How do you solve Count Prefixes of a Given String on LeetCode?

Count Prefixes of a Given String (LeetCode #2255) 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: Understanding prefixes is crucial for string manipulation problems.

What is the time complexity of Count Prefixes of a Given String?

The optimal solution for Count Prefixes of a Given String runs in O(n) time and uses O(n) space. The time complexity is O(n) because we create a set of prefixes in linear time and check each word against this set, which is efficient.

What is the brute force solution for Count Prefixes of a Given String?

The brute force approach for Count Prefixes of a Given String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each word in the list to see if it matches the beginning of the string 's'. This is straightforward but can be inefficient as it compares each word individually. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count Prefixes of a Given String?

Count Prefixes of a Given String uses the following concepts: Array, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Prefixes of a Given String?

Always clarify the problem statement and constraints. Discuss your thought process and possible approaches before coding. Test your solution with different inputs to ensure correctness.

What are common mistakes when solving Count Prefixes of a Given String?

Not considering edge cases like empty strings or duplicates. Overlooking the efficiency of string operations.

#2255

Count Prefixes of a Given String

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)

The optimal approach leverages the fact that we can directly check the prefix of 's' against the words in 'words' without unnecessary comparisons. This reduces the overall time complexity.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Create a set of valid prefixes from 's' by iterating through its characters.
3Step 3: For each word in 'words', check if it exists in the set of prefixes.
4Step 4: If it exists, increment the counter.
5Step 5: Return the counter as the result.

solution.py3 lines

1def countPrefixes(words, s):
2    prefixes = {s[:i] for i in range(1, len(s) + 1)}
3    return sum(1 for word in words if word in prefixes)

ℹ

Complexity note: The time complexity is O(n) because we create a set of prefixes in linear time and check each word against this set, which is efficient.

1Understanding prefixes is crucial for string manipulation problems.
2Using data structures like sets can optimize lookup times.

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