"Coding is just like talking to a unknown girl. Once you talked to her with guts, the next time you'll never afraid",Really? Yes , try it Now !!! Everything here is for education and learning purpose only , individual holds rights on outer posts/links etc. Team Cracking The Code
import java.io.FileReader; class WordCount { public static void main(String args[]) throws Exception { String text="dice gambling las vegas";//"the quick brown fox jumped over the lazy dog"; char ch[]=new char[text.length()]; char c; boolean lastWhite = true; String whiteSpace = " \t"; int i=0; for(i=0;i<text.length();i++) { c=text.charAt(i); int index = whiteSpace.indexOf(c); if (index == -1) { if (lastWhite == true) { ch[i]=c; } lastWhite = false; } else { lastWhite = true; } } String ss = new String(ch); System.out.println(ss); } }
Run Here http://ideone.com/zVquq
B for instance, the program would press 22.
In order to insert two characters in sequence from the same key, the
user must pause before pressing the key a second time. The space
character ' ' should be printed to indicate a pause. For example, 2 2 indicates AA whereas 22 indicates B.
hi Case #1: 44 444Follow Up-Now Think if Question asked reverse e.g. Given a telephone number print or produce all the possible telephone words or combinations of letters that can represent the given telephone number, then can you device an algorothm that will solve this problem efficiently. This is More Simple Then Previous One. |
O 2, B 1, B 2, O 4O 2 means button 2 in Orange's hallway, B 1
means button 1 in Blue's hallway, and so on. The robots can push this
sequence of buttons in 6 seconds using the strategy shown below:
Time | Orange | Blue -----+------------------+----------------- 1 | Move to button 2 | Stay at button 1 2 | Push button 2 | Stay at button 1 3 | Move to button 3 | Push button 1 4 | Move to button 4 | Move to button 2 5 | Stay at button 4 | Push button 2 6 | Push button 4 | Stay at button 2Note that Blue has to wait until Orange has completely finished pushing
O 2 before it can start pushing B 1.
find(N,X) that returns 1 if
there exists a sensor at height X and returns 0 otherwise. Your
procedure can call a procedure query(i) to get
hi. However, you can make at most 50 calls, for each run of procedure find.find. Therefore, you should not assume that two different calls to procedure find share the same height information.
| sensor | height |
|---|---|
| 0 | 10 |
| 1 | 7 |
| 2 | 9 |
| 3 | 15 |
query:| calls | returns |
|---|---|
| query(0) | 10 |
| query(1) | 7 |
| query(2) | 9 |
| query(3) | 15 |
find, when called by the
grader, should return as in the following table.| calls | returns |
|---|---|
| find(4,10) | 1 |
| find(4,2) | 0 |
| find(4,9) | 1 |
| find(4,15) | 1 |
| find(4,100) | 0 |
maxk(N,T,K) that returns the
maximum sum of temperatures of any K consecutive days, where
N is the number of days and T is an array of positive
integers where T[i], for 0 <= i < N, is the temperature of
day i.maxk should return 100.a - z.
In order to do so, you have to encode the data into a sequence of
alphabets, transmit the encoded data, and decode it to obtain the
original integer data.encode(N,D) that encodes the data, where N
denotes the length of data and D is an array of integers
representing the data, where D[i] is the i-th integer in
the data. (0 <= D[i] <= 255.) This procedure must
call procedure send_data(c) for each character c in the
sequence to transmit the encoded data. Each encoded character c
must be lowercase English alphabets, i.e., alphabets a - z.decode(M) that decodes the data, where M
denotes the length of the encoded data. To read the encoded data as
a sequence of characters, this procedure must call procedure
read_data() for each character in the sequence. It may call
read_data for at most M times. To output the decoded
data, it must call procedure output_data(y) for each integer
y in the decoded data.D[i] <= 25.encode may not call send_data more than 100N times.encode may not call send_data more than 100N times.encode may not call send_data more than 2N times.
