Can someone give a test case where my code fails or rather guide me to the correct approach ?
After taking care of the last digit cases, I’m using a greedy approach to find the first number that on removal will lead to (sum of digits) %3=0 and has the next number greater than itself. If such a number doesn’t exist, I remove the lowest rightmost number that will lead to that result.
havent taken care of the case when last%2 == 0 and l2%2 == 0.
why u define soo much macro’s?
That’s when the greedy approach is used.
consider the case 3 960. the answer should be 96
I applied the same technique, passed only subtask2.
Didn’t understand where the approach is wrong.
[1]
[1]: https://www.codechef.com/viewsolution/15992745In your code on line 73:
x < a[i+1]-‘0’ this condition is wrong because it doesn’t guarantee that the resulting number after deleting a digit will be greater than the one you found earlier.
Consider this test case:
Input
2
668
663
Your Output
-1
66
Correct Output
66
66
Can anyone tell me due to which testcase it was showing wrong answer?
Input:
1
6666666666
Correct Output:
666666666
Your Output:
If you want another approach, Have a look here.
If you wannna find bug in your approach, explain it in a bit detail. 
I’m noob in java so I can’t help you with why your code gives no output but I can say that you’ve read the question wrong.
I see you’ve taken input as integer which is wrong because the input constraint says that input number can be of 10^5 digits long i.e input number can be 10^(10^5) which your long int input can’t handle.
Try implementing using string.
Hope it helps.
Can anyone give me a test case where my code fails? It’s working well for all the test cases I checked.
PLEASE HELP ME!!!
I’m storing the digits in an array from one’s place. Then I am iterating through all the digits and finding the new number by removing that particular digit. Now, I’m checking whether it is divisible by 6 or not and if it is then I check whether it is largest number possible which I initially set to -1. At last I’m printing the new number with k-1 digits where k is the total number of digits in the initial number. Leading zeros will be printed since I used %0*d and passed k-1 and maxnum to it.
LINK:- My Solution
It would time out.
The solution time complexity is O(N^2) which won’t pass.
Further, the number can have 10^5 digits, thus, can’t be stored in int or even long long.
But it’s giving wrong answer
n can be 100000 so it can only have a maximum of 6 digits right?
n can have 100000 digits .
So you can’t store it in int
But in constraints it is giving n>=2
You’re welcome bro.