💡 Problem Formulation: Imagine you want to process a string in Python to find all the homogenous substrings within it. A homogenous substring is a sequence of identical characters, such as “aaa” or “zz”. Given an input string, the desired output is the count of such substrings present. For instance, the string “aabbaa” contains five such substrings: “aa”, “bb”, “aa”, “aaa”, and “aaa” (with the last “aaa” counted twice for the overlapping substrings).
Method 1: Iterative Approach
This method involves traversing the string iteratively and keeping track of the current homogenous substring. The function counts the homogenous substrings by increasing the counter when the current character is the same as the previous one, otherwise, it resets the counter.
Here’s an example:
def count_homogenous_substrings(s):
count, curr, total = 0, 1, 0
for i in range(1, len(s)):
if s[i] == s[i - 1]:
curr += 1
else:
curr = 1
total += curr
return total
print(count_homogenous_substrings("aabbaa"))Output: 5
This function starts with a counter curr set to 1 and loops through the string. If a character is the same as the previous one, it increments curr; otherwise, it resets it to 1. It adds the value of curr to total on each iteration, resulting in the overall count of homogenous substrings.
Method 2: Using itertools.groupby
Python’s itertools.groupby method is a sophisticated tool that groups consecutive elements in an iterable that are the same. We can use it to group homogenous substrings in our string and then count the number of times these occur.
Here’s an example:
from itertools import groupby
def count_homogenous_substrings(s):
return sum((len(list(g)) * (len(list(g)) + 1)) // 2 for _, g in groupby(s))
print(count_homogenous_substrings("aabbaa"))Output: 5
The function applies groupby to the string and iterates over the groups. For each group, it calculates the count of homogenous substrings using the formula n * (n + 1) / 2 where n is the length of the group, and adds them together to get the total.
Method 3: Using regex
Regular expressions can be used to find repeating character sequences. By utilizing the regex pattern (.)\\1*, we can match and count every sequence of identical characters in a string.
Here’s an example:
import re
def count_homogenous_substrings(s):
return sum((len(m.group(0)) * (len(m.group(0)) + 1)) // 2 for m in re.finditer(r'(.)\\1*', s))
print(count_homogenous_substrings("aabbaa"))Output: 5
Here we use a regex pattern to match homogenous substrings and find all occurrences with re.finditer. We then calculate the count for each match with n * (n + 1) / 2 where n is the length of the match.
Method 4: Dynamic Programming
Dynamic programming can optimize the substring counting process. We can build an array that stores the count of homogenous substrings ending at a particular index, and use this array to calculate the total.
Here’s an example:
def count_homogenous_substrings(s):
dp = [1] * len(s)
total = 1
for i in range(1, len(s)):
if s[i] == s[i-1]:
dp[i] = dp[i-1] + 1
total += dp[i]
return total
print(count_homogenous_substrings("aabbaa"))Output: 5
The dp array initializes with ones, representing the count of homogenous substrings ending at each index, being at least 1 (the character itself). We then accumulate counts for each substring and the total count.
Bonus One-Liner Method 5: Using sum with slicing
Python’s list slicing allows for a succinct one-liner that sums the length comparisons between characters in the string and exploits the arithmetic progression sum formula.
Here’s an example:
print(sum((s[i] == s[i + 1]) for i in range(len(s) - 1)) + len(s))
Output: 5
This one-liner iterates through the string indices and sums 1 whenever two consecutive characters match, which contributes to the count of homogenous substrings, finally adding the length of the string to account for single-character substrings.
Summary/Discussion
- Method 1: Iterative Approach. Easy to understand. Linear time complexity. No import required.
- Method 2: Using itertools.groupby. Elegant and Pythonic. Utilizes Python’s standard library functions efficiently. Slightly more complex to understand.
- Method 3: Using regex. Useful for pattern matching problems. Could be slow for very large strings due to regex processing overhead.
- Method 4: Dynamic Programming. Efficient for large input strings. Requires understanding of dynamic programming concepts.
- Method 5: Bonus One-Liner. Extremely concise. Might not be as readable for beginners. Efficient for all string lengths.