How do you solve Count Words Obtained After Adding a Letter on LeetCode?

Count Words Obtained After Adding a Letter (LeetCode #2135) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n + m log m) time complexity and O(n) space complexity. Key insight: Sorting helps in quickly identifying if a target can be formed from a start word.

What is the brute force solution for Count Words Obtained After Adding a Letter?

The brute force approach for Count Words Obtained After Adding a Letter is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each target word against every start word to see if it can be formed by adding a letter and rearranging. This is straightforward but inefficient as it involves a lot of comparisons. The optimal Optimal Solution approach improves this to O(n log n + m log m).

What algorithm or data structure is used to solve Count Words Obtained After Adding a Letter?

Count Words Obtained After Adding a Letter uses the following concepts: Array, Hash Table, String, Bit Manipulation, Sorting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Count Words Obtained After Adding a Letter?

Always consider the data structures that can optimize your solution. Think about edge cases, such as when startWords or targetWords are empty. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions.

What are common mistakes when solving Count Words Obtained After Adding a Letter?

Failing to account for the length difference when checking if a target can be formed. Not using a set for quick lookups, leading to inefficient solutions.

#2135

Count Words Obtained After Adding a Letter

Medium

ArrayHash TableStringBit ManipulationSortingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

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Intuition

Time O(n log n + m log m)Space O(n)

The optimal solution uses a set to store the sorted versions of start words, allowing for quick checks against target words. By focusing on the sorted version, we can easily see if a target can be formed by adding a letter to a start word.

⚙️

Algorithm

3 steps

1Step 1: Create a set to store the sorted versions of all start words.
2Step 2: For each target word, sort it and check if the sorted version minus one character exists in the set.
3Step 3: Count how many target words can be formed and return this count.

solution.py10 lines

1def countWords(startWords, targetWords):
2    start_set = set(''.join(sorted(word)) for word in startWords)
3    count = 0
4    for target in targetWords:
5        sorted_target = ''.join(sorted(target))
6        for c in 'abcdefghijklmnopqrstuvwxyz':
7            if sorted_target[:-1] in start_set:
8                count += 1
9                break
10    return count

ℹ

Complexity note: The complexity arises from sorting the words in both startWords and targetWords, which is O(n log n) for startWords and O(m log m) for targetWords, where n and m are their respective lengths.

1Sorting helps in quickly identifying if a target can be formed from a start word.
2Using a set allows for O(1) average time complexity checks for existence.

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