if given a 15 digit number whats the best way to find the next palindrome?

Question

in c++ what will be the fastest logic to find next palindrome of a given 15 digit number? for example what will be the next palindrome of: 134567329807541 ?

Answer 1

• Split the number into three parts, head, mid, tail

  1345673 2 9807541

• Reverse head and compare it to tail 3765431

• If reverse(head) <= tail ( if they are equal the initial input is a palindrome, and you want the next )

  • If mid < 9, increment mid
  • Else increment head part and set mid := 0

• result := head mid reverse(head).

  1345673 3 reverse(1345673) => 134567333765431

Answer 2

I believe it's like this

1. Split the number into three parts 1345673 2 9807541
2. Flip the last one 1457089
3. If it's larger than the first part (it is in this case)
   • firstpart++
   • middlepart = 0
4. flip first part and replace last part.

Answer 3

I am not about to implement anything, but I imagine the logic would be:

1. Split the number at the middle of the string: X being the left part and Y being the right part.
2. Let X' = { X + 1 if reverse(X) < Y; X otherwise }
3. The result is then concat(X',reverse(X'));

If the length is uneven, you need to treat the middle digit separately. But that is quite trivial.

Answer 4

I think the following algo should also work .. It is easier to implement also

  i) Divide the given nos into three parts  HEAD MID TAIL 
  ii) Add 1 to number HEAD MID
          (in case of carry, follow basic addition rules)   
  iii) reverse the new HEAD(store it in HEAD_REV)
  iv) required ans is:-  'new HEAD' MID  HEAD_REV

Hoping that following example will help in better understanding of the algo

let nos be:- 23469 9 12367

       So HEAD -> 23469   MID -> 9   TAIL --> 12367

       step 2:-   23469 9 +1 = 23470 0 
              (now HEAD -> 23470 MID -> 0 HEAD_REV -> 07432 )

required Ans:-
23470 0 07432

Plz do infrom me if there exist any flaw in this procedure

Answer 5

Split the number into three parts head, mid and tail

if reverse(head)>tail 
result := head mid reverse(head)
else if reverse(head)= tail && mid<9 
    mid++
    result := head mid tail
else 
mid =0
head++
result := head mid reverse(head)