How do you solve Find Words That Can Be Formed by Characters on LeetCode?

Find Words That Can Be Formed by Characters (LeetCode #1160) 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 frequency counts allows for faster checks compared to character-by-character matching.

What is the brute force solution for Find Words That Can Be Formed by Characters?

The brute force approach for Find Words That Can Be Formed by Characters 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 can be formed using the characters in the given string. This is done by checking each character of the word against the available characters, which can be inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Words That Can Be Formed by Characters?

Find Words That Can Be Formed by Characters uses the following concepts: Array, Hash Table, String, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Words That Can Be Formed by Characters?

Always start with a brute force solution to understand the problem better. Think about how to optimize your solution by reducing unnecessary checks. Practice explaining your thought process clearly, as communication is key in interviews.

What are common mistakes when solving Find Words That Can Be Formed by Characters?

Not considering character frequency, leading to incorrect results for words with repeated characters. Overlooking edge cases where words cannot be formed even if they contain valid characters.

#1160

Find Words That Can Be Formed by Characters

Easy

ArrayHash TableStringCountingHash 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 solution uses a frequency count of characters in both the chars string and each word. This allows us to quickly determine if a word can be formed by comparing counts, which is more efficient than checking character by character.

⚙️

Algorithm

6 steps

1Step 1: Create a frequency map (dictionary) for characters in chars.
2Step 2: Initialize a variable to keep track of the total length of good words.
3Step 3: For each word in the words array, create a frequency map for that word.
4Step 4: Compare the frequency of each character in the word with the frequency in chars.
5Step 5: If all characters in the word can be formed, add the length of the word to the total.
6Step 6: Return the total length of all good words.

solution.py11 lines

1# Full working Python code
2
3def countCharacters(words, chars):
4    from collections import Counter
5    char_count = Counter(chars)
6    total_length = 0
7    for word in words:
8        word_count = Counter(word)
9        if all(word_count[c] <= char_count[c] for c in word_count):
10            total_length += len(word)
11    return total_length

ℹ

Complexity note: The time complexity is O(n) because we traverse each word and chars string once to build frequency maps. The space complexity is O(n) due to the storage of these frequency maps.

1Using frequency counts allows for faster checks compared to character-by-character matching.
2Understanding the constraints of the problem helps in designing efficient algorithms.

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