How do you solve Find Words Containing Character on LeetCode?

Find Words Containing Character (LeetCode #2942) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using built-in string functions can simplify code and improve readability.

What is the brute force solution for Find Words Containing Character?

The brute force approach for Find Words Containing Character is Brute Force, with O(n²) time complexity and O(1) space complexity. We will check each word in the array to see if it contains the character x. This is straightforward but can be slow because we check each character in every word. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Words Containing Character?

Find Words Containing Character uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Words Containing Character?

Always clarify the problem requirements before jumping into coding. Think out loud to show your thought process to the interviewer. Test your solution with different inputs to ensure it works as expected.

What are common mistakes when solving Find Words Containing Character?

Not considering edge cases where the character does not exist in any word. Confusing the indices of the words with their positions in the string.

#2942

Find Words Containing Character

Easy

ArrayStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

We can achieve better efficiency by directly iterating through the words and checking for the character in a single pass, reducing unnecessary checks.

⚙️

Algorithm

5 steps

1Step 1: Initialize an empty list to store indices.
2Step 2: Loop through each word in the words array using its index.
3Step 3: Use the 'in' operator (or equivalent) to check if x is in the current word.
4Step 4: If true, append the index to the list.
5Step 5: Return the list of indices.

solution.py2 lines

1def find_words_containing_char(words, x):
2    return [i for i, word in enumerate(words) if x in word]

ℹ

Complexity note: The time complexity is O(n) because we only loop through the words once. The space complexity is O(n) for storing the indices of the words that contain the character.

1Using built-in string functions can simplify code and improve readability.
2Understanding the difference between time complexity helps in optimizing solutions.

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