The “String Matching in an Array” problem on Leetcode is a fascinating problem that poses an interesting challenge involving string manipulation and search algorithms. This comprehensive article aims to provide a thorough examination of the problem, along with different methodologies to solve it, while diving deep into the Python language’s features and nuances.

## Table of Contents

- Understanding the Problem Statement
- Initial Thoughts and Strategies
- Brute-Force Solution
- Trie-Based Solution
- KMP Algorithm Solution
- Time and Space Complexity Analysis
- Testing the Solution
- Conclusion

## 1. Understanding the Problem Statement

The problem asks us to find all the strings from a given list of strings that are substrings of other strings in that list.

### Constraints and Assumptions

- 1 <= words.length <= 100
- 1 <= words[i].length <= 30
- words[i] contains only lowercase English letters.
- It’s guaranteed that words[i] will be unique.

## 2. Initial Thoughts and Strategies

The first thing that comes to mind is to check each string against all other strings, but this would require nested loops, leading to a quadratic time complexity. However, given the problem’s constraints, this might be acceptable. Other more complex solutions include using a Trie or the Knuth-Morris-Pratt (KMP) algorithm.

## 3. Brute-Force Solution

In the simplest approach, we can use nested loops to compare each string against all other strings.

Python Code:

```
def stringMatching(words):
result = []
for i in range(len(words)):
for j in range(len(words)):
if i != j and words[i] in words[j]:
result.append(words[i])
break
return result
```

## 4. Trie-Based Solution

Tries can be used to store strings in such a way that common prefixes are represented by shared branches. While inserting a string, we can check whether it is a prefix of another string in the Trie, or vice versa.

Python Code:

```
class TrieNode:
def __init__(self):
self.children = {}
self.isEndOfWord = False
def insert(root, word, result):
node = root
isSub = True
for ch in word:
if ch not in node.children:
isSub = False
node.children[ch] = TrieNode()
node = node.children[ch]
node.isEndOfWord = True
if isSub: # if this word is a prefix of another
result.append(word)
else: # check if other words are prefixes of this word
node = root
for ch in word:
if node.isEndOfWord:
result.append(word[:len(word) - len(ch) + 1])
node = node.children[ch]
def stringMatching(words):
root = TrieNode()
result = []
for word in words:
insert(root, word, result)
return result
```

## 5. KMP Algorithm Solution

The Knuth-Morris-Pratt algorithm is a text-searching algorithm that can be adapted to solve this problem efficiently. However, given the constraints of the problem, implementing KMP may be overkill.

## 6. Time and Space Complexity Analysis

**Brute-Force Approach**: O(n^2 * m), where n is the number of words and m is the average length of words. Space Complexity is O(1).**Trie-Based Approach**: Time Complexity is O(n * m), and Space Complexity is also O(n * m).

## 7. Testing the Solution

Key test cases include:

- A list of words where no word is a substring of another.
- A list of words where all words are substrings of one long word.
- A list of words with multiple valid answers.

## 8. Conclusion

The “String Matching in an Array” problem offers a classic case of trading off between time complexity and implementation complexity. For small input sizes, a simple brute-force approach might suffice, while Trie-based solutions could be more efficient for larger input sizes.