Okay, so what does the SET stand for in the second line? Why is the second string in<>, ?
public Weighted(In in, String delimiter) {
st = new ST<String, SET<String>>();
while (!in.isEmpty()) {
String line = in.readLine();
String[] names = line.split(delimiter);
for (int i = 1; i < names.length; i++) {
addEdge(names[0], names[i]);
}
}
}
With the little information you gave, I will assume that SET is an abstract data type. An abstract data type can store any values without any particular order and with no duplicates. By telling <String> after SET you are telling you want to store Strings inside your SET.
You can learn more about SETs here: https://en.wikipedia.org/wiki/Set_(abstract_data_type)
Related
Long story short, I'm working in a system that only works with groovy in its expression editor, and I need to create a function that returns the number of significant figures an integer has. I've found the following function in stack overflow for Java, however it doesnt seem like groovy (or the system itself) likes the regex:
String myfloat = "0.0120";
String [] sig_figs = myfloat.split("(^0+(\\.?)0*|(~\\.)0+$|\\.)");
int sum = 0;
for (String fig : sig_figs)
{
sum += fig.length();
}
return sum;
I've since tried to convert it into a more Groovy-esque syntax to be compatible, and have produced the following:
def sum = 0;
def myint = toString(mynum);
def String[] sig_figs = myint.split(/[^0+(\\.?)0*|(~\\.)0+$|\\.]/);
for (int i = 0; i <= sig_figs.size();i++)
{
sum += sig_figs[i].length();
}
return(sum);
Note that 'mynum' is the parameter of the method
It should also be noted that this system has very little visibility in regards to what groovy functions are available in the system, so the solution likely needs to be as basic as possible
Any help would be greatly appreciated. Thanks!
I think this is the regex you need:
def num = '0.0120'
def splitted = num.split(/(^0+(\.?)0*|(~\.)0+$|\.)/)
def sf = splitted*.length().sum()
It's been a while since I've had to think about significant figures, so sorry if I have the wrong idea. But I've made two regular expressions that combined should count the number of significant figures (sorry I'm no regex wizard) in a string representing a decimal. It doesn't handle commas, you would have to strip those out.
This first regex matches all significant figures before the decimal point
([1-9]+\d*[1-9]|[1-9]+)
And this second regex matches all significant figures after the decimal point:
\.((\d*[1-9]+)+)?
If you add up the lengths of the first capture group (or 0 when no match) for both matches, then it should give you the number of significant figures.
Example:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class SigFigs {
private static final Pattern pattern1 = Pattern.compile("([1-9]+\\d*[1-9]|[1-9]+)");
private static final Pattern pattern2 = Pattern.compile("\\.((\\d*[1-9]+)+)?");
public static int getSignificantFigures(String number) {
int sigFigs = 0;
for (int i=0; i < 2; i++) {
Matcher matcher = (i == 0 ? pattern1 : pattern2).matcher(number);
if (matcher.find()) {
try {
String s = matcher.group(1);
if (s != null) sigFigs += s.length();
} catch (IndexOutOfBoundsException ignored) { }
}
}
return sigFigs;
}
public static void main(String[] args) {
System.out.println(getSignificantFigures("0305.44090")); // 7 sig. figs
}
}
Of course using two matches is suboptimal (like I've said, I'm not crazy good at regex like some I could mention) but its fairly robust and readable
I'm new to coding.
How do I return a false if there is a string being added that's already in the arraylist?
For example, if you have a list of dog names in the class and you add new dog names in the list, but don't add it when the same dog name was already in the list?
The Solution:
You could use a for statement to iterate through your array list:
public static bool checkArray(string dogName)
{
for int i=0; i<arrayName.Length; i++) // basic for loop to go through whole array
{
if (arrayName[i] == dogName) //checks if array value at index i is the dog's name
{
return true; //if it is, return true
}
}
return false; //gone through whole array, not found so return false
}
This means you can call your method via
string Name = "myDogsName";
bool isAlreadyPresent = checkArray(Name);
Note
This is written in C#, and so other coding languages will slightly
differ in their syntax.
isAlreadyPresent will then contain a bool value if the dog is
present or not
I have written this (for learning purposes) in (possibly) an
inefficient way, but should allow you to understand what is happening
at each stage.
the i++
The i++ may confuse new programmers, but effectively it is the same as writing
i = i + 1;
This also works for i--;
i = i - 1;
Or even i*=2;
i = i * 2;
Trying to implement my first Parallel::For loop with a tread local variable to sum results of the loop. My code is based on an example listed in "Visual C++ 2010, by W. Saumweber, D. Louis (German). Ch. 33, P.804).
I get stuck in the implementation with syntax errors in the Parallel::For call. The errors are as follows, from left to right: a) expected a type specifier, b) too many arguments for generic class "System::Func", c) pointer to member is not valid for a managed class, d) no operator "&" matches these operands.
In line with the book, I create a collection with data List<DataStructure^> numbers, which is subject to a calculation performed in method computeSumScore which is called by the Parallel::For routine in method sumScore. All results are summed in method finalizeSumScore using a lock.
Below I paste the full code of the .cpp part of the class, to show what I have. The data collection "numbers" may look a bit messy, but that's due to organical growth of the program and me learning as I go along.
// constructor
DataCollection::DataCollection(Form1^ f1) // takes parameter of type Form1 to give acces to variables on Form1
{
this->f1 = f1;
}
// initialize data set for parallel processing
void DataCollection::initNumbers(int cIdx)
{
DataStructure^ number;
numbers = gcnew List<DataStructure^>();
for (int i = 0; i < f1->myGenome->nGenes; i++)
{
number = gcnew DataStructure();
number->concentrationTF = f1->myOrgan->cellPtr[cIdx]->concTFA[i];
number->stringA->AddRange(f1->myGenome->cStruct[i]->gString->GetRange(0, f1->myGenome->cChars));
number->stringB->AddRange(f1->myGenome->cStruct[i]->pString);
if (f1->myGenome->cStruct[i]->inhibitFunc)
number->sign = -1;
else
number->sign = 1;
numbers->Add(number);
}
}
// parallel-for summation of scores
double DataCollection::sumScore()
{
Parallel::For<double>(0, numbers->Count, gcnew Func<double>(this, &GenomeV2::DataCollection::initSumScore),
gcnew Func<int, ParallelLoopState^, double, double>(this, &GenomeV2::DataCollection::computeSumScore),
gcnew Action<double>(this, &GenomeV2::DataCollection::finalizeSumScore));
return summation;
}
// returns start value
double DataCollection::initSumScore()
{
return 0.0;
}
// perform sequence alignment calculation
double DataCollection::computeSumScore(int k, ParallelLoopState^ status, double tempVal)
{
int nwScore;
if (numbers[k]->concentrationTF > 0)
{
nwScore = NeedlemanWunsch::computeGlobalSequenceAlignment(numbers[k]->stringA, numbers[k]->stringB);
tempVal = Mapping::getLinIntMapValue(nwScore); // mapped value (0-1)
tempVal = (double) numbers[k]->sign * tempVal * numbers[k]->concentrationTF;
}
else
tempVal = 0.0;
return tempVal;
}
// locked addition
void DataCollection::finalizeSumScore(double tempVal)
{
Object^ myLock = gcnew Object();
try
{
Monitor::Enter(myLock);
summation += tempVal;
}
finally
{
Monitor::Exit(myLock);
}
}
Once this problem is solved I need to ensure that the functions called (computeGlobalSequenceAlignment and getLinIntMapvalue) are thread safe and the program doesn't get stalled on multiple treads accessing the same (static) variables. But this needs to work first.
Hope you can help me out.
Hans Passant answered my question in the comments (include full method name, add comma). Yet I cannot mark my question as answered, so this answer is to close the question.
i am currently tried to create a small program were the user enter a string in a text area, clicks on a button and the program counts the frequency of different characters in the string and shows the result on another text area.
E.g. Step 1:- User enter:- aaabbbbbbcccdd
Step 2:- User click the button
Step 3:- a 3
b 6
c 3
d 1
This is what I've done so far....
public partial class Form1 : Form
{
Dictionary<string, int> dic = new Dictionary<string, int>();
string s = "";
public Form1()
{
InitializeComponent();
}
private void button1_Click(object sender, EventArgs e)
{
s = textBox1.Text;
int count = 0;
for (int i = 0; i < s.Length; i++ )
{
textBox2.Text = Convert.ToString(s[i]);
if (dic.Equals(s[i]))
{
count++;
}
else
{
dic.Add(Convert.ToString(s[i]), count++);
}
}
}
}
}
Any ideas or help how can I countinue because till now the program is giving a run time error when there are same charachter!!
Thank You
var lettersAndCounts = s.GroupBy(c=>c).Select(group => new {
Letter= group.Key,
Count = group.Count()
});
Instead of dic.Equals use dic.ContainsKey. However, i would use this little linq query:
Dictionary<string, int> dict = textBox1.Text
.GroupBy(c => c)
.ToDictionary(g => g.Key.ToString(), g => g.Count());
You are attempting to compare the entire dictionary to a string, that doesn't tell you if there is a key in the dictionary that corresponds to the string. As the dictionary never is equal to the string, your code will always think that it should add a new item even if one already exists, and that is the cause of the runtime error.
Use the ContainsKey method to check if the string exists as a key in the dictionary.
Instead of using a variable count, you would want to increase the numbers in the dictionary, and initialise new items with a count of one:
string key = s[i].ToString();
textBox2.Text = key;
if (dic.ContainsKey(key)) {
dic[key]++;
} else {
dic.Add(key, 1);
}
I'm going to suggest a different and somewhat simpler approach for doing this. Assuming you are using English strings, you can create an array with capacity = 26. Then depending on the character you encounter you would increment the appropriate index in the array. For example, if the character is 'a' increment count at index 0, if the character is 'b' increment the count at index 1, etc...
Your implementation will look something like this:
int count[] = new int [26] {0};
for(int i = 0; i < s.length; i++)
{
count[Char.ToLower(s[i]) - int('a')]++;
}
When this finishes you will have the number of 'a's in count[0] and the number of 'z's in count[25].
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I saw this in an interview question ,
Given a sorting order string, you are asked to sort the input string based on the given sorting order string.
for example if the sorting order string is dfbcae
and the Input string is abcdeeabc
the output should be dbbccaaee.
any ideas on how to do this , in an efficient way ?
The Counting Sort option is pretty cool, and fast when the string to be sorted is long compared to the sort order string.
create an array where each index corresponds to a letter in the alphabet, this is the count array
for each letter in the sort target, increment the index in the count array which corresponds to that letter
for each letter in the sort order string
add that letter to the end of the output string a number of times equal to it's count in the count array
Algorithmic complexity is O(n) where n is the length of the string to be sorted. As the Wikipedia article explains we're able to beat the lower bound on standard comparison based sorting because this isn't a comparison based sort.
Here's some pseudocode.
char[26] countArray;
foreach(char c in sortTarget)
{
countArray[c - 'a']++;
}
int head = 0;
foreach(char c in sortOrder)
{
while(countArray[c - 'a'] > 0)
{
sortTarget[head] = c;
head++;
countArray[c - 'a']--;
}
}
Note: this implementation requires that both strings contain only lowercase characters.
Here's a nice easy to understand algorithm that has decent algorithmic complexity.
For each character in the sort order string
scan string to be sorted, starting at first non-ordered character (you can keep track of this character with an index or pointer)
when you find an occurrence of the specified character, swap it with the first non-ordered character
increment the index for the first non-ordered character
This is O(n*m), where n is the length of the string to be sorted and m is the length of the sort order string. We're able to beat the lower bound on comparison based sorting because this algorithm doesn't really use comparisons. Like Counting Sort it relies on the fact that you have a predefined finite external ordering set.
Here's some psuedocode:
int head = 0;
foreach(char c in sortOrder)
{
for(int i = head; i < sortTarget.length; i++)
{
if(sortTarget[i] == c)
{
// swap i with head
char temp = sortTarget[head];
sortTarget[head] = sortTarget[i];
sortTarget[i] = temp;
head++;
}
}
}
In Python, you can just create an index and use that in a comparison expression:
order = 'dfbcae'
input = 'abcdeeabc'
index = dict([ (y,x) for (x,y) in enumerate(order) ])
output = sorted(input, cmp=lambda x,y: index[x] - index[y])
print 'input=',''.join(input)
print 'output=',''.join(output)
gives this output:
input= abcdeeabc
output= dbbccaaee
Use binary search to find all the "split points" between different letters, then use the length of each segment directly. This will be asymptotically faster then naive counting sort, but will be harder to implement:
Use an array of size 26*2 to store the begin and end of each letter;
Inspect the middle element, see if it is different from the element left to it. If so, then this is the begin for the middle element and end for the element before it;
Throw away the segment with identical begin and end (if there are any), recursively apply this algorithm.
Since there are at most 25 "split"s, you won't have to do the search for more than 25 segemnts, and for each segment it is O(logn). Since this is constant * O(logn), the algorithm is O(nlogn).
And of course, just use counting sort will be easier to implement:
Use an array of size 26 to record the number of different letters;
Scan the input string;
Output the string in the given sorting order.
This is O(n), n being the length of the string.
Interview questions are generally about thought process and don't usually care too much about language features, but I couldn't resist posting a VB.Net 4.0 version anyway.
"Efficient" can mean two different things. The first is "what's the fastest way to make a computer execute a task" and the second is "what's the fastest that we can get a task done". They might sound the same but the first can mean micro-optimizations like int vs short, running timers to compare execution times and spending a week tweaking every millisecond out of an algorithm. The second definition is about how much human time would it take to create the code that does the task (hopefully in a reasonable amount of time). If code A runs 20 times faster than code B but code B took 1/20th of the time to write, depending on the granularity of the timer (1ms vs 20ms, 1 week vs 20 weeks), each version could be considered "efficient".
Dim input = "abcdeeabc"
Dim sort = "dfbcae"
Dim SortChars = sort.ToList()
Dim output = New String((From c In input.ToList() Select c Order By SortChars.IndexOf(c)).ToArray())
Trace.WriteLine(output)
Here is my solution to the question
import java.util.*;
import java.io.*;
class SortString
{
public static void main(String arg[])throws IOException
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
// System.out.println("Enter 1st String :");
// System.out.println("Enter 1st String :");
// String s1=br.readLine();
// System.out.println("Enter 2nd String :");
// String s2=br.readLine();
String s1="tracctor";
String s2="car";
String com="";
String uncom="";
for(int i=0;i<s2.length();i++)
{
if(s1.contains(""+s2.charAt(i)))
{
com=com+s2.charAt(i);
}
}
System.out.println("Com :"+com);
for(int i=0;i<s1.length();i++)
if(!com.contains(""+s1.charAt(i)))
uncom=uncom+s1.charAt(i);
System.out.println("Uncom "+uncom);
System.out.println("Combined "+(com+uncom));
HashMap<String,Integer> h1=new HashMap<String,Integer>();
for(int i=0;i<s1.length();i++)
{
String m=""+s1.charAt(i);
if(h1.containsKey(m))
{
int val=(int)h1.get(m);
val=val+1;
h1.put(m,val);
}
else
{
h1.put(m,new Integer(1));
}
}
StringBuilder x=new StringBuilder();
for(int i=0;i<com.length();i++)
{
if(h1.containsKey(""+com.charAt(i)))
{
int count=(int)h1.get(""+com.charAt(i));
while(count!=0)
{x.append(""+com.charAt(i));count--;}
}
}
x.append(uncom);
System.out.println("Sort "+x);
}
}
Here is my version which is O(n) in time. Instead of unordered_map, I could have just used a char array of constant size. i.,e. char char_count[256] (and done ++char_count[ch - 'a'] ) assuming the input strings has all ASCII small characters.
string SortOrder(const string& input, const string& sort_order) {
unordered_map<char, int> char_count;
for (auto ch : input) {
++char_count[ch];
}
string res = "";
for (auto ch : sort_order) {
unordered_map<char, int>::iterator it = char_count.find(ch);
if (it != char_count.end()) {
string s(it->second, it->first);
res += s;
}
}
return res;
}
private static String sort(String target, String reference) {
final Map<Character, Integer> referencesMap = new HashMap<Character, Integer>();
for (int i = 0; i < reference.length(); i++) {
char key = reference.charAt(i);
if (!referencesMap.containsKey(key)) {
referencesMap.put(key, i);
}
}
List<Character> chars = new ArrayList<Character>(target.length());
for (int i = 0; i < target.length(); i++) {
chars.add(target.charAt(i));
}
Collections.sort(chars, new Comparator<Character>() {
#Override
public int compare(Character o1, Character o2) {
return referencesMap.get(o1).compareTo(referencesMap.get(o2));
}
});
StringBuilder sb = new StringBuilder();
for (Character c : chars) {
sb.append(c);
}
return sb.toString();
}
In C# I would just use the IComparer Interface and leave it to Array.Sort
void Main()
{
// we defin the IComparer class to define Sort Order
var sortOrder = new SortOrder("dfbcae");
var testOrder = "abcdeeabc".ToCharArray();
// sort the array using Array.Sort
Array.Sort(testOrder, sortOrder);
Console.WriteLine(testOrder.ToString());
}
public class SortOrder : IComparer
{
string sortOrder;
public SortOrder(string sortOrder)
{
this.sortOrder = sortOrder;
}
public int Compare(object obj1, object obj2)
{
var obj1Index = sortOrder.IndexOf((char)obj1);
var obj2Index = sortOrder.IndexOf((char)obj2);
if(obj1Index == -1 || obj2Index == -1)
{
throw new Exception("character not found");
}
if(obj1Index > obj2Index)
{
return 1;
}
else if (obj1Index == obj2Index)
{
return 0;
}
else
{
return -1;
}
}
}