cakerusk Java one-liner code :
System.out.println(grade < 38 || grade % 5 < 3 ? grade : grade + (5 - (grade % 5)));
eavestn [deleted]eavestn [deleted]eavestn Nice solution. I am writing in C#, myself. I def. had to start with a more fleshed-out solution and work my way down to this. I wonder if a lengthier solution would be more communicative of the imposed requirements. I pushed mine up as a one-liner, though.
cakerusk Thanks. A lengthier solution might improve readability but compromise on lines of code (maintainability). Either way it doesn't affect runtime, so it comes down to a matter of personal preference I guess. :)
sungmkim80 I don't agree that a lengthier answer would compromise maintainability. If you deal with a smaller chunk of code, it's easier to maintain as you can focus on a narrow logic.
cakerusk Probably not in this question but I pointed that out in general from an interview/development perspective. For this particular qs, it's only a matter of preference.
JavaANDHannah It's a terrible solution. Hello? Have you ever heard of the words, "RE-READ YOUR WORK?" Oh, and plus, it doesn't work. :)
cakerusk It works perfectly fine.
https://www.hackerrank.com/challenges/grading/submissions/code/39368045
In case of issues, let me know.
uniyal_guru I have another solution for the same.
static int[] solve(int[] grades){ // Complete this function for(int i=0;i<grades.length;i++) { if(grades[i]>=38) { if(grades[i]+(5-grades[i]%5)-grades[i]<3) grades[i]=(grades[i]+(5-grades[i]%5)); } } return grades; }
doctorj0023 I did this with backward iteration and without your most inner If:
for(int i = grades.Length-1; i >=0; i--) { if(grades[i] >= 38 && grades[i] % 5 >2) { grades[i] = grades[i] + (5 - (grades[i] % 5)); } }
alexb760 in your solution what happen if grade[i] = 67 them grades[i] % 5 = 7 and them 5-7 = -2 so 67 -2 = 65
keickhoff1 No, it will not pass the second condition of the if statement. Because 67 % 5 = 2 not 7.
harshrajan00 [deleted]harshrajan00 As @keickhoff1 pointed out 67 remains 67.
PareshSalunke Your solution is perfectly fine but as a first draft. Whenever possible take out common operations and put it in a variable so next time what if the moderator said that multiple of 6 instead of 5 . you'll have to change at 2 places in your case but it may be a become bigger issue in complex chunk. Just saying. :)
hard_system very good, by the way it has to be grades.Length
avi_ganguly94 in the inner if() , why did you do a grades[i] + blahBlahPivot - grades[i]? i mean, grades[i]-grades[i] is zero anyway so why bother?
ssaulakh2 [deleted]
krishna_kumar_b1 this program have some error check your program
tazpandey Your answer is nice.But why you put -grades[i]<3 in if condition can you explain it.
harshrajan00 (5 - grades[i] % 5) < 3 if this is the condition you are talking about then the question says that If the difference between the grade and the next multiple of 5 is less than 2.
devin99 Your link doesn't work, like the code provided above
rabbyalone my solution
static int[] solve(int[] grades) { int[] res = new int[grades.Length]; for (int i = 0; i < grades.Length; i++) { if (grades[i] % 5 > 2 && !(grades[i] < 38)) res[i] = grades[i] + (5-grades[i]%5); else res[i] = grades[i]; } return res; }
aliriomambo [deleted]van07 Would this be better? For reability only. Im not sure. It's really the same solution.
int[] res = new int[grades.Length];
for (int i = 0; i < grades.Length; i++) { int current = grade[i]; if (grades[i] % 5 > 2 && !(grades[i] < 38)) current = grades[i] + (5-grades[i]%5); res[i] = current; } return res;
aaronederby You are right. There is a pretty big epedimic of poor readability in coding these days. Some of these projects are pretty good proof of that.
munna1991vikram Get the complete information of windows 10 operating system by visiting here on this website. windows 10 free download Because this website provides us complete information of windows 10 operating system in your personal computer.
piyushhajare A simple and perhaps readable solution.. You can use n instead of temp as its just a dummy..
int main() { int n,temp; cin>>temp; while(cin>>n) { if(n<38) cout<<n<<endl; else if(n%5 >= 3) cout<<n+ (5-n%5)<<endl; else cout<<n<<endl; } return 0; }
owenized Why use EOF?
deepesh463 it stands for end of file
agnivabasak1 according to me it shouldn't work but if it passed test cases ,then i dont know how .the reason is that consider a number like 73 ,then n%5 should give 3 ,and hence it would redirect to the else statement and print n, whereas it should have printed 75 as we need to consider the next multiple of 5 ,75-73 =2 (<3) .Correct me ,if i am wrong ,anyone. (might be i am wrong too :P ,i developed my solution in my way ,though!)
steven_tang 73 % 5 = 3 and 3 <= 3, so it'll satisfy the else if condition. Then we have cout << 73 + (5- 73%5) << endl; which simplifies to cout << 73 + 2 << endl;
zeashaikh3306 Why are we checking <=3 and not some other number?
vijayashreeshet1 wow ..thank u
JavaANDHannah [deleted]
borb_ You can do the same thing in C# ->
return grades.Select(grade => grade < 38 || (grade % 5 < 3) ? grade : grade + (5 - grade % 5 )).ToArray();
return grades.Select(grade => grade < 38 || (grade % 5 < 3) ? grade : grade + (5 - grade % 5 )).ToArray();
sidra_mbn Excellent
sidra_mbn Can you please also tell me that ToArray is to be used only when we are returning an array?
moh_so2001 foreach(`$grades as $`grade){grade < 38 || grade : grade % 5))); } return $g;
php code
eavestn Don't know why I got downvoted for a simple comment?
rajatdhoot Great Solution.........
alexandre_ragal1 Got some fun with Java 8:
Arrays.stream(new int[n]).map(i-> in.nextInt()).map(g -> (g >= 38 && g % 5 >=3) ? g+5-g%5 : g).forEach(System.out::println);
jude_niroshan11 nice shortcut for nearest 5th divisible value.
g+5 - g%5
ashishsinghchau1 Does not working out.It did some basics mistakes on the Editor.
alexandre_ragal1 It is working since I passed all tests.
public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); Arrays.stream(new int[n]).map(i-> in.nextInt()).map(g -> g >= 38 && g % 5 >=3 ? g+5-g%5 : g).forEach(System.out::println); }
a_gutsal Arrays.stream(grades).map(grade -> (grade < 38) || (grade % 5 < 3)?grade:((int)(grade / 5) + 1) * 5).toArray()
ismailkuet Get some fun with Python 3::
return [n if n < 38 or n % 5 < 3 else (n + 5 - (n % 5)) for n in grades]
nvinson234 A lot of modulo use when it's not really needed. An alternate version:
return [max(n, (n + 2) // 5 * 5) if n >= 38 else n for n in grades]
auratice it would be really helpful if you could explain this. thank you!
Cuteboy06 Can anyone explain me how it is printing grade for the values like 67.
cakerusk The rules for rounding off is stated as follows :
If the difference between the grade and the next multiple of 5 is less than 3, round grade up to the next multiple of 5. If the value of grade is less than 38, no rounding occurs as the result will still be a failing grade.
Since the value 67 is greater than 38 and the difference between value and next multiple of 5 (i.e. 70) is 3 (70 - 67) which is not less than 3, it will not be rounded. Hence, 67 i.e. 'grade' value is printed.
However, for the value 68, the difference is 2 (70 - 68) < 3, so the grade will be rounded to the next multiple by the calculation :
grade + (5 - (grade % 5)) = 68 + (5 - (68 % 5))
= 68 + (5 - 3)
= 68 + 2
= 70 (next multiple of 5)
Hope this helps! :)
Cuteboy06 Thanks for your time :)
TejanSD17 Thank you.
alternativo_vini thanks
piyushhajare A simple and perhaps readable solution.. You can use n instead of temp as it's dummy..
int main() { int n,temp; cin>>temp; while(cin>>n) { if(n<38) cout<<n<<endl; else if(n%5 >= 3) cout<<n+ (5-n%5)<<endl; else cout<<n<<endl; } return 0; }
bg2407111 Just a tip I've learned from my professors, you should rarely ever have to rewrite the same lines of code, see how your initial if and your default else are doing the same thing? You could condense this by doing this:
if(n>38 && n%5 > 3) { cout << n+(5-n%5) << endl; } else { cout << n << endl; }Makes it still readable, but in less lines of code. Also, in Industry code it's generally good practice to use curly braces on the body of if statements to prevent any confusion.
ssaulakh2 Will I get kicked out if I do this from nothing.
Ardillo Python3 one-liner:
grades = [int(input()) for __ in range(int(input()))]
[print(g+5 - g%5 if g%5 > 2 and g>37 else g) for g in grades]
h201551072 i am new in pyhton can u help me with the mistake?
#!/bin/python import sys def solve(grades): r=[] y=0 d=0 # Complete this function for i in range(0,len(grades)-1): if(grades[i]>=38): y=grades[i]/5 d=5*(y+1)-grades[i] if(d<3): r[i]=5*(y+1) else: r[i]=grades[i] else: r[i]=grades[i] return r n = int(raw_input().strip()) grades = [] grades_i = 0 for grades_i in xrange(n): grades_t = int(raw_input().strip()) grades.append(grades_t) result = solve(grades) print "\n".join(map(str, result))
immanoj16 def solve(grades): result = [] for i in grades: if i >= 38: if i % 5 >= 3: i += (5 - i % 5) result.append(i) return result
JehanJoshi You can club both the if conditions using an and here and neatly get it in one line :)
vishalnirne98 how doesthe second if statement work
utkarshGG oh, that is an equivalent of Ternery Operator ?: in other languages. Do_this if this_is_true else do_this
say for example,
>>> x = 10 >>> y = 21 if x > 12 else 5
y will be equal to 21 if x is greater than 12 else y will be 5.
mahmut_bilgen Good solution! Congratulation !
def gradingStudents(grades): # Write your code here result = [] for grade in grades: rounding_num= ((grade // 5)+1)*5 if grade<38 : result.append(grade) elif rounding_num-grade<3: result.append(rounding_num) else: result.append(grade) return result
Mandar_Inamdar you are doing y=grades[i]/5 which will give you decimal numbers. round division using math functions or simply like: y=int(grades[i]/5)
ditonikola #!/bin/python3 import sys def solve(grades): for k in range(0,n): if grades[k]<38: print(grades[k]) else: l=grades[k]/10 m=round(l) if 0.25<abs(l-m)<0.5 or 0.5<abs(l-m)<0.75: a=round(grades[k]/5) b=a*5 c=b-grades[k] if c<0: if c<-3: print(b) else: print(grades[k]) else: if c<3: print(b) else: print(grades[k]) else: f=m*10 d=f-grades[k] if d<0: if d<-3: print(f) else: print(grades[k]) else: if d<3: print(f) else: print(grades[k]) n = int(input().strip()) grades = [] grades_i = 0 for grades_i in range(n): grades_t = int(input().strip()) grades.append(grades_t) result = solve(grades) print ("\n".join(map(str, result)))
whats wrong with this pls tell me, it gives me runtime error, but passes 5 times
_dugy You have to return array of ints from "solve" function.
mtausif_0003 its so complicated logic
ismailkuet One line solution::
return [n if n < 38 or n % 5 < 3 else (n + 5 - (n % 5)) for n in grades]
ismailkuet @h201551072 You can do it in one line with list comprehensions
return [n if n < 38 or n % 5 < 3 else (n + 5 - (n % 5)) for n in grades]
archit_imsec10 And how about this?
map(lambda x: 5*(1 + x//5) if (x > 37 and ((x%5) > 2)) else x, grades)
ahmettortumlu25 so good.
sidra_mbn Thank you
massimiliano_de1 nice one liner, but write something like that in my project and I'll fire you.
you must spare maintainer/programmer time, not whitespaces in the file.
Spiznak 1 iteration, almost 1-liner. O(n) is slightly better than O(2n):
for _ in range(int(input())): x = int(input()) print(x + (5 - x % 5)) if x >= 38 and 5 - x % 5 < 3 else print(x)
I wouldn't be surprised to learn that there is a better way to do it, though...
Dimple151997 Can you explain the code.. why do we write __ in for loop and 2nd line in [] brackets
MilosMM Hi, because we don't care about the value of the iterator.
rudralittle i ran it on my jupyer notebook. i cant understand the output. input>>>>>>>>>>>>> grades = [int(input()) for __ in range(int(input()))]
[print(g+5 - g%5 if g%5 > 2 and g>37 else g) for g in grades]
output>>>>>> 5 45 66 23 56 98 45 66 23 56 100 [None, None, None, None, None]
chris_dziewa I knew I would find something much shorter than my code here but wow! Nice work!
ashishsinghchau1 Your code does not working out in the editor.
arunesh_sarin [deleted]arunesh_sarin but this code should round off 67 as 67%5 is 2 which is less 3 . Help me undertand this please.
borb_ The problem states that the rounding should occur if the DIFFERENCE is LESS than 3. The difference here is 3 so it is not rounded.
ummahusla Wow! Clearly beatiful and cleanest way I've seen so far
SupreethMC Neat !
vijayashreeshet1 i dont understan java .. bt i understood the logic .. thank you:)
UWUTM8 I've got a pretty sick one-liner too for C++ can your boy get some upvotes =D
for (int i = 0; i < grades.size(); i++){int nearest5Multiple = 5*((grades[i]/5)+1);if (grades[i] >= 38 &&(nearest5Multiple-grades[i]) < 3 ){grades[i] = nearest5Multiple;}}return grades;
sfelde21 haha I hope you're joking
ditonikola No im not joking
WatchandLearn it looks good,but keeping result of (grade%5) in a variable could save you from calculating it two times :)
dipznjoy actually not , as it will work for 41,42 . as 41 % 5 =1 , but we want to have the for 43,44. that goes for all the cases.
Tunvir07 Great.
santhoshkumar_n It works
rishabh0_9 This is my solution
static int[] solve(int[] grades){ // Complete this function for(int i=0; i<grades.length; i++){ if(grades[i]>=38 && grades[i]%5>2) grades[i] = grades[i] + (5 - (grades[i]%5)); } return grades;
ramamoorthypand1 [deleted]
saiful007 if (grade > 37){ int newGrade = grade; for (int i=1; i<=2; i++){ newGrade++; if(newGrade%10 == 5 || newGrade%10 == 0){ grade = newGrade; return grade; } } } return grade;
vikramhpatel12 won't work for grade =73 since 73%5 is 3
saiful007 i dont think so it pass all the test cases ...but thanks for your kindness
tejasree438 I achived useing my own implementaion upto 20 lines to achieve this code.But your solution is just awe.I will adopt for this technique.Thank you.
rajkumarkanjo nice logic ..
m_eli10 [deleted]harshrajan00 Even my mobile has 4GB ram.
coder_aky Python 3 , 3-liner
for _ in range(int(input())): g=int(input()) print(g if g < 38 or g % 5 < 3 else g+(5-(g % 5)))
meghrajpardesi One liner for abouve code
def gradingStudents(grades): return [g if g < 38 or g % 5 < 3 else g + (5 -(g % 5)) for g in grades]
soheil_jahangir1 awesome.
CyberZii Amazing. Bravo!
gabrielbb0306 Your solution is bad for the problem as it is today. You are printing the elements, but the problem, at least today, says you should return the array. If you do the assigning to the array in one line then it is not possible to check when to assign, so you will always assign, which will be much slower. For this reason i will have to give a downvote to your answer. Still pretty cool for a problem that ask to print them instead of returning, tho
vangalasuneel this soluation will not pass all test cases
ismailkuet Get some fun with Python 3::
return [n if n < 38 or n % 5 < 3 else (n + 5 - (n % 5)) for n in grades]
vinayak_prabhut1 This will not work for inputs like 70, 80...
jasonganub A simple and clean solution with only one print statement and minimal conditionals.
if grade >= 38: if grade % 5 == 3: grade += 2 elif grade % 5 == 4: grade += 1 print(grade)
jakemager awesome use of the modulo operation
trungskigoldberg this solution is mathematically cool man
franchetti Could be condensed to a single condition if you wanted.
if grade >= 38: if grade % 5 > 2: grade += 5 - (grade % 5) print(grade)
cuteprince1989 this is such a clever solution great mind...
cuteprince1989 it is not working it is giving me error unsupported operand type for %: list and int. can any help me? there compiler is not working correctly, i have run this code in python 2 in my system it is working fine but not in here
ThaiNguyen You are using grade, which is a list. You have to access the indices in the list using loop like this
for i in grade: if i >= 38: if i % 5 == 3: i += 2 elif i % 5 == 4: i += 1 print (i)
h201551072 new in python can u help me with this code
#!/bin/python import sys def solve(grades): r=[] y=0 d=0 # Complete this function for i in range(0,len(grades)-1): if(grades[i]>=38): y=grades[i]/5 d=5*(y+1)-grades[i] if(d<3): r[i]=5*(y+1) else: r[i]=grades[i] else: r[i]=grades[i] return r n = int(raw_input().strip()) grades = [] grades_i = 0 for grades_i in xrange(n): grades_t = int(raw_input().strip()) grades.append(grades_t) result = solve(grades) print "\n".join(map(str, result))
immanoj16 def solve(grades): result = [] for i in grades: if i >= 38: if i % 5 >= 3: i += (5 - i % 5) result.append(i) return result
tomarshubham49 fgfhyhyrihowhhflhiihiw{} bhai yeah galti hai teri program mmm
dylanjcastillo Great solution!
lukus_2222 neat
heres the same solution in PHP
function solve($grades){ for ($i=0;$i <= sizeof($grades); $i++){ if ($grades[$i] >= 38){ if ($grades[$i] % 5 == 3){ $grades[$i] += 2; } else if ($grades[$i] % 5 == 4){ $grades[$i] += 1; } } } return $grades; }
16wh1a05b7 [deleted]
lvsz_ A difficult and intricate solution with only one line and no conditionals (nor variables).
print . ((uncurry (+) .) . ((. ((*) . fromEnum)) . flip second) <*> (uncurry ((. (>) 3) . (&&) . (<) 37))) . ((,) <*> (flip mod 5 . negate))
sammy9292 fantastic. Such simple logic!!!
max_aka21 You can just group 1st and 2nd if condition into single if condition
imerljak Javascript with ES6 :)
function solve(grades){ return grades.map((n) => { let diff = 5 - (n % 5); if(diff < 3 && n >= 38) { n += diff; } return n; }) }
hroger19 Wow that's short!
jeremymgould I think you can make it even shorter without hurting readability too badly.
return grades.map(g => (g < 38 || g%5 < 3) ? g : Math.ceil(g/5)*5)
This would work too and is probably a little clearer.
return grades.map(g => (g < 38 || g%5 < 3) ? g : g+(5-(g%5)))
sethkal7 I did the same, and was making the same decision between Math.ceil(g/5)5 and g+(5-(g%5)) I ended up with the following:
const gradingStudents = grades => grades.map(grade=> grade <= 37 || grade % 5 < 3 ? grade : grade + (5 - grade % 5));
svl_taylaran Short and sweet unlike mine lol
function gradingStudents(grades) { // Write your code here let finalGrades = new Array(grades.length); for (let x = 0; x <= grades.length - 1; x++) { if (grades[x] >= 38 && (grades[x] + 2) % 5 === 0) { finalGrades[x] = grades[x] + 2; } else if (grades[x] === 99 || (grades[x] >= 38 && (grades[x] + 1) % 5 === 0)) { finalGrades[x] = grades[x] + 1; } else { finalGrades[x] = grades[x]; } } return finalGrades; }
Totally forgot about the map method too.
sidinsomniac My JS solution:
function gradingStudents(grades) { let finalMarks = []; grades.forEach(grade => { grade < 38 || grade % 5 < 3 ? finalMarks.push(grade) : finalMarks.push(Math.ceil(grade/5)*5) }) return finalMarks; }
drpika1134 Can someone explain why he substracts the
(n % 5)from5in line 3?
waweruhmwaurah n%5 will give you the remainder of the numbers to make n divisible by 5..
say n = 78
78%5 = 3 // so we have a remainder of 3 if we divide 78/5..
We however do not need the remainder to convert to a number divisible by 5 say we add 3 to 78 we will end up with 81, which is not divisible by 5.
We want to know how far we are from achieving a 5 from the modulus of 78 which is why we then subtract the modulus from 5..
In our case it would be we have a remainder of 3 and from 78 we require 2 (i.e 5-3(remainder) ) to be able to make our number divisible by 5..
sense?
maggiew61 thanks
accounts47 The output asks us to print each grade on a new line. The function should actually return a list of values.
jmhrovat Thanks, this question should be re-written :/
ikhan77727 python solution
def gradingStudents(grades): return [ i if (i < 38 or i % 5 < 3) else (i + (5 - i%5)) for i in grades]
samarthsegu1999 public static List<Integer> gradingStudents(List<Integer> grades) { int n= grades.size(); List<Integer> result = new ArrayList<>(); for(int i=0;i<n;i++) { int ele=grades.get(i); if((ele%5)>=3 && ele>=38) ele=ele+(5-(ele%5)); else ele=ele; result.add(ele); } return result; }yahyamohamed1991 there is no need for else statment BTW but nice code and true anyway
kevinlai31 C version is broken btw. main function calls "gradingStudents" with undeclared first argument. change the first argument to "n"
melodicdata def gradingStudents(grades): r = [] for grade in grades: diff = 5 - grade%5 if grade >= 38 and diff < 3: grade += diff r.append(grade) return r
16FE1A0592CSE in java
public static List<Integer> gradingStudents(List<Integer> grades) { for(int i=0;i<grades.size();i++){ if(grades.get(i)>=38){ int r=grades.get(i)%5; int diff=Math.abs(r-5); if(diff<3) grades.set(i,grades.get(i)+diff); } } return grades; }
marinskiy Here is Python 3 solution from my HackerrankPractice repository:
for _ in range(int(input())): grade = int(input()) if grade >= 38 and grade % 5 >= 3: grade = (grade + 5) // 5 * 5 print(grade)
Feel free to ask if you have any questions :)
Rekt4 Why do you add 5 to grade?
joshuapfarina The syntax highlighting is off. The '//' represents floor division in python.
They are adding five, Then dividing by five (rounding down to the nearest whole number) Then multiplying by five.
Sort 1299 Discussions, By:
Please Login in order to post a comment