Longest Common Prefix in Array of Strings Optimal Vertical Scanning

Problem Statement❯

Given an array of strings, your task is to find the longest common prefix shared among all the strings.

The longest common prefix is the starting portion of each string that is identical across all strings.
If no common prefix exists, return an empty string "".
If the input array is empty, return "" as well.

This problem is commonly asked in coding interviews to test your understanding of string manipulation and iteration logic.

Examples❯

Input Strings	Longest Common Prefix	Description
["flower", "flow", "flight"]	"fl"	"fl" is the longest starting part common in all three strings
["dog", "racecar", "car"]	""	No common prefix exists
["interview", "internet", "internal"]	"inte"	All strings share "inte" as a common starting sequence
["a"]	"a"	Only one string — the prefix is the entire string
[""]	""	Single empty string, no prefix
[]	""	Empty input array — return empty string
["apple", "apple", "apple"]	"apple"	All strings are exactly the same
["test", "testing", "tester", "testify"]	"test"	"test" is the longest starting match across all strings
["abc", ""]	""	One string is empty, so no common prefix

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Solution❯

The goal is to find the longest portion at the beginning of all strings that is identical. This is called the longest common prefix.

Imagine writing all the words in a vertical column and comparing them letter by letter, from top to bottom. We go through the characters of the first string one at a time and check whether the same character appears in the same position in all other strings.

Let’s explore the different cases you might encounter:

Normal case: If the strings have a common start — like ["flower", "flow", "flight"] — we start comparing character by character. At position 0, all start with 'f'. At position 1, all have 'l'. At position 2, 'flower' and 'flow' have 'o', but 'flight' has 'i'. So the common prefix ends just before the mismatch: "fl".
No match at all: If the first characters don’t even match — like in ["dog", "racecar", "car"] — there is no common prefix, and we return "".
Single string: If there’s only one string, like ["a"], that string itself is the longest common prefix.
Some string is empty: If even one of the strings is empty — like ["abc", ""] — there can’t be a common prefix. Return "".
Empty array: If the input array has no strings at all — return "" right away.
All strings are identical: In cases like ["apple", "apple", "apple"], all characters will match and the whole string is the common prefix.

This method is known as vertical scanning because it checks each character vertically across all strings at a particular index. It stops as soon as it finds a mismatch, making it efficient and easy to understand.

Edge cases like empty arrays or strings are simple to handle with early checks. This way, we avoid unnecessary comparisons and ensure correctness in all scenarios.

Algorithm Steps❯

Check if the input array is empty. If so, return "".
Loop through each character index i of the first string.
For each string in the array, check if the character at position i matches that in the first string.
If a mismatch is found or if a string is shorter than i, return the prefix found up to that point.
If no mismatch is found, the entire first string is the common prefix.

Code

Java

Python

JavaScript

C++

Kotlin

Swift

Php

public class LongestCommonPrefix {
  public static String longestCommonPrefix(String[] strs) {
    if (strs == null || strs.length == 0) return "";

    for (int i = 0; i < strs[0].length(); i++) {
      char ch = strs[0].charAt(i); // Take character from first string
      for (int j = 1; j < strs.length; j++) {
        // If i is out of bounds or character doesn't match, return prefix
        if (i >= strs[j].length() || strs[j].charAt(i) != ch) {
          return strs[0].substring(0, i);
        }
      }
    }
    return strs[0];
  }

  public static void main(String[] args) {
    String[] input = {"flower", "flow", "flight"};
    System.out.println("Longest Common Prefix: " + longestCommonPrefix(input));
  }
}

Time Complexity❯

Case	Time Complexity	Explanation
Best Case	O(n)	When all strings are identical, only the length of one string is scanned.
Average Case	O(n × m)	Where `n` is the number of strings and `m` is the length of the shortest string.
Worst Case	O(n × m)	In the worst case, all strings are compared fully until a mismatch is found at the end.

Space Complexity❯

O(1)

Explanation: Only a few variables are used for iteration and prefix tracking. No extra memory required.

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