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  1. Prepare
  2. Algorithms
  3. Dynamic Programming
  4. Abbreviation
  5. Discussions

Abbreviation

  • Parmera
     + 4 comments

    Trying crazy things to solve it in O(NlogN) My code failed in 3 tests, but i can't think of an small test which it fails, can someone help?

    string abbreviation(string a, string b) {
    int j = 0;
    map<char, int> dp1, dp2;
    for(int i = 0;i<a.size();i++)
    {
        dp1[a[i]]++;
    }
    
    for(int i = 0;i<b.size();i++)
    {
        dp2[b[i]]++;
    }
    
    for(int i = 0;i<b.size();i++)
    {
        dp1[b[i]]--;
        if(dp1[b[i]] < 0)
        {
            dp1[char(b[i]+32)]--;
            if(dp1[char(b[i]+32)] < 0)
                return "NO";
        }
    }
    
    for(int i = 0;i<b.size();i++)
    {
        if(dp1[b[i]]>0)
            return "NO";
    }
    
    for(int i = 0;i<a.size();i++)
    {
        if(dp1[a[i]]>0 && a[i]<='Z')
            return "NO";
    }
    
    for(int i = 0;i<a.size();i++)
    {
        if(a[i] <= 'Z' && dp2.find(a[i]) == dp2.end())
            return "NO";
    }
    
    for(int i = 0;i<a.size();i++)
    {
        if(j < b.size())
        {
            if(a[i] == b[j] || char(a[i]-32) == b[j])
            {
                j++;
            }
        }
    }
    
    if(j == b.size())
    {
        return "YES";
    }
    return "NO";
}