# Mastering "Longest Common Prefix": A LeetCode Guide for Aspiring Software Engineers

## Unravel the nuances of finding the longest common prefix in strings with Python, TypeScript, and Java - an essential skill for coding interviews.

## Table of contents

## Introduction to the Longest Common Prefix Problem

In coding interviews, particularly those you might encounter on platforms like LeetCode, the problem of finding the longest common prefix among an array of strings is a classic. This is LeetCode 14. Longest Common Prefix.

It tests your ability to manipulate strings, understand edge cases, and apply efficient algorithms. For example, given an array `["flower","flow","flight"]`

, the longest common prefix is `"fl"`

. Conversely, for `["dog","racecar","car"]`

, there is no common prefix, so the expected output is an empty string `""`

.

Write a function to find the longest common prefix string amongst an array of strings.

If there is no common prefix, return an empty string

`""`

.

Example 1:`Input: strs = ["flower","flow","flight"] Output: "fl"`

Example 2:`Input: strs = ["dog","racecar","car"] Output: "" Explanation: There is no common prefix among the input strings.`

This problem might seem straightforward at first glance, but it's a wonderful exercise in string manipulation and algorithm optimization. Let's dive deep into understanding the problem and exploring multiple approaches to solve it, their complexities, and implement solutions in Python, TypeScript, and Java.

## Consideration of Different Approaches

When tackling the longest common prefix problem, several strategies come to mind. One might consider:

**Horizontal Scanning**: Starting with the first string as the prefix, compare it with the next string, reducing the prefix length with each mismatch. This approach intuitively mimics how we might manually look for common prefixes but can be inefficient if the first string is significantly longer than the others.**Vertical Scanning**: Instead of comparing strings horizontally, this method examines each character position across all strings sequentially, stopping at the first sign of a mismatch. It's a direct approach but can suffer from unnecessary comparisons, especially if a mismatch occurs early in the strings.**Divide and Conquer**: This technique involves dividing the array of strings into two halves, finding the longest common prefix for each half, and then finding the common prefix between these two results. It leverages recursion and can be more efficient in terms of comparisons made.**Binary Search**: By applying binary search on the length of the shortest string in the array, one can find the longest common prefix by checking mid-length prefixes and adjusting the search space based on whether a common prefix is found.**Sorting**: As discussed earlier, sorting the array first significantly reduces the problem's complexity. By only comparing the first and last strings post-sorting, this method efficiently finds the longest common prefix with the minimum number of character comparisons.

### Big O Notation Analysis

Each method has its merits and drawbacks, primarily differing in their time and space complexity:

**Horizontal Scanning**: The worst-case time complexity is O(S), where S is the sum of all characters in all strings. The space complexity is O(1).**Vertical Scanning**: Similar to horizontal scanning, its worst-case time complexity is O(S), and the space complexity remains O(1).**Divide and Conquer**: This method has a time complexity of O(S), similar to the others, but might perform better in practice due to fewer overall comparisons. Its space complexity can increase to O(m log n) due to recursive calls, where m is the length of the longest string and n is the number of strings.**Binary Search**: The time complexity is O(S log m), where m is the length of the shortest string. The space complexity is O(1).**Sorting**: After sorting, the time complexity for the comparison is O(m), where m is the length of the shortest string. However, sorting itself takes O(N log N), leading to an overall time complexity of O(N log N + m). The space complexity depends on the sorting algorithm, usually O(1) to O(n).

## Solutions in Python, TypeScript, and Java

Let's implement the sorting-based approach in all three languages, considering its efficiency and simplicity.

### Python Solution with Horizontal Scanning

```
def longestCommonPrefix(strs):
if not strs:
return ""
prefix = strs[0]
for s in strs:
while not s.startswith(prefix):
prefix = prefix[:-1]
if not prefix:
return ""
return prefix
```

### Python Solution with Sorting

```
def longestCommonPrefix(strs):
if not strs:
return ""
strs.sort()
prefix = ""
for x, y in zip(strs[0], strs[-1]):
if x == y:
prefix += x
else:
break
return prefix
```

### TypeScript Solution with Sorting

```
function longestCommonPrefix(strs: string[]): string {
if (strs.length === 0) return "";
strs.sort();
let prefix = "";
for (let i = 0; i < strs[0].length; i++) {
if (strs[0][i] === strs[strs.length - 1][i]) {
prefix += strs[0][i];
} else {
break;
}
}
return prefix;
}
```

### Java Solution with Sorting

```
public String longestCommonPrefix(String[] strs) {
if (strs == null || strs.length == 0) return "";
Arrays.sort(strs);
StringBuilder prefix = new StringBuilder();
for (int i = 0; i < strs[0].length(); i++) {
if (strs[0].charAt(i) == strs[strs.length - 1].charAt(i)) {
prefix.append(strs[0].charAt(i));
} else {
break;
}
}
return prefix.toString();
}
```

## Conclusion

Finding the longest common prefix is a deceptively simple problem that offers a rich exploration of string manipulation and algorithm design. By understanding and applying different strategies, you can enhance your problem-solving toolkit and prepare yourself for software engineering interviews.

Each approach has its context where it shines, highlighting the importance of assessing the problem's specifics before diving into coding. Remember, mastering these challenges is not just about finding a solution but understanding the rationale behind each method and its implications on performance and efficiency.

Happy coding!