The past few days, I've been desperately trying to fix an issue with my list count in C++.
This is my function that handles Count & Add new plr in list.
/*Iterator for players*/
list<Player*>::iterator itc;
list<Player*>PlayerCount;
int FindPosition(list<Player*>List, Player* plr)
{
int pos = 1;
for (itc = List.begin(); itc != List.end(); itc++, pos++)
{
if ((*itc) = plr)
break;
}
return pos;
}
What i'm trying to do is implement each time someone Queue in my list.
My output is currently showing this :
Player One Queue. Player Two Queue
it keep saying : Player in queue [1]
like there was no implementation for next row
How it Should be.
Player One : Player in queue [1]
Player Two : Player in Queue [2]
Thanks for helping !
As I mentioned in my comment above, the problem was the if ((*itc) = plr) was assigning plr to *itc instead of performing a comparison if ((*itc) == plr).
One friendly suggestion,
In FindPosition function, list<Player*>List is being passed by value which means that you are creating a copy of the whole list every time that you are calling the function. If that is the intended behavior then you can keep it as is otherwise you could rewrite it as follows:
// pass List by value
// int FindPosition(list<Player*>List, Player* plr)
// pass List by reference
int FindPosition(list<Player*>& List, Player* plr)
{
int pos = 1;
for (itc = List.begin(); itc != List.end(); itc++, pos++)
{
// changed from assigment (*itc) = plr
// to comparison (*itc) == plr
if ((*itc) == plr)
break;
}
return pos;
}
Related
I am new to coding and struggling with a section in my code. I am at the part where i want to remove duplicate int values from my vector.
my duplicated vector contains: 1 1 2 1 4
my goal is to get a deduplicated vector: 1, 2, 4.
This is what I have so far, It also needs to be a rather simple solution. No pointers and fancy stuff as I still need to study those in the future.
for(int i = 0; i < duplicatedVector.size(); i++) {
int temp = duplicatedVector.at(i);
int counter = 0;
if(temp == duplicatedVector.at(i)) {
counter++;
if(counter > 1) {
deduplicatedVector.push_back(temp);
}
}
}
Could anyone tell me what I do wrong ? I genuinly am trying to iterate through the vector and delete duplicated int, in the given order.
Your algorithm is not well-enough thought out.
Break it up:
for each element of the original vector:
is it in the result vector?
yes: do nothing
no: add it to the result vector
You have your (1) loop, but the (2) part is confused. The result vector is not the same as the original vector, and is not to be indexed the same.
To determine whether an element is in a vector, you need a loop. Loop through your result vector to see if the element is in it. If you find it, it is, so break the inner loop. If you do not, you don't.
You can tell whether or not you found a duplicate by the final value of your inner loop index (the index into the result vector). If it equals result.size() then no duplicate was found.
Clearer variable naming might help as well. You are calling your original/source vector duplicatedVector, and your result vector deduplicatedVector. Even hasDuplicates and noDuplicates would be easier to mentally parse.
You could use a set since it eliminates duplicates:
#include <bits/stdc++.h>
using namespace std;
int main () {
vector<int> vec = vector<int>();
vector<int> dedupl = vector<int>();
vec.push_back(2);
vec.push_back(4);
vec.push_back(2);
vec.push_back(7);
vec.push_back(34);
vec.push_back(34);
set<int> mySet = set<int>();
for (int i = 0; i < vec.size(); i++) {
mySet.insert(vec[i]);
}
for (int elem : mySet) {
dedupl.push_back(elem);
}
for (int elem : dedupl) {
cout << elem << " ";
}
}
I'm making an air dictionary and I have a(nother) problem. The main app is ready to go and works perfectly but when I tested it I noticed that it could be better. A bit of context: the language (ancient egyptian) I'm translating from does not use punctuation so a phrase canlooklikethis. Add to that the sheer complexity of the glyph system (6000+ glyphs).
Right know my app works like this :
user choose the glyphs composing his/r word.
app transforms those glyphs to alphanumerical values (A1 - D36 - X1A, etc).
the code compares the code (say : A5AD36) to a list of xml values.
if the word is found (A5AD36 = priestess of Bast), the user gets the translation. if not, s/he gets all the possible words corresponding to the two glyphs (A5A & D36).
If the user knows the string is a word, no problem. But if s/he enters a few words, s/he'll have a few more choices than hoped (exemple : query = A1A5AD36 gets A1 - A5A - D36 - A5AD36).
What I would like to do is this:
query = A1A5AD36 //word/phrase to be translated;
varArray = [A1, A5A, D36] //variables containing the value of the glyphs.
Corresponding possible words from the xml : A1, A5A, D36, A5AD36.
Possible phrases: A1 A5A D36 / A1 A5AD36 / A1A5A D36 / A1A5AD36.
Possible phrases with only legal words: A1 A5A D36 / A1 A5AD36.
I'm not I really clear but to things simple, I'd like to get all the possible phrases containing only legal words and filter out the other ones.
(example with english : TOBREAKFAST. Legal = to break fast / to breakfast. Illegal = tobreak fast.
I've managed to get all the possible words, but not the rest. Right now, when I run my app, I have an array containing A1 - A5A - D36 - A5AD36. But I'm stuck going forward.
Does anyone have an idea ? Thank you :)
function fnSearch(e: Event): void {
var val: int = sp.length; //sp is an array filled with variables containing the code for each used glyph.
for (var i: int = 0; i < val; i++) { //repeat for every glyph use.
var X: String = ""; //variable created to compare with xml dictionary
for (var i2: int = 0; i2 < val; i2++) { // if it's the first time, use the first glyph-code, else the one after last used.
if (X == "") {
X = sp[i];
} else {
X = X + sp[i2 + i];
}
xmlresult = myXML.mot.cd; //xmlresult = alphanumerical codes corresponding to words from XMLList already imported
trad = myXML.mot.td; //same with traductions.
for (var i3: int = 0; i3 < xmlresult.length(); i3++) { //check if element X is in dictionary
var codeElement: XML = xmlresult[i3]; //variable to compare with X
var tradElement: XML = trad[i3]; //variable corresponding to codeElement
if (X == codeElement.toString()) { //if codeElement[i3] is legal, add it to array of legal words.
checkArray.push(codeElement); //checkArray is an array filled with legal words.
}
}
}
}
var iT2: int = 500 //iT2 set to unreachable value for next lines.
for (var iT: int = 0; iT < checkArray.length; iT++) { //check if the word searched by user is in the results.
if (checkArray[iT] == query) {
iT2 = iT
}
}
if (iT2 != 500) { //if complete query is found, put it on top of the array so it appears on top of the results.
var oldFirst: String = checkArray[0];
checkArray[0] = checkArray[iT2];
checkArray[iT2] = oldFirst;
}
results.visible = true; //make result list visible
loadingResults.visible = false; //loading screen
fnPossibleResults(null); //update result list.
}
I end up with an array of variables containing the glyph-codes (sp) and another with all the possible legal words (checkArray). What I don't know how to do is mix those two to make legal phrases that way :
If there was only three glyphs, I could probably find a way, but user can enter 60 glyphs max.
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.
Closed. This question needs to be more focused. It is not currently accepting answers.
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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;
}
}
}