Facebook Hackercup 2013 Qualification Round Solutions

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

Case #1: 152

Case #2: 754

Case #3: 491

Case #4: 729

Case #5: 646

Here goes my solution for the “Beautiful Strings” problem:

import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.LineNumberReader;
import java.util.ArrayList;
import java.util.Collections;

/**
 *
 * @author VIK
 */
public class Problem1 {
        private static ArrayList < Integer >  integers;
    private static File f=new File("Solution1.txt");
    public static void main(String[] args) throws Exception {
        FileReader fr = new FileReader("beautiful_stringstxt.txt");
        LineNumberReader lnr = new LineNumberReader(fr);
        int t = Integer.parseInt(lnr.readLine().trim());
        FileWriter fw=new FileWriter(f);
        for (int i = 0; i  <  t; i++) {
           int sum= getMax(lnr.readLine().trim());
        if ( i < t-1 )
        fw.write ( "Case #" + ( i+1 ) + ": " + sum+ "\n" );
       else 
           fw.write( "Case #" + (i+1) + ": " + sum );
       fw.flush();
        
        }
        }


    private static int getMax ( String string ) throws Exception {
        integers = new ArrayList < Integer > ();
        for ( int i = 65, j = 97; i  <=  90 && j <= 123 ;  i++, j++ ) {
            String s = string.replaceAll ( "[^" + (char) i + (char) j + "]", "" );
            s.trim();
            integers.add(s.length());
        }
        int sum = 0;
        Collections.sort(integers);
        for ( int i = integers.size() - 1, j = 26; i  >= 0 && j  >= 1; --i, --j) {
            sum += integers.get(i) * j;
                        }
      return sum;
    }

}

Download Source Code :  

2-Balanced Smileys
The problem statement is given below:
Your friend John uses a lot of emoticons when you talk to him on Messenger. In addition to being a person who likes to express himself through emoticons, he hates unbalanced parenthesis so much that it makes him go 
Sometimes he puts emoticons within parentheses, and you find it hard to tell if a parenthesis really is a parenthesis or part of an emoticon.
A message has balanced parentheses if it consists of one of the following:
1.  An empty string “”
2. One or more of the following characters: ‘a’ to ‘z’, ‘ ‘ (a space) or ‘:’ (a colon)
3.  An open parenthesis ‘(‘, followed by a message with balanced parentheses, followed by a close parenthesis ‘)’.
4. A message with balanced parentheses followed by another message with balanced parentheses.
5. A smiley face “:)” or a frowny face “:(“
Write a program that determines if there is a way to interpret his message while leaving the parentheses balanced.
Input

The first line of the input contains a number T (1 ≤ T ≤ 50), the number of test cases. 
The following T lines each contain a message of length s that you got from John.

Output

For each of the test cases numbered in order from 1 to T, output “Case #i: ” followed by a string stating whether or not it is possible that the message had balanced parentheses. If it is, the string should be “YES”, else it should be “NO” (all quotes for clarity only)

Constraints
1 ≤ length of s ≤ 100
Sample Input
5
:((
i am sick today (:()
(:)
hacker cup: started :):)
)(
Sample Output
Case #1: NO
Case #2: YES
Case #3: YES
Case #4: YES
Case #5: NO
 
Here goes my solution for the “Beautiful Strings” problem:


 //Problem2.java

import java.io.File;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.LineNumberReader;
import java.util.ArrayList;

/**
 *
 * @author VIK 
 * 
 */
public class Problem2 {

    private static File f = new File("Solution2.txt");

    public static void main(String[] args) throws Exception {
        FileReader fr = new FileReader("balanced_smileystxt.txt");
        LineNumberReader lnr = new LineNumberReader(fr);
        int t = Integer.parseInt(lnr.readLine().trim());
        ArrayList < String > list = new ArrayList < String >();
        FileWriter fw = new FileWriter(f);
        for (int i = 0; i < t; i++) {
            list.add(lnr.readLine().trim());
            String res = printResult(list.get(i), i + 1);
            if ( i < t - 1 ) {
                fw.write("Case #" + (i + 1) + ": " + res + "\n");
            } else {
                fw.write("Case #" + (i + 1) + ": " + res);
            }
            fw.flush();
        }

    }

 private static String printResult ( String string, int index) {
        String s = string.replaceAll ( "[^:\\(\\)]", "" );

        for (int i = 0; i  <  s.length() / 2; i++) {
            String lead = "";
            String trail = "";
            if (s.charAt(i) == '(' && s.charAt(s.length() - i - 1) == ')') {
                if (i > 0) {
                    lead = s.substring(0, i);
                    trail = s.substring (s.length() - i, s.length());
                }
                s = lead + s.substring ( i + 1, s.length() - i - 1 ) + trail;
                --i;
            }
        }
        s = s.replaceAll( ":\\(", "" );
        s = s.replaceAll( ":\\)", "" );

        for (int i = 0; i < s.length(); i++) {
            if (s.charAt(i) == '(' || s.charAt(i) == ')') {
                return "NO";
            }
        }

        return "YES";
     }
}
 
Download Source Code : 

3- Find the Min
After sending smileys, John decided to play with arrays. Did you know  that hackers enjoy playing with arrays? John has a zero-based index  array, m, which contains n non-negative integers. However, only the first k values of the array are known to him, and he wants to figure out the rest.

John knows the following: for each index i, where k <= i < n, m[i] is the minimum non-negative integer which is *not* contained in the previous *k* values of m.

For example, if k = 3, n = 4 and the known values of m are [2, 3, 0], he can figure out that m[3] = 1.

John is very busy making the world more open and connected, as such, he  doesn’t have time to figure out the rest of the array. It is your task  to help him.

Given the first k values of m, calculate the nth value of this array. (i.e. m[n – 1]).

Because the values of n and k can be very large, we use a pseudo-random number generator to calculate the first k values of m. Given positive integers a, b, c and r, the known values of m can be calculated as follows:

m[0] = a
m[i] = (b * m[i – 1] + c) % r, 0 < i < k

Input
The first line contains an integer T (T <= 20), the number of test cases.
This is followed by T test cases, consisting of 2 lines each.
The first line of each test case contains 2 space separated integers, n, k (1 <= k <= 105, k < n <= 109).
The second line of each test case contains 4 space separated integers a, b, c, r (0 <= a, b, c <= 109, 1 <= r <= 109).

Output
For each test case, output a single line containing the case number and the nth element of m.  
Example input:





5
97 39
34 37 656 97
186 75
68 16 539 186
137 49
48 17 461 137
98 59
6 30 524 98
46 18
7 11 9 46

Example output



Case #1: 8
Case #2: 38
Case #3: 41
Case #4: 40
Case #5: 12

Solutions by Vikash verma (@vikashvverma) &Sobre Krolzaum