One of the fundamental aspects of string manipulation problems is to understand the underlying character frequency and distribution. Leetcode’s “Find Common Characters” problem offers a deep dive into this facet, requiring participants to understand and manipulate character frequencies across multiple strings. This article will elucidate the problem, discuss its intricacies, and present an optimized solution in Python.

### Table of Contents

- Problem Statement
- Problem Insights and Observations
- Strategy to Approach the Problem
- Python Implementation
- Time and Space Complexity Analysis
- Conclusion

### 1. Problem Statement

Given an array `A`

of strings made only from lowercase letters, return a list of all characters that show up in all strings within the list (including duplicates). For example, if a character occurs 3 times in all strings but not 4 times, you need to include that character three times in the final answer.

**Example:**

```
Input: ["bella","label","roller"]
Output: ["e","l","l"]
```

### 2. Problem Insights and Observations

The problem requires us to find characters common across all strings in the list. One initial observation is that if a character is not present in even one string, it cannot be part of the output. Also, the frequency of a character in the result is determined by its minimum frequency across all strings.

### 3. Strategy to Approach the Problem

A possible approach can be:

**Initialize an Array for Frequency Count:**Start with an array of size 26 (representing each lowercase letter) initialized with a large number, say`float('inf')`

.**Count Frequencies for Each String:**For each string in the list, count the frequency of each character in it. Update the initial array to reflect the minimum frequency of each character compared to the previously counted strings.**Construct the Result:**Finally, for each character, add it to the result list as many times as its minimum count.

### 4. Python Implementation

```
def commonChars(A):
# Step 1: Initialize an array with a large value for each character
min_frequency = [float('inf')] * 26
for string in A:
# Step 2: Count frequency for each string
current_frequency = [0] * 26
for char in string:
current_frequency[ord(char) - ord('a')] += 1
# Update the min_frequency array to store minimum count for each character
for i in range(26):
min_frequency[i] = min(min_frequency[i], current_frequency[i])
# Step 3: Construct the result
result = []
for i in range(26):
result.extend([chr(i + ord('a'))] * min_frequency[i])
return result
```

### 5. Time and Space Complexity Analysis

**Time Complexity:**- Counting frequency for each character in each string is O(n), where n is the average size of the strings in the list.
- Doing this for all strings in the list makes it O(m×n), where m is the number of strings.
- Constructing the result is O(26)=O(1) as we have a fixed number of characters.
- Thus, the overall time complexity is O(m×n).

**Space Complexity:**- The space used by
`min_frequency`

and`current_frequency`

arrays is O(1) since they have a constant size of 26 irrespective of the input size. - The output size in the worst case can be O(n) if all characters are common in all strings.
- Thus, the overall space complexity is O(n).

- The space used by

### 6. Conclusion

The “Find Common Characters” problem on Leetcode emphasizes the importance of understanding character frequency distribution across multiple strings. The problem, while simple in its essence, requires careful implementation to ensure accuracy and efficiency. It serves as an excellent practice problem for those aiming to hone their skills in string manipulation and frequency count techniques. Moreover, such problems form the foundation for more complex text analysis tasks, making them invaluable for beginners and experts alike.