💡 Problem Formulation: The task at hand is to develop a Python program that can count substrings consisting entirely of the digit ‘1’ within a given binary string. For instance, the binary string “110011010” contains five such substrings: “11”, “1”, “1”, “1”, and “1”. Thus, the desired output for this input would be 5.
Method 1: Naive Approach
The naive approach involves iteratively checking all possible substrings of the binary string to determine if they consist of only ‘1’s and count them. Despite its simplicity, this method suffers from a high time complexity, especially with large strings.
Here’s an example:
def count_substrings(s):
count = 0
for i in range(len(s)):
for j in range(i+1, len(s)+1):
if set(s[i:j]) == {'1'}:
count += 1
return count
print(count_substrings("110011010"))Output: 5
This code snippet defines a function count_substrings which iterates over all possible substrings of the binary string s. For each substring, it converts it into a set to check if it contains only the character ‘1’. If the condition is true, it increments the count.
Method 2: Efficient Grouping
This method leverages grouping consecutive ‘1’s together to reduce the number of iterations required. Consecutive ‘1’s contribute a triangular number series to the count, based on their quantity. This approach is more efficient than the naive approach.
Here’s an example:
def count_substrings(s):
groups = [len(group) for group in s.split('0') if group]
count = sum(length * (length + 1) // 2 for length in groups)
return count
print(count_substrings("110011010"))Output: 5
This function count_substrings first splits the binary string s by ‘0’s, collecting lengths of groups of ‘1’s. Then, it calculates the sum of triangular numbers for each group length, which gives the total count of valid substrings.
Method 3: Dynamic Programming
Dynamic programming can be used to solve this problem efficiently by building up the solution using previously calculated values. This method stores the count of valid substrings ending at each position of the string to dynamically update the total count.
Here’s an example:
def count_substrings(s):
count, curr = 0, 0
for digit in s:
curr = curr + 1 if digit == '1' else 0
count += curr
return count
print(count_substrings("110011010"))Output: 5
In this code snippet, the function count_substrings keeps a running total count and a current subtotal curr of the 1s seen so far. When a ‘1’ is encountered, curr is incremented, contributing its value toward count.
Method 4: Regular Expressions
Regular expressions provide a powerful way to match patterns in strings. In this case, the pattern is consecutive ‘1’s. By matching these patterns and calculating the length of each match, one can quickly determine the number of valid substrings.
Here’s an example:
import re
def count_substrings(s):
return sum(len(match.group(0)) * (len(match.group(0)) + 1) // 2
for match in re.finditer(r'1+', s))
print(count_substrings("110011010"))Output: 5
The function count_substrings uses the re module to find all non-overlapping matches of the regex pattern r'1+', which represents one or more consecutive ‘1’s. For each match, it calculates the number of substrings made up solely of ‘1’s.
Bonus One-Liner Method 5: Functional Programming
The functional programming method takes a concise approach, performing operations on the string with minimal explicit iteration or conditionals. The use of generator expressions within a sum function leads to elegant and compact code.
Here’s an example:
print(sum(len(x) * (len(x) + 1) // 2 for x in "110011010".split('0') if x))Output: 5
This one-liner code utilizes the split method to divide the string by ‘0’s and calculates the sum of the lengths of valid groups multiplied by their length plus one divided by two. It’s a succinct expression of the problem’s solution.
Summary/Discussion
Method 1: Naive Approach. Straightforward but not scalable due to its poor performance with long strings. High time complexity.
Method 2: Efficient Grouping. Reduced iterations by using grouping and triangular numbers for counting. Significantly more efficient than the naive approach.
Method 3: Dynamic Programming. Uses subproblems to build an efficient solution with better time complexity. It’s scalable and practical for long strings.
Method 4: Regular Expressions. Utilizes the power of regex for pattern matching. It is concise and can be very efficient for complex pattern detection, but may not be the fastest for simple patterns due to overhead.
Bonus One-Liner Method 5: Functional Programming. Elegant and concise, but may lack clarity for those unfamiliar with functional paradigms. Good for short, simple scripts.