How do you solve Counting Words With a Given Prefix on LeetCode?

Counting Words With a Given Prefix (LeetCode #2185) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n * m) time complexity and O(1) space complexity. Key insight: Using string methods can simplify code but may add overhead.

What is the brute force solution for Counting Words With a Given Prefix?

The brute force approach for Counting Words With a Given Prefix is Brute Force, with O(n * m) time complexity and O(1) space complexity. We can check each word in the array to see if it starts with the given prefix. This is straightforward and easy to understand, but it may not be the most efficient way. The optimal Optimal Solution approach improves this to O(n * m).

What algorithm or data structure is used to solve Counting Words With a Given Prefix?

Counting Words With a Given Prefix uses the following concepts: Array, String, String Matching. The recommended patterns to study are: String Matching, Prefix Trees.

What are the interview tips for Counting Words With a Given Prefix?

Always clarify the problem and constraints before diving into coding. Discuss your thought process and the complexity of your approach. Consider edge cases, such as empty strings or very short words.

What are common mistakes when solving Counting Words With a Given Prefix?

Not checking the length of the prefix against the word length. Using unnecessary string methods that can slow down execution.

#2185

Counting Words With a Given Prefix

Easy

ArrayStringString MatchingString MatchingPrefix Trees

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n * m)Space O(1)

We can achieve better performance by directly checking the prefix without using additional string methods. This way, we can avoid unnecessary overhead.

⚙️

Algorithm

5 steps

1Step 1: Initialize a counter to zero.
2Step 2: Loop through each word in the words array.
3Step 3: For each word, compare the first characters up to the length of the prefix.
4Step 4: If the characters match the prefix, increment the counter.
5Step 5: After checking all words, return the counter.

solution.py9 lines

1# Full working Python code
2words = ["pay", "attention", "practice", "attend"]
3pref = "at"
4
5count = 0
6for word in words:
7    if word[:len(pref)] == pref:
8        count += 1
9print(count)

ℹ

Complexity note: The time complexity remains O(n * m) since we still iterate through each word, but we avoid the overhead of method calls, making it more efficient in practice.

1Using string methods can simplify code but may add overhead.
2Direct character comparison can improve performance.

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