Need logic for number pattern

Question

12345
23451
24513
24135
24153

Answer 1

Basically, you'll just need some sort of list/array/etc... and you'll create a loop for it. For each iteration of the loop (i), we'll shift the number in that position to the end. This will prevent us from touching the previous positions on each loop through, and it'll stop after we've dealt with the entire list/array.

JAVA EXAMPLE:

https://code.sololearn.com/c13eIMGwuAsw/#java

import java.util.*;

public class Program
{
public static void main(String[] args) {
List numberList = new ArrayList(Arrays.asList(1,2,3,4,5));

for(int i = 0; i < numberList.size(); ++i){
numberList.add(numberList.remove(i));
System.out.print(numberList.get(i));
}

}
}

Answer 2

@Daniel,

Thanks for pointing that out, I possibly misunderstood the question, it's been a while, and the TS didn't say anything, so I didn't notice, so sorry...

Lemme take that bad egg out...

Big Thanks :)

Answer 3

Okay, after thinking on it for a moment, this is my take on it.

Take your first state:
>> 12345

We're going to take the first number, push it to the back, and the new number at the beginning will be "locked in" and no longer able to change.
>> 23451

We'll repeat the same process until all numbers are locked in. 2 is locked in, so go to the second number which is 3. Move it to the end and now the 4 is locked in as the second number.
>> 24513

As before, move the 5 to the end, and now 1 is locked in as the third number.
>> 24135

As before, move the 3 to the end, and 5 is locked in as the 4th number. No room to move now, so 3 is locked in as the last number.
>> 24153

Answer 4

Ipang your code not work like example, check it ;-)

Answer 5

I'm late to the game...?😅
Here's mine in JS [ES6],
works with more characters too:
https://code.sololearn.com/WMWf1Q2UWWo8/?ref=app

Answer 6

i know it theoretically. but i need code.. in any programming language.

Answer 7

https://code.sololearn.com/cXbASVlw1J67/?ref=app

This is the final code from Ipang code ;-)