(n^2) complexity.

Here’s an example:

def count_k_length_substrings(s, k):
    substring_count = {}
    for i in range(len(s) - k + 1):
        substring = s[i:i+k]
        substring_count[substring] = substring_count.get(substring, 0) + 1
    return sum(1 for count in substring_count.values() if count > 1)

print(count_k_length_substrings("ababc", 2))

Output:

1

This code defines a function count_k_length_substrings that iterates through the string s, extracts all possible substrings of length k, and stores their frequency in a dictionary substring_count. It then returns the count of such substrings that have more than one occurrence.

Method 2: Using Collections’ Counter

The Counter class from Python’s Collections module can be used to count the occurrences of unique elements when passed an iterable. This method optimizes the data collection phase by using a specialized counting object.

Here’s an example:

from collections import Counter

def count_k_length_substrings(s, k):
    substrings = (s[i:i+k] for i in range(len(s) - k + 1))
    counts = Counter(substrings)
    return sum(1 for count in counts.values() if count > 1)

print(count_k_length_substrings("ababc", 2))

Output:

1

This snippet uses a generator expression to create all possible substrings of length k from the string s. The Counter object then tallies these substrings, and the sum is calculated for those with more than one occurrence, which is returned as the result.

Method 3: Sliding Window Technique

This method enhances efficiency by using a sliding window to keep track of substring counts, essentially updating counts without re-slicing strings. It is more space and time-efficient, particularly for large strings.

Here’s an example:

def count_k_length_substrings(s, k):
    substring_count = {}
    for i in range(len(s) - k + 1):
        substring = s[i:i+k]
        substring_count[substring] = substring_count.get(substring, 0) + 1
        
    window_start = 0
    unique_count = 0
    for i in range(k, len(s)):
        if substring_count[s[window_start:i]] > 1:
            unique_count += 1
        substring_count[s[window_start:i]] -= 1
        window_start += 1
        substring = s[i:i+k]
        substring_count[substring] = substring_count.get(substring, 0) + 1
    return unique_count

print(count_k_length_substrings("ababc", 2))

Output:

1

The code utilizes a sliding window to update the counts of substrings dynamically. It first counts all substrings of length k as before, then slides the window across the string, updating counts and thus avoiding redundant recounting.

Method 4: Using DefaultDict for Counting

The defaultdict from Collections allows us to automatically initialize dictionary values, simplifying the counting logic. This method is a variation of the first method with a cleaner implementation.

Here’s an example:

from collections import defaultdict

def count_k_length_substrings(s, k):
    substring_count = defaultdict(int)
    for i in range(len(s) - k + 1):
        substring = s[i:i+k]
        substring_count[substring] += 1
    return sum(1 for count in substring_count.values() if count > 1)

print(count_k_length_substrings("ababc", 2))

Output:

1

Here, we use defaultdict to create a dictionary with a default integer value. This means we don’t need to check if a key exists before incrementing its value, making the code cleaner and slightly more efficient than a standard dictionary.

Bonus One-Liner Method 5: Using List Comprehension and Set

A one-liner solution that uses a set for uniqueness combined with list comprehension to count substrings can be both terse and expressive.

Here’s an example:

def count_k_length_substrings(s, k):
    return len({s[i:i+k] for i in range(len(s) - k + 1) if s.count(s[i:i+k]) > 1})

print(count_k_length_substrings("ababc", 2))

Output:

1

This method uses a set comprehension to identify all unique substrings with a count greater than one in the string, and the length of the resultant set gives the desired count.

Summary/Discussion

  • Method 1: Brute Force with Substring Slicing. Simple to understand. Good for small strings. Inefficient for larger strings due to O(n^2) complexity.
  • Method 2: Using Collections’ Counter. More Pythonic with built-in optimizations. Suitable for average-sized strings. Performance gain over brute force.
  • Method 3: Sliding Window Technique. Highly efficient for larger strings. Reduces time complexity. Requires more complex logic to understand.
  • Method 4: Using DefaultDict for Counting. Cleaner implementation of the brute force method. Avoids key existence checks. Can have a slight performance advantage over standard dictionaries.
  • Bonus Method 5: Using List Comprehension and Set. Elegant one-liner. Less efficient for large strings due to repeated count operations. Best for quick small-scale tasks.