How do you solve Maximum Number of Words You Can Type on LeetCode?

Maximum Number of Words You Can Type (LeetCode #1935) 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 a HashSet allows for O(1) average time complexity for lookups.

What is the brute force solution for Maximum Number of Words You Can Type?

The brute force approach for Maximum Number of Words You Can Type is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we will check each word in the text to see if it can be typed without using any broken letters. This is straightforward but can be inefficient for longer texts. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Maximum Number of Words You Can Type?

Maximum Number of Words You Can Type uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Maximum Number of Words You Can Type?

Explain your thought process clearly while coding. Consider edge cases, such as no broken letters or all letters broken. Practice explaining your code as if teaching someone else.

What are common mistakes when solving Maximum Number of Words You Can Type?

Not using a HashSet for broken letters, leading to inefficient checks. Forgetting to split the text correctly into words.

#1935

Maximum Number of Words You Can Type

Easy

Hash TableStringHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n)Space O(n)

The optimal approach uses a HashSet to store broken letters for O(1) lookup time. This allows us to efficiently check each word against the broken letters without nested loops.

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Algorithm

5 steps

1Step 1: Convert brokenLetters into a HashSet for fast lookups.
2Step 2: Split the text into individual words.
3Step 3: For each word, check if it contains any letter from the HashSet of broken letters.
4Step 4: If a word can be fully typed, increment a count.
5Step 5: Return the count of words that can be fully typed.

solution.py10 lines

1# Full working Python code
2
3def canType(text, brokenLetters):
4    broken_set = set(brokenLetters)
5    words = text.split()
6    count = 0
7    for word in words:
8        if not any(char in broken_set for char in word):
9            count += 1
10    return count

ℹ

Complexity note: The time complexity is O(n) because we only traverse the text and check each character against the HashSet. The space complexity is O(n) due to the storage of broken letters in the HashSet.

1Using a HashSet allows for O(1) average time complexity for lookups.
2Checking each word independently is crucial for counting correctly.

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