LeetCode Python/Java/C++/JS  >  14. Longest Common Prefix  >  Solved in Ruby, Python, Java  >  GitHub

LeetCode link: 14. Longest Common Prefix, difficulty: Easy.

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.

Constraints:

  • 1 <= strs.length <= 200
  • 0 <= strs[i].length <= 200
  • strs[i] consists of only lowercase English letters if it is non-empty.
Topics

String

Intuition

  1. To find the longest common prefix, you can scan from left to right, using the characters of the first string as the baseline.
  2. If the current character in the other strings does not match the baseline character, return the result. After each round of iteration, add the baseline character to the common prefix.

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Complexity

Time complexity

O(M * N)

Space complexity

O(1)

Solution in languages: Ruby Python Java

Ruby #

# @param {String[]} strs
# @return {String}
def longest_common_prefix(strs)
  result = ''

  (0...strs[0].size).each do |i|
    char = strs[0][i]

    strs[1..].each do |str|
      if str[i] != char
        return result
      end
    end

    result << char
  end

  result
end

Python #

class Solution:
    def longestCommonPrefix(self, strs: List[str]) -> str:
        result = ''

        for i in range(len(strs[0])):
            char = strs[0][i]

            for string in strs[1:]:
                if i >= len(string) or string[i] != char:
                    return result

            result += char

        return result

Java #

class Solution {
    public String longestCommonPrefix(String[] strs) {
        var result = "";

        for (var i = 0; i < strs[0].length(); i++) {
            var c = strs[0].charAt(i);

            for (var j = 1; j < strs.length; j++) {
                if (i >= strs[j].length() || strs[j].charAt(i) != c) {
                    return result;
                }
            }

            result += c;
        }

        return result;
    }
}
Solution in languages: Ruby Python Java

Other languages

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