navjotahuja92 this :)
i, j, k = [int(x) for x in input().split()] beautifulDays = [1 for day in range(i, j+1) if (day - int(str(day)[::-1])) % k == 0] print(sum(beautifulDays))
kevinmathis08 Nice solution! Here is my not so elegant solution, however maybe easier to understand for beginners..
i,j,k = [int(x) for x in input().split()] count = 0 for x in range(i,j+1): if (x - int(str(x)[::-1])) % k == 0: count += 1 print(count)
himaja97 [deleted]
Harshith1995 [deleted]
alpha52omega41 I have done the same thing, python is amazing.....
amylokh Whoa!
benazus my code exactly
pgppratik Cool, mine is the same!!
ghassan_karwchan def beautifulDays(i, j, k): return sum ([1 for i in range(i, j + 1) if (int(str(i)[::-1]) - i) % k == 0])
joshuajuicehan I don't quite understand what you are saying in the conditional statement. Can you explain?
shashankjain_ec2 I am checking if the difference between number and reversed number is evenly divisible by k or not.
Ninja_kx [deleted]coslic Nice, but I don't like the entaglement in beautifulDays of the looping through range with the algorithm itself that decides to put an 1 or not depending on the (day - reverse of the day) is divisible or not by k. I would disentangle the looping and the filtering by the problem condition in 2 separate functions, and the algorithm will become more readable.
Kostyantin_Kon Like I did?
def reverse(n):
m = 0 while n != 0: m = 10 m += n % 10 n = n // 10 print(m) return mdef beautifulDays(i, j, k):
a = range(i,j+1) return sum([1 for x in a if abs(x-reverse(x)) % k == 0])*nithishjureddi [deleted]
nemertes Same :)
I'm just getting my feet wet with Python and just love list comprehensions so far, one-liners full of magic, aren't they.
TedYav using functional operations :)
i,j,k = map(int,input().strip().split()) def reversed(n): return int(str(n)[::-1]) def is_beautiful(n): return abs(n - reversed(n)) % k == 0 print(len(list(filter(is_beautiful, range(i,j+1)))))
or two line, unintelligible version :p
i,j,k = map(int,input().strip().split()) print(len(list(filter(lambda n: abs(n - int(str(n)[::-1])) % k==0, range(i,j+1)))))
askarovdaulet filter and lambda together look too confusing. Generator expressions would be much better here. Also, you don't need absolute value. Mod operation should work on negative numbers. Like this:
i,j,k = map(int,input().strip().split()) print(sum((x-int(str(x)[::-1]))%k==0 for x in range(i,j+1)))
TedYav Good solution! You're right about negative mod. Wasn't thinking about that. And I'm not super familiar with generator expressions. Your solution has inspired me to look into them more.
boertjie_seun This was my solution in Swift
import Foundation // Complete the beautifulDays function below. func beautifulDays(i: Int, j: Int, k: Int) -> Int { var count = 0 for value in i...j { var r = value var q = 0 while r > 0 { q = q * 10 + r % 10 r /= 10 } abs(value - q) % k == 0 ? count += 1 : () } return count }
nithishjureddi i did the same ,but iam getting runtime error
Tsukuyomi1 [deleted]Tzalumen [deleted]beckwithdylan did essentially the same but summed a generator with bool values converted to ints
i,j,k = [int(x) for x in input().split()] print(sum((int(abs(day - int(str(day)[::-1])) % k == 0) for day in range(i,j+1))))
freebird1195 I have done the exact same thing in C++. Fails all test cases except one, can you please help me figure out? Here is my code: https://www.hackerrank.com/challenges/beautiful-days-at-the-movies/forum/comments/419761
Madhurachanna check this one, Its easy to understand
int beautifulDays(int i, int j, int k) { int num,count=0,gn; int quo, newNum = 0, x = 0; for (int x = i; x <= j ; x++) { num = x; newNum = 0; //For reversing no while (num != 0) { quo = num % 10; newNum = 10 * newNum + quo; num = num / 10; } gn = abs(x- newNum); if (gn%k==0) { count++; } } return count; }
photo9609 i solved problem like this. but i gave "Terminated due to timeout".. Are u give "Terminated due to timeout" too? how too more fast in c++??
bhargavkumarg182 I am also getting same error. Did you find a solution?
abhinavmittra use in built reverse method of c++ to reverse.
17ce027 and which one is that?
umeshpant97 if you are facing time out problem, you can use printf() function here because in c++, streams are slower.
pg4409 thanks ! :-)
bitralahari20171 [deleted]
bitralahari20171 tq ..ur suggestion helped me..
sk7029 i solved same thing in java i got time out
chakri_learner same her, I have used string functions to reverse and converted it to number
shashankjain_ec2 static int beautifulDays(int i, int j, int k) {
int count = 0; for(int m = i ; m <= j ; m++){ int reverse = reverseNumber(m); if((Math.abs(reverse - m) % k) == 0){ count++; } } return count; } static int reverseNumber(int number){ Stack<Integer> digits = new Stack<Integer>(); while(number > 0){ int remainder = number % 10; digits.push(remainder); number /= 10; } int reversedNumber = 0; int count = 0; while(!(digits.isEmpty())){ reversedNumber += digits.pop() * Math.pow(10, count); count += 1; } return reversedNumber; }
bitralahari20171 try this out.it worked for me.and it is in java
static void beautifulDays(int i, int j, int k) { int cnt=0; for(int p=i;p<=j;p++) {int tmp=p,r=0; while(tmp>0) {r=r10+tmp%10; tmp/=10;} if(((p-r)%k)==0) {cnt+=1;} } System.out.println(cnt); }
shashankjain_ec2 static int beautifulDays(int i, int j, int k) {
int count = 0; for(int m = i ; m <= j ; m++){ int reverse = reverseNumber(m); if((Math.abs(reverse - m) % k) == 0){ count++; } } return count; } static int reverseNumber(int number){ Stack<Integer> digits = new Stack<Integer>(); while(number > 0){ int remainder = number % 10; digits.push(remainder); number /= 10; } int reversedNumber = 0; int count = 0; while(!(digits.isEmpty())){ reversedNumber += digits.pop() * Math.pow(10, count); count += 1; } return reversedNumber; }
hasija14 always assign x to different variable or you will get caught in infinite loop
guddudtu same issue
madhusharma22411 m also getting terminated due to timeout
bhargavanishanth I think that u guys are not initialising a new_number(=i) but using 'i' instead .. that changes the value of i to 0 making the loop infinte..
I have done the same mistake..
saivamsi2507 please explain the mistake clearly bro
bhargavanishanth int beautifulDays(int i, int j, int k) { int num,count=0,gn; int quo, newNum = 0, x = 0; for (int x = i; x <= j ; x++) { num = x; //make sure to assign this newNum = 0; //For reversing no while (num != 0) { quo = num % 10; newNum = 10 * newNum + quo; num = num / 10; } gn = abs(x- newNum); if (gn%k==0) { count++; } } return count; }
the correct code for reversing (from above commented code)
num=x //assignment of the temp number while(num!=0) { //code } if u dont assign a new variable(num) that leads to changing the value of for loop initiailizer (x) to 0 at end of the loop ,which makes the loop infinite.
saivamsi2507 thank u bro!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
pg4409 but i made another function for reversing the number so that value of x doesnt change. still i am having same trouble of time out.
saivamsi2507 show me the code which u written
pg4409 int reverse(int n){ int rev=0; int rem=0; while (n!=0){ rem=n%10; rev=rev*10+rem; n/=10; } cout<< rev<< endl; return rev; } // Complete the beautifulDays function below. int beautifulDays(int i, int j, int k) { int ctr=0; for (int x =i;x<=j;x++){
int rev=reverse(x); if (abs(rev-x)%k==0){ ctr+=1; } } return ctr;
1634810124_coe3b import java.io.; import java.math.; import java.security.; import java.text.; import java.util.; import java.util.concurrent.; import java.util.regex.*;
public class Solution {
// Complete the beautifulDays function below. static int beautifulDays(int i, int j, int k) { int c=0,num; int revno=0; int temp,gn; for(int l=i;l<=j;l++) { num=l; temp=0; while(num!=0) { temp=num%10; revno=revno*10+temp; num=num/10; } gn=Math.abs(l-revno); if(gn%k == 0) { c++; } } return c; } private static final Scanner scanner = new Scanner(System.in); public static void main(String[] args) throws IOException { BufferedWriter bufferedWriter = new BufferedWriter(new FileWriter(System.getenv("OUTPUT_PATH"))); String[] ijk = scanner.nextLine().split(" "); int i = Integer.parseInt(ijk[0]); int j = Integer.parseInt(ijk[1]); int k = Integer.parseInt(ijk[2]); int result = beautifulDays(i, j, k); bufferedWriter.write(String.valueOf(result)); bufferedWriter.newLine(); bufferedWriter.close(); scanner.close(); }}
(I followed the same approach then why ist showing wrong answer??)
torpidsnake [deleted]torpidsnake Before the beginning of while loop, make sure you do revno=0;
1634810124_coe3b [deleted]1634810124_coe3b Thank You Sir.I got the problem !
rakeshsingh0018 nice
cbrocks I am also getting the same error.
umangsomtiya8083 hey ,i written same code but it not giving right answer
- int main()
- {
- long int m,n,k,r=0;
- long int dif,div;
- cin>>m>>n>>k;
- long int i,j,p=0;
- for(i=m;i<=n;i++)
- {
- for(j=i;j!=0;j=j/10)
- {
- r=(r*10)+(j%10);
- }
- if(i>r)
- {
- dif=i-r;
- }
- if (i <= r) {
- dif = r- i;
- }
- if(dif%k==0)
- ++p;
- }
- cout<
- } 1.
dvjpy786 learn python 3
b_s_apsara_july Very true...In Python 3 the code is one line.......
dhingra_shubham1 Beauty of python haha!
chris_stevenson You can also use the fact that True == 1 in Python:
sum([not abs(int(str(x)[::-1]) - x) % k for x in range(i, j + 1)])mahender933 def beautifulDays(i, j, k):
return sum( 1 for num in range(i,j+1) if abs(num-int(str(num)[::-1]))%k == 0 )b_s_apsara_july Same logic...Pyhton is superb......
RosieJunghwa print(sum((day - int(str(day)[::-1])) % k == 0 for day in range(i, j+1)))
You can do this :)
venkatsaikiran Solution in Java
import java.io.*; import java.util.*; import java.text.*; import java.math.*; import java.util.regex.*; public class Solution { public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner in = new Scanner(System.in); int i = in.nextInt(); int j = in.nextInt(); int k = in.nextInt(); int count = 0; for (int a=i;j>a; a++){ StringBuilder temp = new StringBuilder(); temp.append(a); temp=temp.reverse(); String temp1 = temp.toString(); int aRev = Integer.parseInt(temp1); if(Math.abs((a-aRev)%k)==0){ count++; } } System.out.println(count); } }
Akash_Mishraa can you pls explain this line
if(Math.abs((a-aRev)%k)==0)
venkatsaikiran it's the main logic of our program. it is for checking if the day is beautiful or not. here we are checking if the difference of the number and its reverse can be evenly divided by k.
vaibhavpkc whenever the difference of a and reverse a is divisible by the k no. then the count will increase and hence we will get the required result.
singirthi_abhin1 Math.abs() is used to avoid negative result . Hope this helps!
crazypan public class Solution {
public static void main(String[] args) { Scanner in = new Scanner(System.in); int i = in.nextInt(); int j = in.nextInt(); int k = in.nextInt(); int numOfBeautifulDays=0; for(int l=i; l<=j;l++) { if((l-reverse(l))%k ==0) numOfBeautifulDays++; } System.out.println(numOfBeautifulDays); } public static int reverse(int number){ int reverse = 0; while(number!=0){ reverse = 10*reverse +number%10; number = number/10; } reverse += number; return reverse; }}
phyolintun87 Same idea here
srikanth_27 MR.CRAZYPAN
why did you do this reverse += number;
Wasim_AbbaCi to get reverse of number
vaibhavpkc "reverse += number;" can you explain me this part?
vaibhavpkc Actually there is no need to introduce "reverse += number".It's working well without the use of it. here is my code: import java.io.; import java.util.; import java.text.; import java.math.; import java.util.regex.*;
public class Solution {
public static int reverse(int a) { int reverse = 0; while(a!=0) { reverse = reverse * 10; reverse = reverse + a%10; a = a/10; } return reverse; } public static void main(String[] args) { /* Enter your code here. Read input from STDIN. Print output to STDOUT. Your class should be named Solution. */ Scanner in = new Scanner(System.in); int i = in.nextInt(); int j = in.nextInt(); int k = in.nextInt(); //int n = i-j+1; // int s[] = new int[n]; int count = 0; for(int a = i;a<j+1;a++) { int d= a-reverse(a); if(d%k==0) count++; } System.out.println(count); }}
Bladefan960 why do you need "reverse += number;"?
samplenhold You don't need to do the absolute value.
crazypan You are right!
goncalves_josepr Very well done!!
rakshaavaidyana1 static int beautifulDays(int i, int j, int k) { int r=0,count=0;
int l=Math.abs(i-j)+1; int[] ar=new int[l]; int[] ar1=new int[l]; int[] sum=new int[l]; int n=i; for(int m=0;m<ar.length;m++) { ar[m]=i; i++; ar1[m]=n; n++; sum[m]=0; while(ar[m]>0) { r=ar[m]%10; sum[m]=sum[m]*10 + r; ar[m]=ar[m]/10; } if(((Math.abs(ar1[m]-sum[m]))%k)==0) count++; } return count; } Performed similar logic but without String builder class.
HyperMystic for loop should be this
for (int a=i;a<=j; a++)
should be less than or equal to
18dcs024 That's a good solution but in the for loop instead of for (int a=i;j>a; a++) you need to use for (int a=i;j>=a; a++) so that it also checks for the value of a itself and will return the correct and required output.In the case what you have done there might be wrong answer for some of the test cases. HOPE THIS HELPS!!!
patrickdekelly That Lily girl is loony. Get out while you still can Logan! :P
condrint hahahahahah I was cracking up for a minute or so when I had that thought
Aswath3167 My C Solution......
#include<stdio.h> int main() { int i=0,j=0,k=0,x=0,rem=0,sum=0,count=0; scanf("%d%d%d",&i,&j,&k); for(i;i<=j;i++) { x=i; while(x!=0) { rem=x%10; sum=(sum*10)+rem; x=x/10; } if(abs(i-sum)%k==0) count=count+1; sum=0; } printf("%d",count); return 0; }
UtkarshSWG my code isnt working..please tell me why
int main() { long int i=0,j=0,k=0,m=0,x=0,d=0,r=0; cin>>i>>j>>k; for(m=i;m<=j;m++) { r=0; while(m!=0) { d=m%10; r=r*10+d; m=m/10; } if((abs(r-m)%k)==0) x++; } cout<<x; return 0; }
non_programmer Dude, you are using same m in your code... which will keep on becoming zero after completion from while loop and it will never be equal to j, better try out with this:
include
include
using namespace std; int main() {
long int i=0,j=0,k=0,m=0,x=0,d=0,r=0,m1=0; cin>>i>>j>>k; for(m=i;m<=j;m++) { r=0; m1=m; while(m1!=0) { d=m1%10; r=r*10+d; m1=m1/10; } if((abs(r-m)%k)==0) x++; } cout<<x; return 0;} I hope you got what's wrong.
UtkarshSWG thanks dude...I got it
nayan_7991 in this when the loop run for the first time the vue of m becomes 0. so we again give the value of i to m and so Infinite loop runs
yashwant97 sum=(sum*10)+rem; i dint understand this line can u pls help me out?
jluisg_23 I did it using conversion to string, but now that I see your solution, yours is simpler and faster. I share mine anyways:
#include <stdio.h> #include <stdlib.h> int reversed(int a) { char c, *ptr_b, *ptr_e; char buff[12]; int b; //Integer to string sprintf(buff, "%d", a); //Reversing string ptr_b = buff; ptr_e = buff; while(*ptr_e){ //Moving ptr_e to point the last char in the string ++ptr_e; } --ptr_e; while(ptr_b < ptr_e){ //Reversing c = *ptr_b; *ptr_b = *ptr_e; *ptr_e = c; ++ptr_b; --ptr_e; } //String to integer sscanf(buff, "%d", &b); return b; } int main() { int i, j, k, rev, diff, count = 0; //Input scanf("%d %d %d", &i, &j, &k); //Counting for(int ii = i; ii <= j; ++ii){ rev = reversed(ii); diff = abs(ii-rev); if(diff%k == 0) ++count; } //Output printf("%d\n", count); return 0; }
AGRAWALTUSHAR27 Thanks man! You have cleared my confusion.
juan_cr Java 8 using lambdas
static int beautifulDays(int i, int j, int k) { return (int) IntStream.rangeClosed(i, j) .map(c -> Math.abs(c - reverse(c))) .filter(c -> c % k == 0) .count(); } private static int reverse(int n) { return Integer.parseInt( new StringBuilder(String.valueOf(n)) .reverse().toString()); }
giovannict73 Beautiful days using beautiful Swift.....
import Foundation let input: [Int] = readLine()!.componentsSeparatedByString(" ").map { Int($0) ?? 0 } let i: Int = input[0] let j: Int = input[1] let k: Int = input[2] var beautifulDays: Int = 0 for integer in i...j { var int2reverse: Int = integer var reversedNum: Int = 0 while int2reverse != 0 { reversedNum = reversedNum * 10 + int2reverse % 10 int2reverse = int2reverse / 10 } let absDiff: Int = max(integer, reversedNum) - min(integer, reversedNum) if absDiff % k == 0 { beautifulDays += 1 } } print(beautifulDays)
jadex1 what happen to i and j? im still learnig swift.
giovannict73 what do you mean ?
jadex1 I see that i, and j are set by the input string. I also see that beautiful days is an int and set to zero. where I get lost is the for integer in i...j. int2reverse and reversedNum don't look like they get set to me. Where do they get their values from? I don't understand how the input value of i and j get into the
for integer in i..j{}giovannict73 for item in items {} is how for cycle is intended for Swift. The block {} will be executed items.count times and 'item' variable will assume each value of 'items' per iteration. In my code I used 'integer' as variable. According to that, int2reverse is initialized with 'integer' that is an int value equals to 'i' at first iteration, i+1, ... j in the last iteration.
sendsons So here does not support Swift 3 yet? I've used componentsSeparated(by:), and get a error :(
giovannict73 unfortunately Swift 3 here is not supported yet.
rafaelkgobara Nice solution!
What do you think about using tuple here?
let params = readLine()!.components(separatedBy: " ").map{ Int($0)! } let (i, j, k) = (params[0], params[1], params[2])
karshankhanil // Complete the beautifulDays function below. static int beautifulDays(int i, int j, int k) { int days = 0; for (int start = i; start <= j; start++) { char[] reverse = start.ToString().ToCharArray(); Array.Reverse(reverse); int rev = Convert.ToInt32(new string(reverse)); if ((Math.Abs(start - rev) % k) == 0) { days += 1; } } return days; }
Timing out for 3 TC, any suggestions please?
ddorshorst Don't use a string to manipulate the data. Use the % and / operators to reverse the input.
Also you don't need to use the absolute value function when checking the difference's equality to 0 since 0 does not have a sign.
edit:
karshankhanil Thanks a lot, it's working now. :)
slava_stackover1 I got time out error too in C#. Resolved it only by adding this code: int result = 0; string rev = ""; int dif = 0; int revNum = 0; for (int n = i; n < j + 1; n++) { if (j > k) { dif = n; rev = ""; while (dif >= 10) { rev = rev + (dif % 10).ToString(); dif = dif / 10; } rev = rev + dif.ToString();
if (rev == n.ToString()) { result++; } else { revNum = Convert.ToInt32(rev); if ((n - revNum) % k == 0) result++; } } else { var poly = n.ToString(); if (poly.Length == 1) result++; for (int l = 0; l < poly.Length / 2; l++) { if (poly[l] != poly[poly.Length - l - 1]) { break; } if (l + 1 == poly.Length / 2) { result++; } } } } return result;
ghassan_karwchan def beautifulDays(i, j, k): return sum ([1 for i in range(i, j + 1) if (int(str(i)[::-1]) - i) % k == 0])
Pythonologist pythonic way to put this...#singleline
def beautifulDays(i, j, k): return len([i for i in range(i,j+1) if (i-int(str(i)[::-1]))%k==0])
Rakesh433jntu Here i am failing many test cases becaue of 49860 205494 155635764 these sample input, here k value is much morw greater than actul values , so how can number can be divisible. in this case result is always 0
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