How do you solve Find Common Characters on LeetCode?

Find Common Characters (LeetCode #1002) 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 us to efficiently track character occurrences across multiple strings.

What is the brute force solution for Find Common Characters?

The brute force approach for Find Common Characters is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check each character in the first string and see if it appears in all other strings. If it does, we add it to our result. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find Common Characters?

Find Common Characters uses the following concepts: Array, Hash Table, String. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Find Common Characters?

Clarify the problem requirements and constraints before jumping into coding. Think about edge cases, such as when there is only one string or when there are no common characters. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Find Common Characters?

Not considering duplicates when counting characters. Failing to initialize frequency arrays correctly.

#1002

Find Common Characters

Easy

ArrayHash TableStringHash 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)

We can count the occurrences of each character in each string and then find the minimum count across all strings. This way, we ensure we only include characters that appear in all strings, considering duplicates.

⚙️

Algorithm

4 steps

1Step 1: Create a frequency array for the first string.
2Step 2: For each subsequent string, update the frequency array to keep the minimum counts.
3Step 3: Construct the result based on the frequency array.
4Step 4: Return the result.

solution.py14 lines

1def commonChars(words):
2    freq = [0] * 26
3    for char in words[0]:
4        freq[ord(char) - ord('a')] += 1
5    for word in words[1:]:
6        current_freq = [0] * 26
7        for char in word:
8            current_freq[ord(char) - ord('a')] += 1
9        for i in range(26):
10            freq[i] = min(freq[i], current_freq[i])
11    result = []
12    for i in range(26):
13        result.extend([chr(i + ord('a'))] * freq[i])
14    return result

ℹ

Complexity note: The time complexity is O(n) because we process each character in each string a constant number of times. The space complexity is O(n) due to the result storage.

1Using frequency counts allows us to efficiently track character occurrences across multiple strings.
2Minimizing counts ensures that we only consider characters that appear in all strings.

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