What if the answer after inverting the digits is 0

#1

like in case of 99 then do we need to print 1 or the number (99 ) only

#2

the most probable error in this question is a corner case where all the digits are 9, or first digit is 9.(solution: first 9 should be untouched)
check your code for following examples:(left side is input, right side is expected output)

  1. 999 ----> 900
  2. 987 -----> 912
  3. 9 -------> 9
  4. 99 -------> 90

thanks

#3

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#4

import java.util.; public class Main { public static void main(String args[]) { // Your Code Here Scanner sc= new Scanner(System.in); int n= sc.nextInt(); int sum=0,multiplier=1; while(n>0){ int ld = n%10; n/=10; if(ld>4){ ld=9-ld; } else if(n==0 && ld==9) ld=9; sum+=ldmultiplier; multiplier*=10; } System.out.println(sum); } }

#5

Hi @Vishu_1801,
Your code is giving wrong output … for a test case say 9999 the output should be 1000 but your code is giving 1111… Please correct this and further errors.

#6

And please save your code on ide.codingblocks.com as this code is very hard to understand … There is a ldmultiplier in the code what is that … check the code you posted

#7

switch the if and else if part… if(n==0 && ld==9) should come before that ld>4 part.

thanks

1 Like
#8

Thanks for that but still test cases 3 and 4 are not covered they are showing a run error

#9
#10

the input can be large, take long long instead of int data type.

thanks

#11

Still the test cases 3 and 4 are not covered

#12

Did that as well but still unable to cover test case 3 and 4

#13

updated code: https://ide.codingblocks.com/s/212306

thanks