Difficulty in k ordered lcs

#1

Sir ,i know how to do lcs and also done the lcs with 3 strings ,but sir for k ordered lcs i am facing difficulty.
please explain the approach and the code

#2

hello @siddhant_samal

see logic is similar to lcs only .

suppose we have s1and s2 as string and k.
i=0 (for string s1)
j=0(for string s2)

if s1[i]==s2[j]
then ans=1+solvefor(i+1,j+1,k)
if(s1[i]!=s2[j] (we have two options)

  ans=solvefor(i+1,j,k) or solvefor(i,j+1,k) take maximum among them // we did similar thing in lcs
  if k>0 then i can use 1 k to make them equal and lcs in that case will be

     ans=1+solvefor(i+1,j+1,k-1) // k-1 because i have used 1 k to make them equal

among these two options whichever will give maximum answe we will consider that.

for optimisation use 3d dp table

code->

#3

this code is not working

#4
#5

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#6

this code is not working