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)

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.

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 :)

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)))))

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))))

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

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)])

def beautifulDays(i, j, k):

 return sum( 1 for num in range(i,j+1) if abs(num-int(str(num)[::-1]))%k == 0 )

Same logic...Pyhton is superb......

print(sum((day - int(str(day)[::-1])) % k == 0 for day in range(i, j+1)))

You can do this :)

Python

return len([z for z in [abs(i-int(str(i)[::-1])) for i in list(range(i,j+1))] if z%k==0])

can you pls explain this line

if(Math.abs((a-aRev)%k)==0)

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;
}

}

You don't need to do the absolute value.

Very well done!!

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.

for loop should be this

for (int a=i;a<=j; a++)

should be less than or equal to

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!!!

hahahahahah I was cracking up for a minute or so when I had that thought

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;
}

sum=(sum*10)+rem; i dint understand this line can u pls help me out?

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;
}

Thanks man! You have cleared my confusion.