## Longest Palindromic Substring Part II

November 20, 2011 in string

Given a string S, find the longest palindromic substring in S.

**Note:**

This is Part II of the article: Longest Palindromic Substring. Here, we describe an algorithm (Manacher’s algorithm) which finds the longest palindromic substring in linear time. Please read Part I for more background information.

In my previous post we discussed a total of four different methods, among them there’s a pretty simple algorithm with O(N^{2}) run time and constant space complexity. Here, we discuss an algorithm that runs in O(N) time and O(N) space, also known as Manacher’s algorithm.

**Hint:**

Think how you would improve over the simpler O(N^{2}) approach. Consider the worst case scenarios. The worst case scenarios are the inputs with multiple palindromes overlapping each other. For example, the inputs: “aaaaaaaaa” and “cabcbabcbabcba”. In fact, we could take advantage of the palindrome’s symmetric property and avoid some of the unnecessary computations.

**An O(N) Solution (Manacher’s Algorithm):**

First, we transform the input string, S, to another string T by inserting a special character ‘#’ in between letters. The reason for doing so will be immediately clear to you soon.

For example: S = “abaaba”, T = “#a#b#a#a#b#a#”.

To find the longest palindromic substring, we need to expand around each T_{i} such that T_{i-d} … T_{i+d} forms a palindrome. You should immediately see that *d* is the length of the palindrome itself centered at T_{i}.

We store intermediate result in an array P, where P[ i ] equals to the length of the palindrome centers at T_{i}. The longest palindromic substring would then be the maximum element in P.

Using the above example, we populate P as below (from left to right):

T = # a # b # a # a # b # a # P = 0 1 0 3 0 1 6 1 0 3 0 1 0

Looking at P, we immediately see that the longest palindrome is “abaaba”, as indicated by P_{6} = 6.

Did you notice by inserting special characters (#) in between letters, both palindromes of odd and even lengths are handled graciously? (Please note: This is to demonstrate the idea more easily and is not necessarily needed to code the algorithm.)

Now, imagine that you draw an imaginary vertical line at the center of the palindrome “abaaba”. Did you notice the numbers in P are symmetric around this center? That’s not only it, try another palindrome “aba”, the numbers also reflect similar symmetric property. Is this a coincidence? The answer is yes and no. This is only true subjected to a condition, but anyway, we have great progress, since we can eliminate recomputing part of P[ i ]‘s.

Let us move on to a slightly more sophisticated example with more some overlapping palindromes, where S = “babcbabcbaccba”.

Above image shows T transformed from S = “babcbabcbaccba”. Assumed that you reached a state where table P is partially completed. The solid vertical line indicates the center (C) of the palindrome “abcbabcba”. The two dotted vertical line indicate its left (L) and right (R) edges respectively. You are at index i and its mirrored index around C is i’. How would you calculate P[ i ] efficiently?

Assume that we have arrived at index i = 13, and we need to calculate P[ 13 ] (indicated by the question mark ?). We first look at its mirrored index i’ around the palindrome’s center C, which is index i’ = 9.

The two green solid lines above indicate the covered region by the two palindromes centered at i and i’. We look at the mirrored index of i around C, which is index i’. P[ i' ] = P[ 9 ] = 1. It is clear that P[ i ] must also be 1, due to the symmetric property of a palindrome around its center.

As you can see above, it is very obvious that P[ i ] = P[ i' ] = 1, which must be true due to the symmetric property around a palindrome’s center. In fact, all three elements after C follow the symmetric property (that is, P[ 12 ] = P[ 10 ] = 0, P[ 13 ] = P[ 9 ] = 1, P[ 14 ] = P[ 8 ] = 0).

Now we are at index i = 15. What’s the value of _{15}, it forms the palindrome “a#b#c#b#a”, which is actually shorter than what is indicated by its symmetric counterpart. Why?

Colored lines are overlaid around the center at index i and i’. Solid green lines show the region that must match for both sides due to symmetric property around C. Solid red lines show the region that might not match for both sides. Dotted green lines show the region that crosses over the center.

It is clear that the two substrings in the region indicated by the two solid green lines must match exactly. Areas across the center (indicated by dotted green lines) must also be symmetric. Notice carefully that P[ i ' ] is 7 and it expands all the way across the left edge (L) of the palindrome (indicated by the solid red lines), which does not fall under the symmetric property of the palindrome anymore. All we know is

Let’s summarize the key part of this algorithm as below:

**if**P[ i' ] ≤ R – i,

**then**P[ i ] ← P[ i' ]

**else**P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].

See how elegant it is? If you are able to grasp the above summary fully, you already obtained the essence of this algorithm, which is also the hardest part.

The final part is to determine when should we move the position of C together with R to the right, which is easy:

In each step, there are two possibilities. If P[ i ] ≤ R – i, we set P[ i ] to P[ i' ] which takes exactly one step. Otherwise we attempt to change the palindrome’s center to i by expanding it starting at the right edge, R. Extending R (the inner while loop) takes at most a total of N steps, and positioning and testing each centers take a total of N steps too. Therefore, this algorithm guarantees to finish in at most 2*N steps, giving a linear time solution.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 | // Transform S into T. // For example, S = "abba", T = "^#a#b#b#a#$". // ^ and $ signs are sentinels appended to each end to avoid bounds checking string preProcess(string s) { int n = s.length(); if (n == 0) return "^$"; string ret = "^"; for (int i = 0; i < n; i++) ret += "#" + s.substr(i, 1); ret += "#$"; return ret; } string longestPalindrome(string s) { string T = preProcess(s); int n = T.length(); int *P = new int[n]; int C = 0, R = 0; for (int i = 1; i < n-1; i++) { int i_mirror = 2*C-i; // equals to i' = C - (i-C) P[i] = (R > i) ? min(R-i, P[i_mirror]) : 0; // Attempt to expand palindrome centered at i while (T[i + 1 + P[i]] == T[i - 1 - P[i]]) P[i]++; // If palindrome centered at i expand past R, // adjust center based on expanded palindrome. if (i + P[i] > R) { C = i; R = i + P[i]; } } // Find the maximum element in P. int maxLen = 0; int centerIndex = 0; for (int i = 1; i < n-1; i++) { if (P[i] > maxLen) { maxLen = P[i]; centerIndex = i; } } delete[] P; return s.substr((centerIndex - 1 - maxLen)/2, maxLen); } |

**Note:**

This algorithm is definitely non-trivial and you won’t be expected to come up with such algorithm during an interview setting. However, I do hope that you enjoy reading this article and hopefully it helps you in understanding this interesting algorithm. You deserve a pat if you have gone this far!

**Further Thoughts:**

- In fact, there exists a sixth solution to this problem — Using suffix trees. However, it is not as efficient as this one (run time O(N log N) and more overhead for building suffix trees) and is more complicated to implement. If you are interested, read Wikipedia’s article about Longest Palindromic Substring.
- What if you are required to find the longest palindromic subsequence? (Do you know the difference between substring and subsequence?)

**Useful Links:**

» Manacher’s Algorithm O(N) 时间求字符串的最长回文子串 (Best explanation if you can read Chinese)

» A simple linear time algorithm for finding longest palindrome sub-string

» Finding Palindromes

» Finding the Longest Palindromic Substring in Linear Time

» Wikipedia: Longest Palindromic Substring

wayne said on November 21, 2011

i think my solution is simpler than this one, the key point of my solution is:

The center point of palindromic substring is always follow this pattern, either is “…..XyX…..” or “….XX….”.

so you can scan once and then find those center point of palindromic substring and then expand it on each center points to find the one with maxium length.

i ve already posted my java solution in the comments of Longest Palindromic Substring Part I

-121337c0d3r said on November 22, 2011

Yes your solution is simpler but runs in O(N^2) worst case. It is already discussed in my previous post.

0wayne said on November 22, 2011

i agreed, it is O(N^2) worst case, thanks.

-41337c0d3r said on November 22, 2011

No problem.

Basically this algorithm is an improvement over your method. It is using the symmetric property of a palindrome to eliminate some of the recomputations of palindrome’s length, and amazingly improve it to a linear time solution.

0Andres said on September 24, 2012

Excellent post, I learnt a lot

Thanks!

-1GuojiaAgain said on January 20, 2014

Your implementation is simpler but cost more time.

0Sreekar said on November 21, 2011

Looks like this is O(N^2) algorithm as there is a while loop in for loop. Could you please clarify?

-41337c0d3r said on November 22, 2011

Even with the extra while loop inside, it is guaranteed in the worst case the algorithm completes in 2*n steps.

Think of how the i and right edge (R) relates. In the loop each time, you look if this index is a candidate to re-position the palindrome’s center. If it is, you increment the existing R one at a time. See? R could only be incremented at most N steps. Once you incremented a total of N steps, it couldn’t be incremented any more. It’s not like you will increment R all the time in the while loop. This is called amortized O(1).

+5Vikas said on January 5, 2013

How is it amortised O(1) ? I am confused from the definition from amortise.

Amortised to me means , for a worst case of runs of some operations its amortised cost over all of them

-2yy said on March 20, 2014

Great, I see why it’s amortized O(1) now. Thanks!

-2Jason said on September 16, 2014

It seems a O(N^2) alg. For example, “abcdcba”, go through the string needs N step, and the while loop needs N/2 step when meets character ‘d’, so O（N*N/2）=O(N^2)?

+1Sreekar said on November 22, 2011

Got it. Thanks for the clarification. Great solution

+21337c0d3r said on November 22, 2011

Thanks! Hope you understands it. Let me know if you have any more questions!

-1coderrush said on February 29, 2012

I am wondering if the algorithm is O(NlgN) or O(N)?

+1flo said on November 22, 2011

Great write-up. Thanks for the article

+11337c0d3r said on November 22, 2011

Thanks!

My goal of writing this article is to provide an intuitive way to understand the algorithm. I hope you really appreciate the beauty of this algorithm.

+3J said on May 5, 2012

I would if I could understand it.

0vaibhav said on November 24, 2011

while explaining how to fill P[i] you mentioned

”

if P[ i' ] ≤ R – i,

then P[ i ] ← P[ i' ]

else P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].

”

is the else statement right?? shouldnt be “else P[ i' ] ≥ P[ i ]“

+6ironmanz said on May 17, 2013

I agree with you.

0bleu said on November 24, 2011

It should rather be:

else P[ i ] ≥ (R-i) (Which we have to expand past the right edge (R) to find P[ i ])

.. Also note how coherent the reasoning in the bracket sounds now.

Explanation :

When P[i'] > R-i then all we know, by symmetry about C, is :

P[i'] > R-i .. by obviousness

and

P[i] ≥ R-i .. by the meaning of R

From this we clearly cannot conclude upon max(P[i'], P[i])

+6J said on May 5, 2012

Don’t see the coherency.

0ClaireQu said on September 17, 2014

I agree with you

0Fei said on November 25, 2011

Using suffix tree can do this in O(n). And building suffix tree can be also done in O(n): http://blog.csdn.net/g9yuayon/article/details/2574781

But this algorithm is pretty cool too!

+21337c0d3r said on November 29, 2011

Thanks Fei! I will look into that.

0LB said on November 29, 2011

Clear explanation. It could hardly be any better

Thumbs up for elucidating this magic O(n) solution in such intuitive manner. You got talent to clearly expressing an algorithm, which I even missed in books like Cormen’s Algorithm!

+11337c0d3r said on November 29, 2011

Thanks! Good to know I’ve done my job — to introduce tricky but interesting algorithms in an intuitive manner.

+1Bahlul Haider said on December 1, 2011

Really beautiful algorithm.

0Googmeister said on December 6, 2011

Wonderful writeup with great illustrations! I think there is one minor bug in your code: if s itself is a palindrome, then the following line accesses the array out of bounds.

01337c0d3r said on December 6, 2011

ahh… you are right! Thanks for your sharp observation.

I thought that I feel something is not right when I decided to add ‘$’ both to the begin and the end of the input string. (It should be adding two different sentinels ‘^’ and ‘$’ to the begin and the end of the string. This should avoid bounds checking and the out of bounds problem)

0online chesstle said on February 15, 2014

s.substr((centerIndex – 1 – maxLen)/2, maxLen); is there a bug here as the rest of the code is treating P[i] as the length of the palindrome on either side of the center and not as the total length of the palindrome.

0online chesstle said on February 15, 2014

a fault in the code here? s.substr((centerIndex – 1 – maxLen)/2, maxLen);

the core algortihm is handling P[i] as the length of the palindrome on either side of the center and not as the total length of the palindrome.

0online chesstle said on February 15, 2014

nevermind, accounting for the hashes.

0Karthick said on December 12, 2011

Thanks for such a lucid explanation. I had already visited all the references that you had suggested at the end. I was not able to understand the essence of it until I read yours.

Well, I have one question.

Is it possible to run the algorithm without using the ‘#’,'^’,'$’ symbols?

0Karthick said on December 12, 2011

As an after-thought, I have one doubt.

Why are we using the line

P[i] = (R > i) ? min(R-i, P[i_mirror]) : 0;

Can you please clarify this?

0MSJ said on January 22, 2012

there are another two solutions

1) suffix array version preprocess requires N*logN query is O(N) http://www.mashengjun.info/?p=901

2) another O(N) solution http://www.mashengjun.info/?p=464 using up&down pointer

01337c0d3r said on January 22, 2012

Did you try running your code through Online Judge? http://www.leetcode.com/onlinejudge

It did not pass all test cases.

0caopeng said on March 2, 2012

The seconde is not O(N) it’s O(N^2) i think…

0Caopeng said on March 1, 2012

The first i_mirror is -1 which is less than 0 so there may be run time error?

0Anonymous said on March 19, 2012

It’s really O(N)?

0Vyas said on April 9, 2012

A Very nice explanation!!!!:)

One thing I could not understand… “In this case, since P[ 21 ] ≠ P[ 1 ], we conclude that P[ i ] = 5.”.. In this statement, from what I have understood, I think it should be P[21] ≠ P[8]. Please correct me if I’m wrong…..

0Kareem said on May 1, 2012

The conclusion of the algorithm above states that

if P[ i' ] ≤ R – i,

then P[ i ] ← P[ i' ]

else P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].

The first check should be P[ i' ] < R – i, as when they are equal the proper value of p[i] can not be fully determined with P[ i' ] only but needs to expand.

for example: string #b#b#a#b#a#b#a# with i = 9, c = 7, R = 12

+1Li said on August 2, 2012

Agree with you.

0J said on May 5, 2012

It is clear that the two substrings in the region indicated by the two solid green lines must match exactly. Areas across the center (indicated by dotted green lines) must also be symmetric.

i is in green area (index 15). So I should have the same value as i[7]. but it doesn’t.

So what is going on?

0J said on May 5, 2012

The first two lines are quotes from the written explanation.

0Chev said on May 27, 2012

Is there a problem in the following code? i_mirror will be -1 and P[i_mirror] will be out of boundary when C=0 and i =1, or do I miss something? Thanks.

…

int C = 0, R = 0;

for (int i = 1; i i) ? min(R-i, P[i_mirror]) : 0;

…

}

0Chev said on May 27, 2012

I see! It is my mistake. Thanks!

0Chev said on May 27, 2012

Here is the piece of code where I am confused:

int C = 0, R = 0;

for (int i = 1; i i) ? min(R-i, P[i_mirror]) : 0;

0Chev said on May 27, 2012

Got it! I am clear now. Thanks!

0Chuanyou Li said on July 6, 2012

Should P[i] R – i ??

0Noone said on September 5, 2012

Have to disagree with the others. You have actually managed to complicate a simple algorithm!

The key idea is quite simple actually.

0Subramanian Ganapathy said on October 14, 2012

Recurrence:

L[i] = max{L[i-1] + 1, #Arr[i-1 - L[i-1]…i] is univalue.

L[i-1]+2 #Arr[i] == Arr[i-1-L[i-1]]}

L[i]=1 otherwise.

Am i missing something here? this recurrrence solves it and it is a lot simpler.

0Subramanian Ganapathy said on October 14, 2012

My bad my recurrence messes up the overlapping palindromes case, awesome solution and nice explanation. you deserve a pat on your back thank you

0Karthick said on November 1, 2012

The following is the implementation of the Manacher’s algorithm without pre-processing the input string. It is a bit clumsy – sorry for that. I tested it using the online judge here and it seems to be working fine.

Language : java

-2Karthick said on November 1, 2012

Sorry,there seems to be some problem – some part of the code seems to be omitted when I post my code using the

tag. So, I am posting it without the code tag.

public String longestPalindrome(String s) {

if(s==null)

return “”;

int len=s.length();

int[] p=new int[2*len-1];

p[0]=1;

int R=0,C=0;

int curLen,l,r,start;

for(int i=1;ii) ? Math.min(curLen,p[iMirror]) : ( i%2==0 ? 1 : 0) );

if(i%2==0){

l=(i/2-p[i]/2-1);

r=(i/2+p[i]/2+1);

}

else{

l=(i/2-p[i]/2);

r=(i/2+p[i]/2+1);

}

while(l>=0 && rR){

C=i;

R=r-1;

}

}

int maxIndex=getMaxIndex(p);

if(maxIndex%2==0)

start=(maxIndex/2-p[maxIndex]/2);

else

start=(maxIndex/2-p[maxIndex]/2 +1);

return s.substring(start,start+p[maxIndex]);

}

int getMaxIndex(int p[]){

int len=p.length;

int maxIndex=0;

for(int i=1;ip[maxIndex])

maxIndex=i;

}

return maxIndex;

}

0abhay said on January 5, 2013

nice explaination thanks for article..

0Naveen Kumar said on January 16, 2013

I am stuck at summarized part of this algo.

if P[ i' ] ≤ R – i,

then P[ i ] ← P[ i' ]

else P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].)

how could we say which one be large for else part.?

even in example for i=15,p[i]=5,i’=7 p[i']=7;

i m confused here. plz help me.

Thanks..

0Mony Sim said on January 20, 2013

try simple solution.

bool is palindrome(string s)

{ int len=s.length(), a=0, b=len-1;

while(a<b)

{

if(s[a]!=s[b])

return false;

}

return true;

}

-2Mony Sim said on January 20, 2013

try simple solution.

bool is palindrome(string s)

{ int len=s.length(), a=0, b=len-1;

while(a<b)

{

if(s[++a]!=s[--b])

return false;

}

return true;

}

0rajat rastogi said on February 2, 2013

Dude your solution is O(n^2) take a case of a string aaaaaa, palindrome length is 6 and solution for this string confirms O(n^2)

0William Gozali said on April 14, 2013

I don’t think so.

Take a look at the explanation, it is guaranteed that the operation needed will not exceed O(N). Maybe you should try to simulate the algorithm with that input

0Žilvinas said on February 4, 2013

Thanks so much, exelent tutorial!

0Nipun Poddar said on February 6, 2013

really nice one..helpedme to learn a lot..

0Kuldeep Yadav said on March 15, 2013

GOOD WORK

0Ayaskant Swain said on April 25, 2013

Wonderful post. Thanks for posting these kind of problems and their solutions which will help in interviews. I have a question in this part-II solution.

I think the complexity will be still O(n * n), since you are traversing the string twice actually. One for modifying the original input string to insert characters ^,#,$ and then again you will do another traversal from the beginning to end of the string to search for actual palindromes in the string.

0lagrange said on June 2, 2013

我觉得那个中文的blog, p[i]表示向左/右延展的长度, 比p[i]表示整个substr的长度要更容易理解一些

+1zhuyao said on September 2, 2014

aglee!

0shivali said on June 17, 2013

can you please xplain how you calculated the complexity of the above algorithm

0shivali said on June 18, 2013

//longest pallindrome in a string(c++)

#include

#include

#define lsfor(i,a,b) for(i=a;i<b;i++)

using namespace std;

char str[100];

int n,curr_len=0,max_len=1,l,r,t;

int main()

{

int i;

cout<<"enter no.of test cases:"<>t;

while(t–)

{

cout<<"please enter your string"<>str;

n=strlen(str);

if(n==1)

{cout<<"1"<<endl;

return(0);}

if(n==2)

{cout<<"2"<=0&&r<=n-1)

{

if(str[l]==str[r])

{

curr_len+=2;

if(max_len<curr_len)

max_len=curr_len;

l–,r++;

}

else

break;

}

}

if(curr_len==1)

cout<<"sorry no palindrome"<<endl;

else

cout<<max_len<<endl;

}

return(0);

}

0shivali said on June 18, 2013

//edited://longest pa

#include

#include

#define lsfor(i,a,b) for(i=a;i<b;i++)

using namespace std;

char str[100];

int n,curr_len=0,max_len=1,l,r,t;

int main()

{

int i;

cout<<"enter no.of test cases:"<>t;

while(t–)

{

cout<<"please enter your string"<>str;

n=strlen(str);

if(n==1)

{cout<<"1"<<endl;

return(0);}

if(n==2)

{cout<<"2"<=0&&r<=n-1)

{

if(str[l]==str[r])

{

curr_len+=2;

if(max_len<curr_len)

max_len=curr_len;

l–,r++;

}

else

break;

}

}

if(max_len==1)

cout<<"sorry no palindrome"<<endl;

else

cout<<max_len<<endl;

}

return(0);

}

0jackyhou said on July 3, 2013

int find_long_palindrome_line(char * str,char *substr)

{

if(str==NULL)

return -1;

int len=strlen(str);

if(len==0)

return 1;

int i=0;

int j=len-1;

int end = len-1;

int curindex=0;

int tmp_len=len*2+1;

char * tmp_allstr=(char *)malloc(sizeof(char)*(tmp_len));

int *i_arr=(int *)malloc(sizeof(int)*(tmp_len));

int index4_tmp_allstr=0;

for(int i=0;i<len;i++)

{

tmp_allstr[index4_tmp_allstr++]='#';

tmp_allstr[index4_tmp_allstr++]=str[i];

}

tmp_allstr[index4_tmp_allstr]='#';

memset(i_arr,0,sizeof(int)*(tmp_len));

for(int i=2;i0 && right <(tmp_len))

{

if(tmp_allstr[left]==tmp_allstr[right])

{

i_arr[i]++;

}

else

{

break;

}

}

else

{

break;

}

}

}

int findI=0;

int findMaxLen=0;

for(int i=2;ifindMaxLen)

{

findMaxLen=i_arr[i];

findI=i;

}

}

int realSubLen=0;

if(findMaxLen>0)

{

for(int i=findI-(findMaxLen)+1;i<=findI+(findMaxLen)-1;i=i+2)

{

substr[realSubLen++]=tmp_allstr[i];

}

substr[realSubLen]='';

}

free (tmp_allstr);

tmp_allstr=NULL;

free (i_arr);//=(int *)malloc(sizeof(int)*(tmp_len));

i_arr=NULL;

return 1;

}

0JS said on July 10, 2013

Won’t the preprocess() take quadratic time? substr() is linear is time.

+1Saber said on July 23, 2013

The algorithm you write is wrong. I believe it is because u type it faster than what u think:)

if P[ i' ] ≤ R – i,

then P[ i ] ← P[ i' ]

else P[ i ] ≥ P[ i' ]. (Which we have to expand past the right edge (R) to find P[ i ].

should be changed to:

if P[ i' ] < R – i,

then P[ i ] ← P[ i' ]

else P[ i ] ≥ R – i. (Which we have to expand past the right edge (R) to find P[ i ].

0shuaiyangyang said on August 20, 2013

if p[i] > R -i then p[i] should equal to R-i.

only if p[i] = R-i then you have to expand

-1shuaiyangyang said on August 20, 2013

Hey 1337, The relation should be:

if P[ i' ] < R – i,

then P[ i ] ← P[ i' ]

else if P[ i ] = R – i. (Which we have to expand past the right edge (R) to find P[ i ].

else p[i] = R – i

-1Fentoyal said on September 1, 2013

This problem could be solved in O(n) time and O(1) space.

The test case attached here is what leetcode claims “wrong answer”, but I run it on my computer and it is exactly the same as the “expected answer”. I do not know why tho.

#include

#include

using namespace std;

class Solution {

int longestPalindromSubstrHelper(const string & str, bool is_even, int &cur_max_pivot, int & cur_max_radius)

{

int cur_radius = 0, cur_pivot = 0;

for (size_t i = 0; i < str.size(); ++i)

{

cur_radius = i – cur_pivot;

// cout<<cur_radius<<" "<<cur_pivot<<" "<<i<= 0 && str[cur_pivot - cur_radius + is_even] == str[i] &&cur_radius >= cur_max_radius)

{

cur_max_radius = cur_radius;

cur_max_pivot = cur_pivot;

}

while ((cur_pivot – cur_radius + is_even < 0 ||str[cur_pivot - cur_radius +is_even] != str[i] ) && cur_pivot < i)

{

cur_pivot++;

cur_radius–;

//cout<<cur_radius<<" "<<cur_pivot<<" "<<i<<endl;

}

}

return 2 * cur_max_radius + !is_even;

}

public:

string longestPalindrome(string str) {

// Start typing your C/C++ solution below

// DO NOT write int main() function

int even_radius = 0, even_pivot = 0, odd_radius = 0, odd_pivot = 0;

int even_len = longestPalindromSubstrHelper(str, 1, even_pivot, even_radius );

int odd_len = longestPalindromSubstrHelper(str, 0, odd_pivot, odd_radius);

//cout<<even_len<<odd_len<= odd_len)

{

return str.substr(even_pivot – even_radius + 1, 2 * even_radius);

}

return str.substr(odd_pivot – odd_radius, 2 * odd_radius+1);

}

};

int main()

{

Solution s;

//cout<<longestPalindromSubstrHelper("abababab",1)<<endl;

cout<<s.longestPalindrome("321012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210012321001232100123210123210012321001232100123210123")<<endl;

// cout<<longestPalindromSubstr("ababababa")<<endl;

}

+1STANLEY said on October 28, 2013

if the string is atabcccbatabccccbata the cause the longest p. string’s middle is ‘t’ in central. But as your algorithm, the when the i =t R is larger than i; so the t is gonna to equals to the t at the second place. is 3. how could this figure out?

0stanley said on October 28, 2013

I’m not sure that cause you mean the R is the position which generated by the pivot’s left bound of previous D. And when you met i < R it will equals to the min(), so If the string is atabcccbatabcccbata the when D at the position of the second c, the R is gonna to be the a behind the second t. But so when the i turn to the t, t should equals to the min(), but t is the pivot of the longest substring, so how this work in the algorithm? I've got a little confuse.

0Atul Kumar said on November 15, 2013

Great !!!

0Mark said on November 30, 2013

A misleading post.

+2Albert Chen said on February 2, 2014

Building suffix tree is O(n) and preprocessing a general tree for O(1) LCA queries is O(n).

We can build an extended suffix tree by inserting suffixes of reversed text into the same tree.

After these constructions, we enumerate the mid point in text (and we know the corresponding point in the reversed text). By looking at the LCA of the corresponding points in text and reversed text. We can decide in constant time that the longest palindrome from this mid point.

So we will be able to solve the problem in linear time (for odd length text.)

For more details, look into the book Algorithms on Strings, Trees and Sequences.

0Leet said on February 3, 2014

this itself wil give the solution..

Whats the need of center and right part.. plz expalin me

0online chesstle said on February 15, 2014

s.substr((centerIndex – 1 – maxLen)/2, maxLen); is there a bug here as the rest of the code is treating P[i] as the length of the palindrome on either side of the center and not as the total length of the palindrome. maxlen is derived from P…

0Acmerblog said on March 22, 2014

great job.I’m not sure that cause you mean the R is the position which generated by the pivot’s left bound of previous D. And when you met i < R it will equals to the min(), so If the string is atabcccbatabcccbata the when D at the position of the second c, the R is gonna to be the a behind the second t. But so when the i turn to the t, t should equals to the min(), but t is the pivot of the longest substring, so how this work in the algorithm? I've got a little confuse.

+1Donne said on April 9, 2014

Why is the algoritmn O(n). The while loop where we attempt to expand the palindrome would need O(n) in worst case. eg. in case of b of the center. We are indirectly traversing all the nodes in that loop. So wont the order be O(n^2) ?

0Alex said on April 11, 2014

Excellent post…. i don’t find any other posts which is as good and elaborate as this !! Good job.

0programmingmonkey said on April 29, 2014

perhaps this part needs a little more index checking to avoid segment fault?

0Bob said on June 3, 2014

Thank you for your clear explanation！！

0ash said on August 4, 2014

for (int i = 1; i i) ? min(R-i, P[i_mirror]) : 0;

//There is a problem here for i=1 and first run(C=0) through the loop i_mirror = -1;

P[-1] … you are accessing wrong address..

0killa said on August 13, 2014

You’ve no idea how difficult it is for a chinese to understand your method. But you are brilliant. Thanks.

0ccen said on August 20, 2014

learn a lot, thank you

0Dinesh said on August 30, 2014

Excellent O(N) time complexity solution. Thanks for the post

0B_Khan said on November 2, 2014

Great Explanation. Thanks !

0adiggo said on November 6, 2014

it is so brilliant, but so hard to come up with.

0sagiv said on November 20, 2014

great post! thanks

0Leo said on November 28, 2014

Using symmetry to avoid recalculation. Simple and elegant.

0