Approach #1: Expand Around Center [Accepted]
Intuition
Let N
be the length of the string. The middle of the palindrome could be in one of 2N  1
positions: either at letter or between two letters.
For each center, let's count all the palindromes that have this center. Notice that if [a, b]
is a palindromic interval (meaning S[a], S[a+1], ..., S[b]
is a palindrome), then [a+1, b1]
is one too.
Algorithm
For each possible palindrome center, let's expand our candidate palindrome on the interval [left, right]
as long as we can. The condition for expanding is left >= 0 and right < N and S[left] == S[right]
. That means we want to count a new palindrome S[left], S[left+1], ..., S[right]
.
Complexity Analysis

Time Complexity: where is the length of
S
. Each expansion might do work. 
Space Complexity: .
Approach #2: Manacher's Algorithm [Accepted]
Intuition
Manacher's algorithm is a textbook algorithm that finds in linear time, the maximum size palindrome for any possible palindrome center. If we had such an algorithm, finding the answer is straightforward.
What follows is a discussion of why this algorithm works.
Algorithm
Our loop invariants will be that center, right
is our knowledge of the palindrome with the largest rightmost boundary with center < i
, centered at center
with rightboundary right
. Also, i > center
, and we've already computed all Z[j]
's for j < i
.
When i < right
, we reflect i
about center
to be at some coordinate j = 2 * center  i
. Then, limited to the interval with radius right  i
and center i
, the situation for Z[i]
is the same as for Z[j]
.
For example, if at some time center = 7, right = 13, i = 10
, then for a string like A = '@#A#B#A#A#B#A#＄'
, the center
is at the '#'
between the two middle 'A'
's, the right boundary is at the last '#'
, i
is at the last 'B'
, and j
is at the first 'B'
.
Notice that limited to the interval [center  (right  center), right]
(the interval with center center
and rightboundary right
), the situation for i
and j
is a reflection of something we have already computed. Since we already know Z[j] = 3
, we can quickly find Z[i] = min(right  i, Z[j]) = 3
.
Now, why is this algorithm linear? The while loop only checks the condition more than once when Z[i] = right  i
. In that case, for each time Z[i] += 1
, it increments right
, and right
can only be incremented up to 2*N+2
times.
Finally, we sum up (v+1) / 2
for each v in Z
. Say the longest palindrome with some given center C has radius R. Then, the substring with center C and radius R1, R2, R3, ..., 0 are also palindromes. Example: abcdedcba
is a palindrome with center e
, radius 4: but e
, ded
, cdedc
, bcdedcb
, and abcdedcba
are all palindromes.
We are dividing by 2 because we were using halflengths instead of lengths. For example we actually had the palindrome a#b#c#d#e#d#c#b#a
, so our length is twice as big.
Complexity Analysis

Time Complexity: where is the length of
S
. As discussed above, the complexity is linear. 
Space Complexity: , the size of
A
andZ
.
Analysis written by: @awice.