I just had an interview and I have been asked a question: How would you print all the ASCII table characters without using a loop. The language doesn't matter.
The only method for doing so that comes to my mind is by using recursion instead of loops. An algorithm for doing this will be something like:
void printASCII(int i){
if(i == 128)
return;
print(i + " " + ((char)i) + "\n");
printASCII(i + 1);
}
You should call the previous function using:
printASCII(0);
This will print the complete ASCII table, where each line contains the index followed by a space and the actual ASCII character.
I don't think you can find any other way to do so, specially that it clearly says:
The language doesn't matter
This usually means that the question is about an algorithmic idea, rather than being specific for any language.
Two other approaches that weren't mentioned:
The obvious:
#include <stdio.h>
int main() {
printf("%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c%c", 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127);
}
The power of two tree (probably intended by the interviewer):
#include <stdio.h>
int c;
#define a128 a64; a64;
#define a64 a32; a32;
#define a32 a16; a16;
#define a16 a8; a8;
#define a8 a4; a4;
#define a4 a2; a2;
#define a2 a; a;
#define a printf("%c", c++);
int main() {
c = 0;
a128
}
Related
This question is about converting a gzip deflate message from a websocket message and convert it to array OR raw text that I can apply JSON.parse on it...
*** to be also clear: In this question : i use a websocket from a crypto exchange.... but the question is about the received message NOT about crypto exchange
in the documentation they say "please use zlib deflate"
HERE THE JAVASCRIPT
digifinexopen = '{"id":12312,"method":"trades.subscribe","params":["btc_usdt"]}';
digifinex_market_ws = new WebSocket("wss://openapi.digifinex.com/ws/v1/");
digifinex_market_ws.binaryType = "arraybuffer";
digifinex_market_ws.onmessage = event => digifinex_trades(event.data);
digifinex_market_ws.onopen = event => digifinex_market_ws.send(digifinexopen);
function fu_bitmex_trades (jsonx) { console.log(jsonx); }
I have this in the log
object=>[[Int8Array]]: Int8Array(1129) 0 … 99]
0: 120
1: -38
I tried with <script src="https://cdnjs.cloudflare.com/ajax/libs/pako/2.0.4/pako.min.js" ...></script>
if I do pako.deflate(jsonx);
I get
object=> Uint8Array(78) [120, 156, 1, 67, 0, 188, 255, 120, 218, 4, 192, 177, 13, 196, 32, 12, 133, 225, 93, 254, 154, 6, 174, 243, 54, 39, 66, 17, 201, 74, 36, 63, 187, 66, 236, 158, 111, 179, 34, 222, 192, 158, 114, 111, 196, 82, 121, 98, 27, 229, 63, 75, 24, 170, 57, 151, 196, 105, 220, 23, 214, 199, 175, 143, 243, 5, 0, 0, 255, 255, 32, 108, 18, 108, 62, 68, 31,
If I add decoder = new TextDecoder("utf8"); and log(decoder.decode(jsonx)); I get
string=> x�E��xڜ��n\7����'5���*
���$pƋ Ȼ�*��g��#����|�����������������v\�//�_������������
but, HOW TO RETREIVE the array or raw data that I could json.parse ????
If I decompress your data twice, I get:
{"error":null,"result":{"status":"success"},"id":12312}
It looks like you compressed instead of decompressed. Use pako.inflate().
I was wondering if there is a faster way to randomly choose a value from a list and deleting this value from this list so it cannot be chosen again. This drawing of a value will continue until there aint no values left anymore.
The way I did it soved the problem but it takes almost 8 seconds. So I'm wondering if there is a faster way. I am using Jupyter notebook through the Anaconda software. Since this goes through a server, could it be the problem?
This is what I did:
TotalNumbcol = 266
Column_Numbers = list(np.arange(1,TotalNumbcol+1,1)) # creating a list with all column numbers in it from which can be drawn.
#print Column_Numbers
ABC = Column_Numbers # Creating a variable for the len command in the for loop below, since the Column Numbers length will change.
Chosen_Columns = [[0] for i in range(0,len(Column_Numbers))]
for i in range(len(ABC)):
RandChoiceCol = int(random.choice(Column_Numbers)) # chosing a random number from the Column_Numbers range
Chosen_Columns[i]=(RandChoiceCol) # adding each randomly chosen column number to a list in list showing which column has been chosen.
Column_Numbers = [x for x in Column_Numbers if x not in Chosen_Columns] # delete chosen_column from RandChoiceCol
print Chosen_Columns
print Column_Numbers
[21, 131, 145, 218, 153, 60, 201, 15, 158, 189, 230, 210, 18, 103, 69, 76, 226, 180, 67, 187, 238, 20, 157, 24, 48, 11, 47, 117, 101, 51, 122, 155, 109, 225, 86, 243, 146, 30, 58, 7, 66, 132, 22, 110, 1, 142, 234, 245, 266, 129, 232, 39, 184, 49, 114, 182, 162, 144, 92, 126, 5, 254, 150, 102, 135, 173, 36, 52, 42, 26, 228, 63, 17, 8, 163, 40, 78, 174, 222, 205, 183, 140, 221, 70, 125, 72, 247, 237, 64, 246, 185, 130, 248, 90, 197, 53, 107, 77, 108, 256, 207, 139, 176, 192, 2, 164, 4, 124, 241, 113, 188, 178, 235, 265, 190, 212, 99, 175, 79, 231, 257, 202, 50, 242, 181, 46, 161, 133, 104, 28, 251, 213, 204, 59, 149, 252, 179, 43, 137, 195, 160, 220, 119, 74, 87, 255, 98, 208, 105, 239, 170, 203, 167, 136, 250, 134, 32, 165, 229, 9, 258, 13, 141, 240, 262, 34, 227, 148, 41, 111, 54, 71, 61, 94, 249, 29, 75, 10, 193, 152, 73, 123, 65, 6, 116, 68, 91, 56, 25, 233, 156, 261, 35, 171, 211, 215, 186, 154, 138, 200, 44, 112, 57, 166, 120, 147, 89, 31, 106, 118, 199, 198, 81, 223, 83, 12, 214, 45, 121, 244, 95, 168, 55, 37, 206, 263, 93, 196, 115, 169, 217, 236, 82, 143, 96, 33, 209, 14, 100, 216, 128, 259, 219, 151, 16, 177, 159, 23, 38, 84, 80, 27, 19, 264, 62, 85, 127, 97, 224, 172, 191, 88, 253, 3, 260, 194]
[]
If there is a more efficient way saving time please let me know.
Regards,
You may just shuffle the list in place instead of creating a new list and a new number every time:
Column_Numbers = list(np.arange(1,TotalNumbcol+1,1))
random.shuffle(Column_Numbers)
while Column_Numbers:
rand = Column_Numbers.pop()
print(rand)
I am sorry to be dumb, but I have been struggling with my own quicksort implementation for quiet a while. To be more specific, I can't get my partition procedure to work properly. Ridiculously enough, but I've also tried to almost directly copy an implementation from Sedjewick's book, with no success, however.
Here is my code:
void partition(int a[], int size)
{
int i, j = size - 1;
int t, pivot = a[j / 2];
i = 0;
for (;;)
{
while (a[i] < pivot)
i++;
while (a[j] > pivot)
j--;
if (i >= j)
break;
t = a[i];
a[i++] = a[j];
a[j] = t;
}
}
Here is the example of input:
82, 65, 59, 10, 35, 51, 81, 47, 25, 64, 34, 38, 12, 38, 58, 74, 37, 42, 63, 18,
75, 67, 36, 77, 47, 48, 13, 91, 94, 52
The pivot here is 58, but I get wrong output:
52, 13, 48, 10, 35, 51, 47, 47, 25, 36, 34, 38, 12, 38, 18, 58, 37, 42, 63, 74,
75, 67, 64, 77, 81, 59, 65, 91, 94, 82
It looks almost correct, with a little exception of 37 and 42 going right after 58. I've tried a lot of variations of partitioning procedure, but they all get me similar results.
EDIT
My previous answer seemed to fix the issue but wasn't correct.
Your output is fine. Let me visually indicate the partition with vertical bars:
52, 13, 48, 10, 35, 51, 47, 47, 25, 36, 34, 38, 12, 38, 18, 58, 37, 42 || 63, 74, 75, 67, 64, 77, 81, 59, 65, 91, 94, 82
Everything to the left of the vertical bars is <= 58, and everything to the right is >= 58. This is what's expected from the partition step in a quicksort.
You do, however, need to decrement j in addition to incrementing i:
t = a[i];
a[i++] = a[j];
a[j--] = t; // added a decrement here
Other than that, the only thing you're missing is returning the partition index. Simply return the value of i at the end of the function and use that as the array boundary in your recursive step.
Replace a[j] = t; with a[j--] = t
I have just downloaded windows phone 7.1 sdk for windows phone 7.5.
I have two problems:
1. When i create a new project I doesn't see the option to choose windows phone 7 or 7.5
2. Windows phone 7.5 should be based on Silver light 4 but when I create label at my application and write hebrew words it's backwords.
Can I solve the problem at this version?
If you have installed the Mango tools, once you select to create a WP7 application you are prompted on a secondary screen to select whether to target 7.0 or 7.1.
Note that the version number of the SDK and tools is 7.1 but phones running Mango are marketed as version 7.5.
Note that WP7[.1|5] does not yet currently provide native support for RTL languages. You may find a workaround at http://www.danielmoth.com/Blog/RTL-Arabic-And-Hebrew-Support-For-Windows-Phone-7.aspx
For Windows Phone 7 i wrote a mapping function which maps the Arabic encoding[windows-1256] into default WP7 encoding.
public static string ConvertToArabic(string s) { short[] mapping = { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 8364, 1662, 8218, 402, 8222, 8230, 8224, 8225, 710, 8240, 1657, 8249, 338, 1670, 1688, 1672, 1711, 8216, 8217, 8220, 8221, 8226, 8211, 8212, 1705, 8482, 1681, 8250, 339, 8204, 8205, 1722, 160, 1548, 162, 163, 164, 165, 166, 167, 168, 169, 1726, 171, 172, 173, 174, 175, 176, 177, 178, 179, 180, 181, 182, 183, 184, 185, 1563, 187, 188, 189, 190, 1567, 1729, 1569, 1570, 1571, 1572, 1573, 1574, 1575, 1576, 1577, 1578, 1579, 1580, 1581, 1582, 1583, 1584, 1585, 1586, 1587, 1588, 1589, 1590, 215, 1591, 1592, 1593, 1594, 1600, 1601, 1602, 1603, 224, 1604, 226, 1605, 1606, 1607, 1608, 231, 232, 233, 234, 235, 1609, 1610, 238, 239, 1611, 1612, 1613, 1614, 244, 1615, 1616, 247, 1617, 249, 1618, 251, 252, 8206, 8207, 1746 };
string str = string.Empty;
for (int ix = 0; ix < s.Length; ++ix)
{
str = str + (char)mapping[s[ix]];
}
return str;
}
This worked for me well.
This is just a curiosity - I don't have a real question.
The output of AbsoluteTiming has a definite pattern; can anyone confirm/explain ?
xxx = RandomReal[NormalDistribution[0, 1], 10^6];
Sin[#] & /# xxx; // AbsoluteTiming
(* {0.0890089, Null} *)
Max[Exp[#] - 0.5, 0] & /# xxx; // AbsoluteTiming
(* {0.1560156, Null} *)
$Version
8.0 for Microsoft Windows (64-bit) (February 23, 2011)
According to the Documentation, "AbsoluteTiming is always accurate down to a granularity of $TimeUnit seconds, but on many systems is much more accurate." So evaluating $TimeUnit probably can elucidate this issue.
Yep. Let´s check if the time quantum is consistent:
Differences#
Round[10^5 Sort#
Union[AbsoluteTiming[
Sin[#] & /#
RandomReal[NormalDistribution[0, 1], #];][[1]] & /#
RandomInteger[10^6, 100]]]
(*
-> {1562, 1563, 1563, 1562, 1562, 1563, 1563, 1562, 1562, 1563, 1563, \
1562, 1562, 1563, 1563, 1562, 1562}
*)
Edit
Better code
Differences#
Sort#Union[
Round[10^5 AbsoluteTiming[
Sin[#] & /#
RandomReal[NormalDistribution[0, 1], #];][[1]] & /#
RandomInteger[10^6, 100]]]
Presumably your system's clock only has granularity to some fraction of a second that happens to produce a repeating decimal. I have never noticed this on my Macs.
It's cool, though.
EDIT
Now that I am home I can confirm this must be system-specific: here is my output from the code in belisarius's answer:
{56, 119, 28, 25, 33, 397, 35, 82, 185, 67, 41, 67, 218, 192, 115, \
28, 74, 16, 187, 222, 194, 8, 129, 399, 68, 75, 71, 34, 5, 37, 62, \
64, 137, 173, 24, 98, 135, 308, 63, 155, 208, 861, 22, 72, 72, 184, \
609, 564, 112, 1011, 118, 81, 158, 90, 351, 33, 35, 68, 10, 126, 39, \
194, 7, 108, 278, 75, 37, 214, 34, 166, 119, 10, 335, 141, 4, 988, \
90, 121, 71, 130, 117, 186, 33, 123, 111, 110, 57, 64, 213, 217, 210, \
204, 98, 247, 20, 1421, 28, 2003, 353}