Hi, I was reading the solution for this problem and one of the transitions is

dp(k,j)=opt(dp(k−1,j)+(0,ck),dp(k−1,j−1)).

Shouldn’t it be

dp(k,j)=opt(dp(k-1,j),dp(k-1,j-1)+(0,ck))?

If it is, then why when s[k] is not equal to the jth character of “bessie”, dp(k,j)=dp(k−1,j)+(0,ck)? If j is 1 and we add (0, ck), then if at k+1, if we use the first transition to calculate dp(k+1, 2) it might be wrong because ck is included in the answer, right?

Thanks!