Big O of my solution


  • 0
    R

    Is my accepted code below big O(n)? Thanks in advance!

    string longestCommonPrefix(vector<string>& strs) {
            if(strs.size() == 0) return "";
            
            sort(strs.begin(), strs.end());
            
            string prefix = strs[0];
            int length = prefix.length();
            for (int i = 0; i < strs.size(); ++i) {
                if (prefix == strs[i].substr(0, length)) continue;
                if (prefix != strs[i].substr(0, length)) {
                    prefix = prefix.substr(0, --length);
                    length = prefix.length();
                    i = 0;
                }
                if (prefix == "") break;
            }
            
            return prefix;
        }
       
    

  • 0

    @rtmitchell5 said in Big O of my solution:

    Is my accepted code below big O(n)?

    Depends on what you mean with n.


  • 0
    R

    @StefanPochmann For n I mean the size of the strs vector. I'm a little shaky with algorithm analysis, so I'm not too sure on this one.


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