bbg-Eq: longest abbreviation #easier

Given a long word with N non-unique characters, we know there can be 2^N abbreviations (including the original word, and including the empty string).

Requirement: I give you a list of strings and you suspect some of them could be an abbreviation of your word. Find the longest string among them that’s a valid abbreviation. Optimize for time complexity.

I feel this problem appears simple but not easy to complete quickly is my python solution, different from the c++ solution below. Not sure which has more bugs.

I started with string::find() and hit the problem of non-unique characters. Interviewer hinted char-matching — the key point I needed for this type of problem.

#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;

typedef string candidate;

vector<candidate> v{"monkey","aaaaaaaaaaaaaaaaaa", "mamalon", "monk"};
string const haystack{"mamalonkey"};
size_t hsize=haystack.size();

//Tip: can use typdef alias in argument:)
bool lenComp(candidate const & a, candidate const & b){
  return a.size()<=b.size();
ostream & operator<<(ostream & os, vector<candidate> const & c){
  size_t const sz = c.size();
  for (int i=0;i<sz; ++i){
        os<<c[i]<<" ";
  os<<endl; return os; } //single-pass 🙂 

bool check1candate(candidate const & c){ 
  if (c.size() > hsize) return false;

  for(int i=0, h=0; h<hsize; ++h){
        if (c[i] == haystack[h]){
          if (i==c.size()-1) return true;
  return false;
int main(){
  sort(v.begin(), v.end(), lenComp);
  cout<<v; for (int i=v.size()-1; i>=0; --i){
    if (check1candate(v[i])){
        cout<<"winner is "<<v[i];
        return 0;



Fill in your details below or click an icon to log in: 徽标

您正在使用您的 账号评论。 登出 /  更改 )

Google photo

您正在使用您的 Google 账号评论。 登出 /  更改 )

Twitter picture

您正在使用您的 Twitter 账号评论。 登出 /  更改 )

Facebook photo

您正在使用您的 Facebook 账号评论。 登出 /  更改 )

Connecting to %s