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14. Longest Common Prefix

  • Time: $O(|\texttt{strs[0]}| \cdot |\texttt{strs}|)$
  • Space: $O(|\texttt{strs[0]}|)$
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class Solution {
 public:
  string longestCommonPrefix(vector<string>& strs) {
    if (strs.empty())
      return "";

    for (int i = 0; i < strs[0].length(); ++i)
      for (int j = 1; j < strs.size(); ++j)
        if (i == strs[j].length() || strs[j][i] != strs[0][i])
          return strs[0].substr(0, i);

    return strs[0];
  }
};

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class Solution {
  public String longestCommonPrefix(String[] strs) {
    if (strs.length == 0)
      return "";

    for (int i = 0; i < strs[0].length(); ++i)
      for (int j = 1; j < strs.length; ++j)
        if (i == strs[j].length() || strs[j].charAt(i) != strs[0].charAt(i))
          return strs[0].substring(0, i);

    return strs[0];
  }
}

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class Solution:
  def longestCommonPrefix(self, strs: list[str]) -> str:
    if not strs:
      return ''

    for i in range(len(strs[0])):
      for j in range(1, len(strs)):
        if i == len(strs[j]) or strs[j][i] != strs[0][i]:
          return strs[0][:i]

    return strs[0]