11081 - Strings

[Nazmul Quader Zinnuree]

O(n^3) dp ? Then, what is the recurance?
or, sth other ?
Can it solved using n^2 ?

Thanks!

[asif_rahman0]

First take a string like: abc
Then compare it with third string like: abc
then ^^^^

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a b c
| | |
a b c

So you can check it recursively with FOR LOOP. If match(above) then go if not check the second string just the same way.
And you must count the number of return value.

Hope this will help.

[Nazmul Quader Zinnuree]

If match(above) then go if not check the second string just the same way

What does it mean ? Would you explain more elaborately.....

what is the output for

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aa aa aa
a a aaa
ab ac abc

Plz, explain with an example.

Thanks.

[asif_rahman0]

For your input, my output is:

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10
0
2

Here i explain one example for you. Ok.

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aa aa aa
1)
s1[0]->s3[0]
	s1[1]->s3[1] so at the end of s3 return 1;
	s2[0]->s3[1] so at the end of s3 return 1;
	s2[1]->s3[1] so at the end of s3 return 1;
s1[1]->s3[0]
	s2[0]->s3[1] so at the end of s3 return 1;
	s2[1]->s3[1] so at the end of s3 return 1;

Now we got 5 from aaaa(aa aa). But if we combine them differently then it will be 10. So in memoization with indexing(e.g, memo[j]) you will get overlapping value so just return it then it will be 10. Now is it clear?

[tywok]

It is strange... during the contest i got ACed an n^4 algorithm, that of course, i can't get ACed now..

[Riyad]

is there any O(n^3 ) dp for this problem . i used O(n^4) dp to get accepted during the contest which is getting TLE in the judge now . so can some one point out the O(n^3) algorithm or O(n^4) is way to go .........
thanx in advance

[chrismoh]

Well, my rather inefficient ACed solution (I mod 10007 everywhere, instead of just doing a mod where its needed) is O(n^3).

So yes, there is an O(n^3) solution... you just need to look for it

[Riyad]

would u care to tell us your DP solution which runs in O(n^3) iLL be really glad to know one if u dont want to give a spoiler u can PM me ur idea , anywayz thanx for ur reply ...

[Martin Macko]

would u care to tell us your DP solution which runs in O(n^3) iLL be really glad to know one if u dont want to give a spoiler u can PM me ur idea , anywayz thanx for ur reply ...

Denote a = a1a2...ak, b = b1b2...bm and c = c1c2...cn. My DP counts the function f(t,x,y,z) that expresses the number of ways to create string czcx+1...cn from strings axax+1...ak and byby+1...bm under assumption that the character cz is made by a character from b if t=1, or by a character from a if t=0.

To not write too much spoilers here I let the rest of the idea on you

[vinay]

I have a DP solution (hope n^3 ) where I hava a function f( i , j, k) which returns the number of possible ways to form the string by taking the kth character with any character in first string from ith position or with a character from second string starting from jth positon ...
I use memoization to store this value..

Unfortunately it TLE 's...

I no one has problem I may paste my code here...

can anybody suggest me something??

[vinay]

ohh..
I realise that my function is same as
Marko's

then why is it TLE???
Marko can I send u my code

[Martin Macko]

ohh..
I realise that my function is same as
Marko's

then why is it TLE???
Marko can I send u my code

I've no idea why you're getting TLE. My solution is just a straight forward memoization counting the function mentioned above. But it's rather a short recursive function with no loops and just few recursive calls.

During the contest the solution got ACed in 0:00.834. Now it gets ACed in 0:05.256. Probably, they have a much bigger input set, now.

You can post your solution here, if you want, maybe, I'll look on it

[vinay]

here it goes...

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#include<iostream>
#include<cstdio>
#include<string>

using namespace std;
char s1[61],s2[61],s3[61];
int sz1,sz2,sz3,dp[61][61][61];
int fun(int i,int j,int k){
	if(sz1-i+sz2-j<sz3-k) return 0;
	
	if(dp[i][j][k]!=-1) return dp[i][j][k];
	dp[i][j][k]=0;
	for(int index=i;index<sz1;index++){
		if(s1[index]==s3[k]){
		        if(k==sz3-1) dp[i][j][k]=(dp[i][j][k]+1)%10007;
			else
				dp[i][j][k]=(dp[i][j][k]+fun(index+1,j,k+1))%10007;
		}
	}
	for(int index=j;index<sz2;index++){
		if(s2[index]==s3[k]){
			if(k==sz3-1) dp[i][j][k]=(dp[i][j][k]+1)%10007;
			else 
			  dp[i][j][k]=(dp[i][j][k]+fun(i,index+1,k+1))%10007;
		}
	}
	return dp[i][j][k]%10007;
}
int main(){
	int t;
	scanf("%d",&t);
	while(t--){
		scanf("%s%s%s",s1,s2,s3);
		sz1=strlen(s1);
		sz2=strlen(s2);
		sz3=strlen(s3);
		for(int i=0;i<=sz1;i++){
			for(int j=0;j<=sz2;j++){
				for(int k=0;k<=sz3;k++){
					dp[i][j][k]=-1;
				}
			}
		}
		printf("%d\n",fun(0,0,0)%10007);
		
		
	}
	return 0;
}

[Martin Macko]

here it goes...

IMO, your solution is O(N^4). Note the loop in your function.

Try the following input yourself:

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1
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa

If it takes more than 0.2s (on 1G machine), it won't get accepted, definitely. Now it takes 0.8s.

[tywok]

thanks Martin for your hint!