Online algorithm for checking palindrome in a stream

Given a stream of characters (characters are received one by one), write a function that prints 'Yes' if a character makes the complete string palindrome, else prints 'No'. 

Examples:

Input: str[] = "abcba"
Output: a Yes // "a" is palindrome
 b No // "ab" is not palindrome
 c No // "abc" is not palindrome
 b No // "abcb" is not palindrome
 a Yes // "abcba" is palindrome
Input: str[] = "aabaacaabaa"
Output: a Yes // "a" is palindrome
 a Yes // "aa" is palindrome
 b No // "aab" is not palindrome 
 a No // "aaba" is not palindrome 
 a Yes // "aabaa" is palindrome 
 c No // "aabaac" is not palindrome 
 a No // "aabaaca" is not palindrome 
 a No // "aabaacaa" is not palindrome 
 b No // "aabaacaab" is not palindrome 
 a No // "aabaacaaba" is not palindrome 
 a Yes // "aabaacaabaa" is palindrome 

Let input string be str[0..n-1]. 

A Simple Solution is to do following for every character str[i] in input string. Check if substring str[0...i] is palindrome, then print yes, else print no. 

A Better Solution is to use the idea of Rolling Hash used in Rabin Karp algorithm. The idea is to keep track of reverse of first half and second half (we also use first half and reverse of second half) for every index. Below is complete algorithm.

 1) The first character is always a palindrome, so print yes for 
 first character.
 2) Initialize reverse of first half as "a" and second half as "b". 
 Let the hash value of first half reverse be 'firstr' and that of 
 second half be 'second'.
 3) Iterate through string starting from second character, do following
 for every character str[i], i.e., i varies from 1 to n-1.
 a) If 'firstr' and 'second' are same, then character by character 
 check the substring ending with current character and print 
 "Yes" if palindrome.
 Note that if hash values match, then strings need not be same.
 For example, hash values of "ab" and "ba" are same, but strings
 are different. That is why we check complete string after hash.
 b) Update 'firstr' and 'second' for next iteration. 
 If 'i' is even, then add next character to the beginning of 
 'firstr' and end of second half and update 
 hash values.
 If 'i' is odd, then keep 'firstr' as it is, remove leading 
 character from second and append a next 
 character at end.

Let us see all steps for example 

string "abcba" Initial values of 'firstr' and 'second'      

firstr' = hash("a"), 'second' = hash("b") Start from second character, i.e., i = 1      

• Compare 'firstr' and 'second', they don't match, so print no.     
• Calculate hash values for next iteration, i.e., i = 2      

Since i is odd, 'firstr' is not changed and 'second' becomes hash("c") i = 2      

• Compare 'firstr' and 'second', they don't match, so print no.     
• Calculate hash values for next iteration, i.e., i = 3      

Since i is even, 'firstr' becomes hash("ba") and 'second' becomes hash("cb") i = 3      

• Compare 'first' and 'second', they don't match, so print no.      
• Calculate hash values for next iteration, i.e., i = 4      

Since i is odd, 'firstr' is not changed and 'second' becomes hash("ba") i = 4     

• 'firstr' and 'second' match, compare the whole strings, they match, so print yes      
• We don't need to calculate next hash values as this is last index The idea of using rolling hashes is, next hash value can be calculated from previous in O(1) time by just doing some constant number of arithmetic operations. Below are the implementations of above approach. 

a Yes
a Yes
b No
a No
a Yes
c No
a No
a No
b No
a No
a Yes

The worst case time complexity of the above solution remains O(n*n), but in general, it works much better than simple approach as we avoid complete substring comparison most of the time by first comparing hash values. 

The worst case occurs for input strings with all same characters like "aaaaaa". 

Space Complexity: O(1)
The algorithm uses a fixed number of variables (which is independent of the size of the input string). Hence, the space complexity of the algorithm is O(1).