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+26 votes

Consider two strings $A$="qpqrr" and $B$="pqprqrp". Let $x$ be the length of the longest common subsequence (not necessarily contiguous) between $A$ and $B$ and let $y$ be the number of such longest common subsequences between $A$ and $B$. Then $x +10y=$ ___.

asked in Algorithms by Veteran (115k points)
retagged by | 2.9k views

LCS of length 4 ==> x=4

LCS={qpqr,qprr,pqrr} ==>y=3


6 Answers

+21 votes
Best answer
Answer is $34$.

In first string, if we want to get $4$ as maximum length then LCS should end with either "$rr$" or "$qr$".
Only $4$ combinations are possible for LCS with length $4$:





Now, check for matching sequences in second string, except for "$qqrr$" all are possible.
answered by Loyal (8.2k points)
edited by
Do we always need to construct a table(Dynamic Programming Technique) ? or is there some other short method also...for calculating length of LCS?
how do u get qqrr ?? is it in sequence ??
we cant get qqrr..
Why should we dont considr permutation of rwo differnt qpqr

That way answer would become 44
plz clearify

Here in question it is explicitly mentioned (Not necessarily Contiguous) but if Only Longest Common Subsequence is mentioned what should be our approach?Will it be same???
+20 votes

I hope this helps. :)

answered by Boss (12.4k points)
edited by
PLEASE Check in 5th row 4th colum their is a match of R  and you have not incremented it ... but still you are getting  through answer ... how ??
Thank you @bharti ji. Typo is corrected.
+14 votes
Its a big problem if we solve it with DP or recursion.So we should use some intuition to get answer.

A=5 length and B= 7 length.

Maximum x can be 5 but 5 is not possible so try with 4 and we can easily find that 4 is satisfying for x.Now we need to find y.Here we know x=4 ,so let us write all the 4 length subsequences of A.As A is 5 length and we need 4 length so remove 1 character at a time and get the subsequence.

1. qpqr





Now look for these in B and we will see except qqrr all are satisfying.So y=4 ,but 1 and 2 are same ,so y=3.

Now x+10y=34
answered by Boss (27.2k points)
No, we can't common.../// whr u r facing [email protected]
slight mistake ....
Nice approach !! Although successful in certain cases only !

Thanks rahul sharma 5 for such a nice approach


There is $qprr$ in place of $aprr$ @rahul sharma 5

+3 votes

step 1- LCS(i,j) depends on LCS(i+1,j+1) when match else max{ LCS(i+1,j) and LCS(i,j+1)}

step 2- Starts at LCS(m,n) and fill by row,column or diagonal 

step 3- When there is match then value of that cell 1+ LCS(i+1,j+1)  else step 1


Black Arrow - Fill columns and other arrow for backtracking 

so x= 4 and y = 3 

x+10y = 4+10 * 3 = 34 

Reference: -

answered by Boss (11k points)
edited by
$pqrr,qpqr,qprr$ are three strings acc to figure
0 votes
How to calculate the number of largest common subsequences..?
answered by Active (1.4k points)

Go to above given link

This will help you to calculate longest sub sequences by drawing a table and from there it will be easier for you to figure out number of longest subsequence

@Abhishek video is fine . But, still how to find the number of LCSs ??
@VS.Check my answer.It might give a little help
–1 vote
34 should be the answer
answered by Active (3.6k points)

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