Understanding Longest Common Prefix with a Simple Approach

The longest common prefix is a problem of finding the shared starting characters among a set of strings. To solve this, the vertical scanning approach can be used, which involves comparing characters at the same index position across multiple strings. This method is straightforward and easy to implement.

To understand vertical scanning, consider a few example strings: String 1: A B C D, String 2: X B Y D, String 3: P B Q D. Comparing the characters at the same index positions reveals a common prefix of "B" at index 1 and "D" at index 3. Therefore, the longest common prefix across these strings would be "B".

The vertical scanning algorithm works by first using the first string in the array as a reference. The algorithm then iterates over each character of this reference string. For each character, it checks if the corresponding character in all other strings matches. If any character does not match at a particular index, the algorithm stops comparing and returns the prefix found so far. If the comparison reaches the end of any string, the algorithm also stops as the common prefix cannot extend further.

For example, given the array ["flower", "flow", "flight"], the algorithm would compare the first character 'f' across all strings, then 'l', then 'o', and at the third index, it would encounter a mismatch between 'i' and 'o', thus concluding that the longest common prefix is "fl".

The implementation of the vertical scanning algorithm in Java involves taking the first string as the reference and iterating through its characters. At each step, a character is compared with the corresponding character in the other strings. If a mismatch is found or if the end of a string is reached, the function returns the substring of the first string from the beginning up to the current index, which represents the longest common prefix.

The time complexity of this approach is O(n × m), where n represents the number of strings and m represents the length of the reference string. This is because, in the worst case, each character of the reference string is compared with the corresponding characters in all other strings. The space complexity is O(1) since the algorithm only uses a constant amount of extra space, independent of the input size. It does not create any additional storage structures to hold the strings or characters being compared.