LeetCode 1684 Count the Number of Consistent Strings in Python | Easy Coding Tutorial for Beginners

Solve LeetCode 1684 "Count the Number of Consistent Strings" in Python with this beginner-friendly coding tutorial! This problem asks you to count how many strings in a list consist only of characters from a given `allowed` string (e.g., allowed = "ab", words = ["ad","bd","aaab","baa","badab"], return 2). We’ll use a list comprehension to check each word, and explore a more efficient solution using a set for faster lookups. Perfect for Python learners, coding beginners, or anyone prepping for coding interviews!

🔍 **What You'll Learn:**
- Understanding LeetCode 1684’s requirements
- Using list comprehension to check string consistency
- Optimizing with a set for faster character lookups
- Testing with example cases

💻 **Code Used in This Video:**
class Solution(object):
def countConsistentStrings(self, allowed, words):
"""
:type allowed: str
:type words: List[str]
:rtype: int
"""
return len([word for word in words if all(char in allowed for char in word)])

# Test cases
solution = Solution()

# Test case 1: Mix of consistent and inconsistent strings
# "aaab" and "baa" consist only of 'a' and 'b', which are in allowed

# Test case 2: All consistent strings
# All words consist only of characters in allowed

# Test case 3: No consistent strings
# No words consist only of 'a'

# Test case 4: Empty words list
# No words to check

# Optimized: Using a set for faster lookups
class SolutionOptimized(object):
def countConsistentStrings(self, allowed, words):
"""
:type allowed: str
:type words: List[str]
:rtype: int
"""
allowed_set = set(allowed) # Convert allowed string to a set
return sum(1 for word in words if all(char in allowed_set for char in word))

solution_opt = SolutionOptimized()
print("\nOptimized solution:")

🌟 **Why Solve LeetCode 1684?**
This problem is a great introduction to string manipulation and set operations in Python, a common topic in coding interviews! The list comprehension solution has a time complexity of O(n * m * k) where n is the number of words, m is the average word length, and k is the length of allowed (due to string lookups). The optimized solution using a set reduces it to O(n * m) with O(k) space for the set. We’ll explore both methods to count consistent strings efficiently. Master this, and you’ll be ready for more advanced LeetCode challenges!

📚 **Who’s This For?**
- Python beginners learning coding
- Coding enthusiasts tackling LeetCode problems
- Developers prepping for technical interviews

👍 Like, subscribe, and comment: What LeetCode problem should we solve next? Next up: More LeetCode string problems—stay tuned!

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