From 2011 facebook is organizing a Facebook hacker cup every year around feb, visit Facebook hacker cup homepage for more details
This question is the first question in the qualification round of 2013 hacker cup.
When John was a little kid he didn’t have much to do. There was no internet, no Facebook, and no programs to hack on. So he did the only thing he could… he evaluated the beauty of strings in a quest to discover the most beautiful string in the world.
Given a string s, little Johnny defined the beauty of the string as the sum of the beauty of the letters in it.
The beauty of each letter is an integer between 1 and 26, inclusive, and no two letters have the same beauty. Johnny doesn’t care about whether letters are uppercase or lowercase, so that doesn’t affect the beauty of a letter. (Uppercase ‘F’ is exactly as beautiful as lowercase ‘f’, for example.)
You’re a student writing a report on the youth of this famous hacker. You found the string that Johnny considered most beautiful. What is the maximum possible beauty of this string?
Input
The input file consists of a single integer m followed by m lines.
Output
Your output should consist of, for each test case, a line containing the string “Case #x: y” where x is the case number (with 1 being the first case in the input file, 2 being the second, etc.) and y is the maximum beauty for that test case.
Constraints
5 ≤ m ≤ 50
2 ≤ length of s ≤ 500
Example Input
5 ABbCcc Good luck in the Facebook Hacker Cup this year! Ignore punctuation, please :) Sometimes test cases are hard to make up. So I just go consult Professor Dalves
Example Output
Case #1: 152 Case #2: 754 Case #3: 491 Case #4: 729 Case #5: 646
Solution:
Key is to read the question clearly, first convert string to Lower Case, then iterate over the characters of the string, if a character is in the range ‘a’ to ‘z’ put it in a hashmap (with key as character and value as number of occurrences). Now take the values in the hashmap in a list and sort the list in descending/ascending order. multiply first/last number by 26 then next number by 25 and so on. I think it will be more clear by this Java code.
Complexity of this solution is O(n) in time and O(1) in space.
import java.io.BufferedReader;
import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.Reader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
public class BeautifulStrings {
public Long getBeauty(String s) {
int i = -1;
Map<Character, Integer> lettersOccurencesMap = new HashMap<Character, Integer>();
try {
while (true) {
i++;
Character c = s.charAt(i);
c = Character.toLowerCase(c);
if (c < 'a' || c > 'z') {
continue;
}
// Increasing the number of occurence by 1
Integer occurence = lettersOccurencesMap.get(c);
if (occurence == null) {
occurence = 1;
} else {
occurence++;
}
lettersOccurencesMap.put(c, occurence);
}
} catch (StringIndexOutOfBoundsException e) {
System.out.println("Reached @ the end of the string");
}
ArrayList<Integer> integerList = new ArrayList<Integer>(
lettersOccurencesMap.keySet().size() * 4 / 3);
for (Character letter : lettersOccurencesMap.keySet()) {
integerList.add(lettersOccurencesMap.get(letter));
}
// Collection.sort sorts in ascending order
Collections.sort(integerList);
Long beauty = 0L;
int nextHighestBeauty = 26;
for (i = integerList.size() - 1; i >= 0; i--) {
Integer occurence = integerList.get(i);
beauty = beauty + (occurence * nextHighestBeauty);
nextHighestBeauty--;
}
return beauty;
}
}
