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;
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;
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()
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 :)
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.
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
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.
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 wow!! nice
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!!!
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; }) }
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 :)
rakesh_shah19971 C Function -
int* solve(int grades_size, int* grades, int *result_size){
// Complete this function int i, g, r; for (i=0; i<grades_size; i++) { g = grades[i]; if (g < 38) continue; if ((r = g%5) > 2) grades[i] = g + 5 -r; } *result_size = grades_size; return grades;}
All comments most welcome, thanks.
gobisvivek 1)int* solve(int grades_size, int* grades, int *result_size) -> In this line, why are we using "result size" variable? 2)*result_size = grades_size; -> What is the meaning of this line?
kodandapani_v import java.util.Scanner; public class gradingProcess {
public static void main(String args[]) { Scanner sc = new Scanner(System.in); int size = sc.nextInt(); int[] grades = new int[size]; for(int i=0;i<size;i++) { grades[i]=sc.nextInt(); grades[i] = (grades[i]>=38)?((grades[i]%5>=3)?(grades[i]+(5-(grades[i]%5))):grades[i]):grades[i]; } for(int i=0;i<size;i++) System.out.println(grades[i]); }}
bijeli_papir nice
arunboopathi2 [deleted]
feling87 Javascript solution:
function solve(grades){ var index = 0; while(index < grades.length) { if (grades[index]%5 >=3 && grades[index] >37) { grades[index] = grades[index] - grades[index]%5 + 5; } ++index; } return grades; } function main() { var n = parseInt(readLine()); var grades = []; for(var grades_i = 0; grades_i < n; grades_i++){ grades[grades_i] = parseInt(readLine()); } var result = solve(grades); result.forEach(e => console.log(e)) // console.log(result.join(" "));
enilaydagdemir Here is my solution with C#:
static int solve(int grade){ int result = grade; if (grade >= 38) { if ((5 - (grade % 5) + grade) - grade < 3) result = 5 - (grade % 5) + grade; } return result; } static void Main(String[] args) { int n = Convert.ToInt32(Console.ReadLine()); int[] grades = new int[n]; for(int grades_i = 0; grades_i < n; grades_i++){ grades[grades_i] = Convert.ToInt32(Console.ReadLine()); Console.WriteLine(solve(grades[grades_i])); } }
alok66661082 hahahahha lets copy
kevinq93 static int[] solve(int[] grades, int n){ // Complete this function for(int i = 0; i < n;i++) { if(grades[i] >= 38 && grades[i] % 5 > 2) { grades[i] = grades[i]/5*5+5; } } return(grades); } static void Main(String[] args) { int n = Convert.ToInt32(Console.ReadLine()); int[] grades = new int[n]; for(int grades_i = 0; grades_i < n; grades_i++){ grades[grades_i] = Convert.ToInt32(Console.ReadLine()); } int[] result = solve(grades,n); Console.WriteLine(String.Join("\n", result));
Similar answer, slightly different logic.
abdullah_mamun11 Superb
prodip2416 may be this also Goooddddddd
int[] result = grades;
for (int i = 0; i < grades.Length; i++) { if (grades[i] >= 38) { if (5 - (grades[i]%5) < 3) { result[i] = (5-grades[i]%5) + grades[i]; } } } return result;
jaramillob93 Answer in Java 8
package solution; import java.io.*; import java.lang.annotation.ElementType; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { Scanner in = new Scanner(System.in); int n = in.nextInt(); if (n >= 1 && n <= 60) { for(int a0 = 0; a0 < n; a0++){ int grade = in.nextInt(); // your code goes here if (grade >= 0 && grade <= 100 && grade >= 38) { if (multipleOf5(grade) - grade < 3) { grade = multipleOf5(grade); System.out.println(grade); } else { System.out.println(grade); } } else { System.out.println(grade); } } } } public static int multipleOf5(int G) { int aux = 0; if (G % 5 == 0) { return G; } else { while (G % 5 != 0) { G++; if (G % 5 == 0) { aux = G; } } } return aux; } }
Thanks!
goodman_d_erik I would suggest trying to consolidate those if/else blocks a bit. Using the modulus you can simplify rounding quite a bit. After reading in the grade I used a helper method as follows
public static int rounder(int input){ if (input >= 38 && 5-input%5 < 3){ return input += (5-input%5); } return input; }
It's not the cleanest, but hopefully that's pretty legible
RodneyShag Java solution - passes 100% of test cases
From my HackerRank solutions.
import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner scan = new Scanner(System.in); int n = scan.nextInt(); while (n-- > 0) { int grade = scan.nextInt(); System.out.println(roundGrade(grade)); } } private static int roundGrade(int grade) { if (grade >= 38) { int gradeMod5 = grade % 5; if (gradeMod5 > 2) { grade += 5 - gradeMod5; } } return grade; } }
Let me know if you have any questions.
Sukhnoor if (gradeMod5 > 2) { grade += 5 - gradeMod5; } explain please
RodneyShag That's the code to do: "If the difference between the grade and the next multiple of 58 is less than 3, round grade up to the next multiple of 5."
The "If the difference between the grade and the next multiple of 58 is less than 3" corresponds to
if (gradeMod5 > 2)
The "round grade up to the next multiple of 5" corrsponds to
grade += 5 - gradeMod5;
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