How do you solve Count the Number of Consistent Strings on LeetCode?

Count the Number of Consistent Strings (LeetCode #1684) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(k) space complexity. Key insight: Using a set for allowed characters allows for O(1) lookups, significantly speeding up the consistency checks.

What is the brute force solution for Count the Number of Consistent Strings?

The brute force approach for Count the Number of Consistent Strings is Brute Force, with O(n * m) time complexity and O(1) space complexity. The brute force approach checks each word individually to see if all its characters are present in the allowed string. It's straightforward but can be slow for larger inputs. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Count the Number of Consistent Strings?

Count the Number of Consistent Strings uses the following concepts: Array, Hash Table, String, Bit Manipulation, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count the Number of Consistent Strings?

Always start with a brute force solution to demonstrate your thought process. Explain your reasoning for optimizations clearly; interviewers appreciate understanding your thought process. Practice coding problems with varying constraints to get comfortable with different approaches.

What are common mistakes when solving Count the Number of Consistent Strings?

Not using a set for allowed characters, leading to inefficient character checks. Overlooking edge cases, such as empty words or words with characters not in allowed.

#1684

Count the Number of Consistent Strings

Easy

ArrayHash TableStringBit ManipulationCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n * m)	O(n * m)
Space	O(1)	O(k)

💡

Intuition

Time O(n * m)Space O(k)

We can use a set to store the allowed characters for O(1) lookups, making the check for consistency much faster.

⚙️

Algorithm

5 steps

1Step 1: Convert the allowed string into a set of characters for fast lookup.
2Step 2: Initialize a counter to zero for consistent strings.
3Step 3: For each word in the words array, check if all characters are in the set of allowed characters.
4Step 4: Increment the counter for each consistent word.
5Step 5: Return the counter after processing all words.

solution.py7 lines

1def countConsistentStrings(allowed, words):
2    allowed_set = set(allowed)
3    count = 0
4    for word in words:
5        if all(char in allowed_set for char in word):
6            count += 1
7    return count

ℹ

Complexity note: The time complexity remains O(n * m) but with faster character checks due to the set. Space complexity is O(k) where k is the number of distinct characters in allowed.

1Using a set for allowed characters allows for O(1) lookups, significantly speeding up the consistency checks.
2Brute force is often a good starting point, but understanding how to optimize can lead to better performance.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.