consider that
0 -- is the first
1 -- is the second
2 -- is the third
.....
9 -- is the 10th
11 -- is the 11th
what is an efficient algorithm to find the nth palindromic number?
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Merna
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@SingerOfTheFall I could find an algo to find the nth palindromic number with k digits, for example I could find that the 113th palindromic number with 8 digits is 11122111. – Merna Feb 22 '13 at 08:30
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2Use that algorithm and add an iteration over the number of digits. – Oswald Feb 22 '13 at 08:31
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@Oswald, agreed. But the OP should also clarify if he needs just an algorithm, or a _quick_ algorithm. – SingerOfTheFall Feb 22 '13 at 08:32
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@Oswald excuse me, can u give an example ? – Merna Feb 22 '13 at 08:32
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ofcourse if a qucik algorithm exists , that'll be better :) – Merna Feb 22 '13 at 08:34
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1@SingerOfTheFall OP did say "efficient algorithm". – Bernhard Barker Feb 22 '13 at 08:45
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[Here](http://www.ardendertat.com/2011/12/01/programming-interview-questions-19-find-next-palindrome-number/) are some thoughts on getting the next palindromic number for any number. – hsan Feb 22 '13 at 08:59
2 Answers
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I'm assuming that 0110 is not a palindrome, as it is 110.
I could spend a lot of words on describing, but this table should be enough:
#Digits #Pal. Notes
0 1 "0" only
1 9 x with x = 1..9
2 9 xx with x = 1..9
3 90 xyx with xy = 10..99 (in other words: x = 1..9, y = 0..9)
4 90 xyyx with xy = 10..99
5 900 xyzyx with xyz = 100..999
6 900 and so on...
Sjoerd
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The (nonzero) palindromes with even number of digits start at p(11) = 11, p(110) = 1001, p(1100) = 100'001,.... They are constructed by taking the index n - 10^L, where L=floor(log10(n)), and append the reversal of this number: p(1101) = 101|101, p(1102) = 102|201, ..., p(1999) = 999|999, etc. This case must be considered for indices n >= 1.1*10^L but n < 2*10^L.
When n >= 2*10^L, we get the palindromes with odd number of digits, which start with p(2) = 1, p(20) = 101, p(200) = 10001 etc., and can be constructed the same way, using again n - 10^L with L=floor(log10(n)), and appending the reversal of that number, now without its last digit: p(21) = 11|1, p(22) = 12|1, ..., p(99) = 89|8, ....
When n < 1.1*10^L, subtract 1 from L to be in the correct setting with n >= 2*10^L for the case of an odd number of digits.
This yields the simple algorithm:
p(n) = { L = logint(n,10);
P = 10^(L - [1 < n < 1.1*10^L]); /* avoid exponent -1 for n=1 */
n -= P;
RETURN( n * 10^L + reverse( n \ 10^[n >= P] ))
}
where [...] is 1 if ... is true, 0 else, and \ is integer division.
(The expression n \ 10^[...] is equivalent to: if ... then n\10 else n.)
(I added the condition n > 1 in the exponent to avoid P = 10^(-1) for n=0. If you use integer types, you don't need this. Another choice it to put max(...,0) as exponent in P, or use if n=1 then return(0) right at the start. Also notice that you don't need L after assigning P, so you could use the same variable for both.)
