SRM 494 DIV2 Easy(250) - InterestingParty

こちらの問題もMapを知っていれば簡単に実装が可能。このレベルならきっちり1問解くことができそう。

Problem Statement

Mr. White is a very versatile person - absolutely everything is interesting to him. Perhaps this is why he has many friends. Quite unfortunately, however, none of his friends are versatile at all. Each of them is interested only in two topics and refuses to talk about anything else. Therefore, each time Mr. White organizes a party, it's a big problem for him to decide whom to invite so that the party is interesting to everybody. Now that Mr. White has a lot of experience in organizing parties, he knows for sure that a party will be interesting if and only if there's a topic interesting to each of the invited friends. You will be given Strings first and second. The i-th friend of Mr. White is interested in topics first[i] and second[i]. Return the largest number of friends that Mr. White can invite to his party so that the party will be interesting.

Definition

• Class: InterestingParty
• Method: bestInvitation
• Parameters: String, String
• Returns: int
• Method signature: int bestInvitation(String first, String[] second) (be sure your method is public)

Limits

• Time limit (s): 2.000
• Memory limit (MB): 64

Constraints

• first will contain between 1 and 50 elements, inclusive.
• second will contain the same number of elements as first.
• Each element of first and second will contain between 1 and 15 characters, inclusive.
• Each character in first and second will be a lowercase letter ('a'-'z').
• For each valid i, first[i] will not be the same as second[i].

Solution

```import java.util.HashMap;
import java.util.Map;

public class InterestingParty {

public int bestInvitation(String[] first, String[] second) {
Map<String, Integer> interests = new HashMap<String, Integer>();
for (int i = 0; i < first.length; i++) {
interests.put(first[i], 0);
interests.put(second[i], 0);
}

for (int i = 0; i < first.length; i++) {
interests.put(first[i], interests.get(first[i]) + 1);
interests.put(second[i], interests.get(second[i]) + 1);
}

int max = 0;
for (Map.Entry<String, Integer> i : interests.entrySet()) {
max = Math.max(max, i.getValue());
}
return max;
}

}
```