kreznykov It should have a super easy difficulty.
mubparekh Please share your approach
[deleted] struct Player { string name; int score; }; bool comp(Player p1, Player p2){ if(p1.score>p2.score) return true; if(p1.score==p2.score) return p1.name<p2.name; return false; } vector<Player> comparator(vector<Player> players) { sort(players.begin(),players.end(), comp); return players; }
JChiquin You can simplify the function
compif you convert it into this:bool comp(Player &A, Player &B){ return A.score==B.score and A.name<B.name or A.score>B.score; }
Also and just for fun, it can use a lambda expression in its place:
vector<Player> comparator(vector<Player> players) { sort(players.begin(),players.end(),[](Player &A,Player &B){return A.score==B.score and A.name<B.name or A.score>B.score;}); return players; }
Matthewpdias just for fun with ternary
bool compare(const Player& p1,const Player& p2) { return p1.score > p2.score ? true : (p1.score == p2.score ? p1.name < p2.name : false); } vector<Player> comparator(vector<Player> players) { sort(players.begin(), players.end(), compare); return players; }
raduri Absolutely. If this is "Medium", wonder what "Easy" could look like :P.
Paul_Denton Agree if you just implement the comparator thingy and use sort from the standard lib.
aarduain That defeats the purpose completely.
vataseluandrei1 It's super easy without any comparator thingy in Python
def comparator(a, b): if a.score > b.score: return -1 elif a.score < b.score: return 1 elif a.score == b.score: if a.name > b.name: return 1 elif a.name < b.name: return -1 else: return 0
profnandaa This should do too:
def comparator(a, b): if a.score > b.score: return -1 elif a.score < b.score: return 1 elif a.name > b.name: return 1 return -1
andikadevs The resource is Quicksort video. Then I see sample comparator code then I use it and not using Quicksort. Lol...
aarduain Really guys? Come on, this isn't easy for everyone
tatianaensslin definitely depends on the langage youre using, i suppose
gussy In Python 3, I think the template code is incorrect. We can solve it by writing this
def comparator(a, b): if a.score < b.score: return 1 if a.score > b.score: return -1 if a.name < b.name: return -1 if a.name > b.name: return 1 return 0
However if the line
data = sorted(data, key=cmp_to_key(Player.comparator))
was changed to
data = sorted(data, key=Player.key)
we could solve it with the better
def key(self): return(-self.score, self.name)
We can use the second approach by dummying out the cmp_to_key function.
def cmp_to_key(key): return key class Player: def __init__(self, name, score): self.name = name self.score = score def comparator(self): return(-self.score, self.name)
jaj_jaj if in other place you will need to sort players using different criteria, will you change class Player again?
fisher6 You, my friend, are a true hacker.
taoyu_forever for the code: if a.name < b.name: return -1
why here return -1, but not 1?
morita657 cmp_to_key function is nice idea! I have learnt new thing from yours! Thank you!
nacho2 Hi gussy, as documentation says, Python 3 no longer supports comparer functions for classes, it uses a tool function to map between old comparer (that i think HackerRank is used to put in their excercises) and Python 3 comparers:
"functools.cmp_to_key(func) Transform an old-style comparison function to a key function. Used with tools that accept key functions (such as sorted(), min(), max(), heapq.nlargest(), heapq.nsmallest(), itertools.groupby()). This function is primarily used as a transition tool for programs being converted to Python 3 where comparison functions are no longer supported.
A comparison function is any callable that accept two arguments, compares them, and returns a negative number for less-than, zero for equality, or a positive number for greater-than. A key function is a callable that accepts one argument and returns another value to be used as the sort key."
So it takes a func and turns that into something that Python 3 understands.
Source: link
sachin4u4u can anyone explain what is differnce between return 1, return 0 and return -1
nacho2 -1 = a < b
0 = a == b
1 = a > b
a comparer need to tell you if a number is equal, grater or lower than other number, thats why you have 3 possible outputa
artbindu wow your comparator function work wonderful.
i did not understand about your function: ''' def comparator(a, b): if a.score < b.score: return 1 if a.score > b.score: return -1 if a.name < b.name: return -1 if a.name > b.name: return 1 return 0 ''' can you send me any link to understand it well
iunir_iakupov Hi, I solved it in Python 3 without changing anything outside Player class:
def comparator(a, b): if a.score != b.score: return b.score - a.score return (a.name>b.name)-(a.name<b.name)And I agree template is incorrect. We don't need to implement repr
jimmyxbw Simple Java implementation:
class Checker implements Comparator<Player> { @Override public int compare(Player a, Player b) { if(a.score != b.score) { return b.score - a.score; } return a.name.compareTo(b.name); } }
shweta78 could you please explain this?
tylerFowler Subtracting the two scores will either give a negative (indicating less than) or positive (indicating greater then) number. Roughly equivalent to the standard:
if (b.score < a.score) return -1; if (b.score > a.score) return 1;
samirbaaz class Player: def __init__(self, name, score): self.name = name self.score = score def __repr__(self): pass def comparator(a, b): val = b.score - a.score if val == 0: return -1 if a.name < b.name else 1 return val
nik_roby I ended up with nearly identical structure.
erdenezul def comparator(a, b): name_score = int(b.name > a.name) return b.score - a.score - name_score
dimple97verma Please help me in understanding the comaparator method.How is it taking 2 objects of the class Player as argument?Also i am unable to understand the logic in the statements of comparator method.
pratyush773 my c++ code:
bool compare(Player a,Player b){ if(a.score!=b.score){ return a.score>b.score; } else return a.name<b.name; } vector<Player> comparator(vector<Player> players){ sort(players.begin(),players.end(),compare); return players; }
it's simple. isn't it?
jvargas714 For cleaner comparisions could do some op overloading for the fun of it.
// something like this for p1 < p2 bool operator < (Player& p1, Player& p2){ if( p1.score != p2.name ){ return p1.score < p2.score; } else{ // compare names then return p2.name < p1.name; } }
That way you dont even need a compare function, you can just compare the player object using the '<' operator.
pratyush773 yes got it! thanks.
i think
if(p1.score!=p2.name)
should be there in place of
if(p1.score!=p2.score)jvargas714 Yes that is what I meant, thanks for catching that! Sorry about that I was rushing. Here it is corrected...
// something like this for p1 < p2 bool operator < (Player& p1, Player& p2){ if( p1.score != p2.score ){ return p1.score < p2.score; } else{ // compare names then return p2.name < p1.name; } }
MIC0DE Can you explain your code in detail? i am very confused about how dose work. ts
ashutoshtripath3 I couldn't understand, why in first check you're returning result of a.score>b.score, and in the else condition you're returning result of a.nameb.name ???
boatsoap Java: don't make class Checker public.
alperensahincs Yes you are rigth but why is so?
rcgonzalezf In Java every file should have at most one public class, and the public class name has to match with the name of the file. E.g:
public class X {}// the file should be named X.java
The file X.java can define multiple classes, but only X class can be public as a "top level class". Having said that. The compiler allows inner classes to be public, E.g:
// file name should be TopLevel.java public class TopLevel { public static class NotTopLevel{} public class AnotherNotTopLevel{} private class X{} } class TopLevelDifferent { // can't be public }
martyjlewis Hardest part of this challenge was figuring out how the two pieces of java code (editable and non-editable) are put together before compiling. My original assumption was that they were separate classes/files. Obviously not.
solidwater Hi Team,
Can you also enable C# in editor for this problem
vipul_gos In this problem where we have implemented the Quick Sort, i think we have just used the alorithm but not implemented.
It would be better if we pass the arrays of player and we implement our method.
What is your view @GayleLaakmannMcDowell and other developers.
tantk90 I agree. The problem is to implement the Quick Sort, not to use it.
seansrk Here is my solution implementing the quicksort algorithm in C++. I also decided to not use any STL algorithms (wrote my own swap as well only because I feel you should approach these problems without using the STL)
struct Player { string name; int score; }; void swap(vector<Player> &players, int a, int b) { Player temp = players[b]; players[b] = players[a]; players[a] = temp; } int Partition(vector<Player> &players, int begin, int end) { int x = players[end].score; int i = begin-1; for(int j = begin; j<end; j++){ if(players[j].score <= x ){ i++; if(players[i].score != players[j].score){ swap(players, i,j); } if(players[i].score == players[end].score){ if(players[i].name < players[end].name){ swap(players,i,end); } } } } swap(players,i+1,end); i += 1; return i; } void QuickSort(vector<Player> &players, int begin, int end){ if(begin < end) { int q = Partition(players,begin,end); QuickSort(players,begin,q-1); QuickSort(players,q+1,end); } } vector<Player> comparator(vector<Player> players) { QuickSort(players, 0, players.size()-1); return players; } int main() { int n; cin >> n; vector< Player > players; string name; int score; for(int i = 0; i < n; i++){ cin >> name >> score; Player p1; p1.name = name, p1.score = score; players.push_back(p1); } vector<Player > answer = comparator(players); for(int i = answer.size()-1; i >=0; i--) { cout << answer[i].name << " " << answer[i].score << endl; } return 0; }
sylvanc94 I think if they didn't want us to use sort, they would not have the
#include <algorithm>
header.
ohachami public int compare(Player a, Player b) { if(a.score == b.score) { return a.name.compareTo(b.name); }else { return Integer.valueOf(b.score).compareTo(a.score); } }
Sun_Han Java
class Checker implements Comparator<Player>{ String name; int score; public int compare(Player p1, Player p2){ if(p1.score == p2.score) return p1.name.compareTo(p2.name); return p2.score - p1.score; } }
MT2016117 I have the same code, but I declared Checker class to be public and it didn't run, removing the modifier worked. Can you explain why that happened>
ishtiaque_h Had the same issue. Looking forward to see if anyone has any explanation. Thanks.
zeroTYJ Because Solution is declared to be public and a java file only have a public class.
MT2016117 Solution is not declared to be public. I think I figured out why Checker class is not allowed to be public, A java file can have only one public class, the compilation unit is saved as .class file. Example : class A{ public static void main(String[] args){ //any code } }
class B{ any method(){
} }
Now, the compilation unit is saved as a class file names A.class, therefore class B cannot be public. A( the class containing main) may or may not be public.
enricogiurin You don't need the attributes "name" and "score" in the Checker class. Again, better to annotate the method compare with @Override.
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