11475 - Extend to Palindrome

[sonyckson]

Hi!, its not the first time i see this problem, and its not the first time i dont see how to solve it... any hints? thanks! Eric.

[hamedv]

Yes, i don't know too

There were problems like this an i solved them using DP, because the input size was less than 1000 but in this problem it's so huge.

[Ivan]

Hint: Rolling hash might be a viable option here.

[Leonid]

Knuth-Morris-Pratt algorithm is an option to consider if you're trying to solve this task.

[baodog]

Hi,

I use base N representation mod M, reading forward and backward, starting from
the last symbol. I keep get WA. I have tried many cases.
Help is appreciated! Thanks.

Code: Select all

Accepted... silly index error.  Thanks!

[MAK]

@baodog:
Your code prints an extra unprintable character at the end of each line on my machine. But it looks otherwise ok.

@hamedv:
I'm curious about the O(n^2) DP you mentioned. The simple brute force way of solving this problem is also O(n^2). Could you please post an outline of your idea?

[hamedv]

@baodog:
Your code prints an extra unprintable character at the end of each line on my machine. But it looks otherwise ok.

@hamedv:
I'm curious about the O(n^2) DP you mentioned. The simple brute force way of solving this problem is also O(n^2). Could you please post an outline of your idea?

I mean this:
http://www.comp.nus.edu.sg/~stevenha/pr ... Palindrome

[MAK]

Thanks Hamedv. That is a much harder problem. The present problem asks that you only consider adding characters at the end of the string, which simplifies matters a lot.

[emotional blind]

I think that is not that hard. because the length of the string is only 1000. So O(N^2) solution will be enough.

[MAK]

IMHO the idea behind the solution (and not the implementation) of this problem is somewhat simpler that the one Hamedv referred to.

Of course, what I find find hard (or easy) can quite naturally seem easy (or hard) to someone else .

[sijal]

i solved this using kmp string matching algorithm in 0.020 seconds

[Sylpharion]

Anybody can help me?
I've use the KMP and I'm sure that my program run on the largest testcases. And it works.. but I have a TLE on my code.

Code: Select all

#include<iostream>
#include<stack>

using namespace std;

char text[100001], pat[100001];
int f[100001];
stack<char> tempS;

bool isPalindrome(char* T)
{
    int i;
    for(i=0;i<strlen(T)/2;i++)
        if(T[i]!=T[strlen(T)-1-i])
            return false;
    return true;
}

void reverse(char T[], char* P)
{
    int i;
    char temp;
    for(i=0;i<strlen(T);i++)
        P[strlen(T)-1-i]=T[i];
    P[strlen(T)]='\0';
}

void fFunction(char*P)
{
	int i,j,m;
	i=1;
	j=0;
	m=strlen(P);
	while(i<=m-1)
	{
		if(P[j]==P[i])
		{
			f[i]=j+1;
			i++;
			j++;
		}
		else if(j>0)
			j=f[j-1];
		else
		{
			f[i]=0;
			i++;
		}
	}
}

void KMPMatch(char* T, char*P)
{
    bool done=false;
	int i,j,m,trace=0,k;
	i=0;
	j=0;
	while(i<=strlen(T) && !done)
	{
       	m=strlen(P);
		if(P[j]==T[i])
		{
			if(j==m-1)
			{

                cout<<text;
                while(!tempS.empty())
                {
                    cout<<tempS.top();
                    tempS.pop();
                }
                cout<<endl;              
                done=true;
            }
			i++;
			j++;
		}
		else if (j>0)
		{
			trace=j;
			j=f[j-1];
			for(k=0;k<trace-j;k++)
			{
			    tempS.push(P[strlen(P)-1]);
			    P[strlen(P)-1]='\0';
            }
        }
		else
		{
			i++;
			tempS.push(P[strlen(P)-1]);
			P[strlen(P)-1]='\0';
        }
    }
}

int main()
{
	int i;
	while(cin>>text)
	{
        if(isPalindrome(text))
          cout<<text<<endl;
        else
        {
           reverse(text,pat);
    	   fFunction(pat);
	       KMPMatch(text,pat);
        }
    }
	return 0;
}

[MAK]

Try an input with a large string of the same character with a different character near the end.
e.g.
xxxxxx......xxxy
xxxxxx........xyxx

[Sylpharion]

yes, it works very fast, but it was TLE again when i submitted...
oh.. help me...

[serur]

Consider the dual of the problem: find the largest suffix of the input text that is a palindrome.
text = abB, where B is the reverse of b. Then textTEXT = abBbBA, then you find the longest tandem repeat...
This leads to O(nlogn) solution which gets TLE + MLE.