Palindrome count in a string - ruby

Question

So, I explored www.hackerearth.com today and was solving my first problem statement in ruby: http://www.hackerearth.com/problem/algorithm/palindrome-count-1/
Palindrome count Problem:
Given a string S, count the number of non empty sub strings that are palindromes.
A sub string is any continuous sequence of characters in the string.
A string is said to be palindrome, if the reverse of the string is same as itself.
Two sub strings are different if they occur at different positions in S
Input: Input contains only a single line that contains string S.
Output: Print a single number, the number of sub strings that are palindromes.
Constraints
1 <= |S| <= 50
S contains only lower case latin letters, that is characters a to z.
Sample Input (Plaintext Link): dskjkd
Sample Output (Plaintext Link): 7
Explanation -
The 7 sub strings are d, s, k, j, k, d, kjk.
Time limit 3 sec(s)
Memory limit 256 MB
Source limit 1024 KB
Here is what I did:
chars = gets.chomp.gsub(' ', '').split('')
counts = chars.count
(2..chars.count).each do |len|
chars.combination(len).each do |comb|
string = comb.inject(:<<)
counts += 1 if string.reverse == string
end
end
puts counts
However, this approach seems to be inefficient in terms of the time execution and memory usage. Is there any way to optimize this? Or have any other approach to this solution, algorithm is also welcome as solution! Thanks.
Edit
Since, all the answers are correct. I had to choose the one which is efficient. So, I ran benchmark and here is the result: https://gist.github.com/suryart/7577481
Based on the result you can see this answer is much faster. Thank you for the new approaches/ solution guys! :)

Answer 1

This approach -- in pseudo-code -- should work.
input: String s
// each single letter is palindrome itself
palindromsCount = length(s)
// let count all odd-length palindromes first (palindrome of length 1 already counted)
// we will be checking from the very middle of a sub-string, if it is symmetric
for(int i = 1; i < length(s)-1; i++)
for(int j = 1; ; j++)
if (i - j < 0 || i + j >= length(s) || s[i-j] != s[i+j])
break
else
palindromsCount += 1
// let count in similar way all even-length palindromes
for(int i = 0; i < length(s)-1; i++)
for(int j = 0; ; j++)
if (i - j < 0 || i + j + 1 >= length(s) || s[i-j] != s[i+j+1])
break
else
palindromsCount += 1
EDIT Of course both loops can be combined into a single one -- I did not want to do it for better readability.

Answer 2

using the algorithm to get all subsets of the string from What is the best way to split a string to get all the substrings by Ruby?
count = 0
(0..len-1).each do |i|
(i..len-1).each do |j|
temp = s[i..j]
count = count + 1 if temp == temp.reverse
end
end
puts "found #{count} palindromes"

Answer 3

Enumerable#each_cons is handy here:
str = "momanddadpaddledthekayak"
b = str.chars
(1..b.size).reduce(0) {|t,n| t + b.each_cons(n).reduce(0) \
{|r,e| w = e.join; w==w.reverse ? r + 1 : r}} # => 30
If we want to see the palendromes:
b = str.chars
pals = (1..b.size).each_with_object([]) {|n, a| b.each_cons(n).each \
{|e| w = e.join; a << w if w==w.reverse}}
p pals.size # => 30
p pals # => ["m", "o", "m", "a", "n", "d", "d", "a", "d", "p", "a",\
"d", "d", "l", "e", "d", "t", "h", "e", "k", "a", "y",
"a", "k", "dd", "dd", "mom", "dad", "aya", "kayak"]
Edit: #squiguy made the useful observation that we may not want to count duplicates. If that's the case, my first calculation above could not be used and the second would have to be changed as squiguy suggests (e.g., p a.uniq.size) or changed to build a hash rather than an array:
b = str.chars
pals = (1..b.size).each_with_object({}) {|n,h| b.each_cons(n).each \
{|e| w = e.join; h[w] = 0 if w==w.reverse}}.keys
p pals.size # => 17
p pals# => ["m", "o", "a", "n", "d", "p", "l", "e", "t",\
"h", "k", "y", "dd", "mom", "dad", "aya", "kayak"]
[Edit: replaced each with each_with_object. On rereading the question, it appears that dups are to be counted.]