# 14. Longest Common Prefix

• Time: $O(|\texttt{strs[0]}| \cdot |\texttt{strs}|)$
• Space: $O(|\texttt{strs[0]}|)$
  1 2 3 4 5 6 7 8 9 10 11 12 13 14 class Solution { public: string longestCommonPrefix(vector& 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]; } }; 
  1 2 3 4 5 6 7 8 9 10 11 12 13 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]; } } 
  1 2 3 4 5 6 7 8 9 10 11 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]