How do you solve Check if the Sentence Is Pangram on LeetCode?

Check if the Sentence Is Pangram (LeetCode #1832) 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: Every letter must be present at least once.

What is the brute force solution for Check if the Sentence Is Pangram?

The brute force approach for Check if the Sentence Is Pangram is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute force approach, we can check each letter of the alphabet against the characters in the given sentence. If we find all 26 letters, we return true; otherwise, false. This method is straightforward but not efficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Check if the Sentence Is Pangram?

Check if the Sentence Is Pangram uses the following concepts: Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Check if the Sentence Is Pangram?

Explain your thought process clearly. Consider edge cases, like the shortest and longest sentences. Discuss time and space complexity as you develop your solution.

What are common mistakes when solving Check if the Sentence Is Pangram?

Not checking for all 26 letters. Using inefficient data structures for membership checks.

#1832

Check if the Sentence Is Pangram

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 set to store unique characters from the sentence. By adding each character to the set, we can easily check if we have all 26 letters by the end. This is efficient because checking membership in a set is fast.

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Algorithm

3 steps

1Step 1: Initialize an empty set to store unique characters.
2Step 2: Iterate through each character in the sentence and add it to the set.
3Step 3: After processing the sentence, check if the size of the set is 26. If yes, return true; otherwise, return false.

solution.py2 lines

1def checkIfPangram(sentence):
2    return len(set(sentence)) == 26

ℹ

Complexity note: The time complexity is O(n) because we traverse the sentence once, and the space complexity is O(n) due to the storage of unique characters in the set.

1Every letter must be present at least once.
2Using a set helps track unique characters efficiently.

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