Question

I have got ‘n’ rows of data sets, each data set having two components separated by space. First is card number and second is name. A person is same, if he has same card number or the name. How to find the total number of unique person from the data set?

Example:

1 A

1 B

2 B

3 C

This data set has 2 unique person. This is because the first and second row card number is same and second and third row name is same.

What kind of algorithms can be used to solve this kind of problem?

Answer 1

The solution I came up with is using some sort of partitioning: Most operation are done in O(1) or O(logn) and it is done once by user, therefor the time complexity is about O(n),O(logn) depending on how Dictionary are implemented.

int CountUnique(Person persons)
{
        Dictionary<string, int> phones = new Dictionary<string, int>(); //Keep a dictionary where each phone number is mapped to a partition
        Dictionary<string, int> email = new Dictionary<string, int>();  //Keep a dictionary where each email is mapped to a partition
        bool[,] linked = new bool[n, n]; //Lookup table used to tell if 2 partitions are linked (represents the same person)
        int count = 0;
        int max = 0;
        for (int i = 0; i < n; i++)
        {
            Person p = persons[i];
            int pA = -1, pB = -1; // Partition found using the phone number, Partition found using email
            if (phones.ContainsKey(p.Phone))
            {
                pA = phones[p.Phone];
            }
            if (emails.ContainsKey(p.Email))
            {
                pB = emails[p.Email];
            }

            if (pA == -1 && pB == -1) // First case, not found: Add both phones and email and create a new partition. Number of unique persons is also incremented.
            {
                phones.Add(p.Phone.Trim(), max);
                emails.Add(p.Email.Trim().ToLower(), max);
                max++;
                count++;
            }
            else
            {
                if (pA != -1 && pB != -1 && pA != pB) // Found using both parameters on different partitions
                {
                    if (!linked[pA, pB] && !linked[pB, pA]) // If the partition are not linked, link them
                    {
                        count--; // We'lost one partition => one unique person less
                        linked[pA, pB] = linked[pB, pA] = true;
                    }
                }
                if (pA == -1) // We did find an existing email but no phone
                {
                    phones.Add(p.Phone.Trim(), pB); // Add the phone number
                    max++;
                }
                if (pB == -1) // We did find an existing phone but no email
                {
                    emails.Add(p.Email.Trim().ToLower(), pA); // Add the email number
                    max++;
                }
            }

        }



return count;
}