10252 - Common Permutation

All about problems in Volume CII. If there is a thread about your problem, please use it. If not, create one with its number in the subject.

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Postby Jan » Sat Oct 20, 2007 10:27 pm

Nope. If the solution is 'ab' then of course 'ba' is a solution, too. right?
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Postby jackpigman » Wed Mar 19, 2008 10:04 pm

I've got all the test cases I could find right but still get's Wa....
Here's my code.
Code: Select all
removed after AC

Thanks for your help :D
Last edited by jackpigman on Thu Mar 20, 2008 9:48 pm, edited 1 time in total.
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Postby Jan » Thu Mar 20, 2008 12:22 am

There can be multiple cases. Your code works for only one case.
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Postby jackpigman » Thu Mar 20, 2008 6:21 pm

I changed my code, but got a RTE :cry: :cry: :cry:
Please help me.......
Code: Select all
removed after AC
Last edited by jackpigman on Thu Mar 20, 2008 9:49 pm, edited 1 time in total.
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Postby emotional blind » Thu Mar 20, 2008 7:19 pm

when length of a = 1000, and strlen(a) = 1000 then you can't access a[strlen(a)] ie a[1000], coz your size of array is 1000 so that you can access 0 to 999. try increasing size by 1 ie set the max size to 1001.
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Postby jackpigman » Thu Mar 20, 2008 9:50 pm

Thanks a lot emotional blind!!
finally got an AC :D :D :D :D
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Re: 10252 - Common Permutation

Postby palash » Mon Jun 16, 2008 9:09 pm

pls help WA. I didn't find my bug in my code.every time i got WA here is my code
Code: Select all
At last Acc
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Re: 10252 - Common Permutation

Postby maruf » Mon Jun 30, 2008 12:07 pm

Given two strings of lowercase letters, a and b, print the longest string x of lowercase letters such that there is a permutation of x that is a subsequence of a and there is a permutation of x that is a subsequence of


what is meant hear by "permutation" && "subsequence"??
i am totally in dark :cry:
please help :oops:
lives for eternity......
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Re: 10252 - Common Permutation

Postby Jan » Mon Jun 30, 2008 4:46 pm

Subsequence: Suppose you are given two strings A and B. B will be said the subsequence of A, if you can make B from A by deleting some (or none) characters.
Permutation: Suppose you are given two strings A and B. B will be said the permutation of A, if you can make B from A by re-arranging some (or none) characters.

So, 'abc' is a subsequence of 'adbfc' and 'aabc' is a permutation of 'caba'.
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Re: 10252 - Common Permutation

Postby Galileo » Wed Jul 16, 2008 1:33 pm

Can someone tell me what is the output for?

aAaAAa
aaaaaa
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WA Plz help

Postby Galileo » Thu Jul 17, 2008 8:23 am

nothing
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Re: 10252 - Common Permutation

Postby ExUCI » Wed Dec 31, 2008 5:01 am

Hi, I just got AC without caring about such cases with uppercases like aaAAAaaa or whatever and there is no such thing. And yes, LCD can be used, I used it but I gonna change it for the simpler alg now!!

I read too many posters for nothing :evil:
Remove your code after AC :)
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Re: 10252 - Common Permutation

Postby wsb7metal » Mon Mar 30, 2009 3:06 am

I got AC!
I read just letters a..z and <space>
And ignore the last output, I dont know why, but work. =/
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Re: 10252 - Common Permutation

Postby banglacity » Thu Jun 25, 2009 5:43 am

After suffering a lot, finally i got Ac :D

The only problem is- take input by gets() rather than scanf()

Hope it will work !
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Re: 10252 - Common Permutation

Postby amishera » Wed May 12, 2010 1:33 am

My problem is rather simple. I used the print_LCS from the cormen book:

Code: Select all
PRINT-LCS(b, X, i, j )
1 if i = 0 or j = 0
2 then return
3 if b[i, j ] = “&”
4 then PRINT-LCS(b, X, i − 1, j − 1)
5 print xi
6 elseif b[i, j ] = “↑”
7 then PRINT-LCS(b, X, i − 1, j )
8 else PRINT-LCS(b, X, i, j − 1)


and implemented in 2 ways:

recursive:
Code: Select all
void print(char a[], int i, int j)
{
   if (i == 0 || j == 0)
   {
      return;
   }
   if (op[i][j] == 3)
   {
      print(a, i-1,j-1);
      printf("%c", a[i-1]);
   }
   else if (op[i][j] == 1)
   {
      print(a, i-1,j);
   }
   else
   {
      print(a, i,j-1);
   }
}


and iterative:
Code: Select all
void print(char a[], int i, int j, int len)
{
   char pr[MAX];
   int k = len-1;

   while ((i != 0) && (j != 0))
   {      
      if (op[i][j] == 3)
      {
         i--,j--;
         pr[k] = a[i];
         k--;
      }
      else if (op[i][j] == 1)
      {   
         i--;
      }
      else
      {
         j--;
      }
   }
   
   pr[len] = '\0';
   printf("%s", pr);
}


The recursive is giving AC, but iterative is giving WA. While calling I pass the same parameters and for len pass the maximum length. The test cases passes for both implementations. Isn't this peculiar that it still gives WA for iterative one?
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