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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.
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:
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.
importFoundation// Complete the beautifulDays function below.funcbeautifulDays(i:Int,j:Int,k:Int)->Int{varcount=0forvalueini...j{varr=valuevarq=0whiler>0{q=q*10+r%10r/=10}abs(value-q)%k==0?count+=1:()}returncount}
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.
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;
// 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??)
Beautiful Days at the Movies
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this :)
Nice solution! Here is my not so elegant solution, however maybe easier to understand for beginners..
I have done the same thing, python is amazing.....
Whoa!
my code exactly
Cool, mine is the same!!
I don't quite understand what you are saying in the conditional statement. Can you explain?
I am checking if the difference between number and reversed number is evenly divisible by k or not.
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.
Like I did?
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.
using functional operations :)
or two line, unintelligible version :p
Which course/book are you following on python?
Learned Python as part of Discrete Math course and mostly self taught since :)
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:
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.
This was my solution in Swift
i did the same ,but iam getting runtime error
did essentially the same but summed a generator with bool values converted to ints
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
check this one, Its easy to understand
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++??
I am also getting same error. Did you find a solution?
use in built reverse method of c++ to reverse.
and which one is that?
if you are facing time out problem, you can use printf() function here because in c++, streams are slower.
thanks ! :-)
tq ..ur suggestion helped me..
i solved same thing in java i got time out
same her, I have used string functions to reverse and converted it to number
static int beautifulDays(int i, int j, int k) {
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); }
static int beautifulDays(int i, int j, int k) {
always assign x to different variable or you will get caught in infinite loop
same issue
m also getting terminated due to timeout
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..
please explain the mistake clearly bro
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.
thank u bro!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
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.
show me the code which u written
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++){
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 {
}
(I followed the same approach then why ist showing wrong answer??)
Before the beginning of while loop, make sure you do revno=0;
Thank You Sir.I got the problem !
nice
I am also getting the same error.
hey ,i written same code but it not giving right answer
learn python 3
Very true...In Python 3 the code is one line.......
Beauty of python haha!
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)])Same logic...Pyhton is superb......