Count number of possible rhythm in a poem.
Explanation:
Twinkle, twinkle, little star, [ A ]
How I wonder what you are! [A]
Up above the world so high, [B]
Like a diamond in the sky. [B]
In the above poem, we see the first two line ( ending with "star" and "are" ) is in the rhythm that's why they are given as character "A" and similarly last two lines (ending with "high" and "sky" ] s in the rhythm that's why they are given as character "B".
Questions: Given "n" number of lines in a poem, Count number of possible rhythm in a poem.
P.S. output should be in lexicographical order only
Example:
n=1
only one possible "A"
Answer: A, count =1
n=2 [ possible chars are A,B]
AA
AB
BA <- This is not possible as it collide with AB. How? Look at the pattern
AB says first line has different rhythm and second line has different rhythm then first line.
Similarly BA is also shows the same ; first line has different rhythm and second line has different rhythm then first line.
Hence discard
n=2
AA
AB , count=2
n=3 [ possible chars are A,B, C]
AAA
AAB
ABA
ABB
ABC, count=5
Look we can't have AAC as it collide with AAB (first two are same and last is different in both)
Similarly other BCA/ BAC etc we'll discard them because of collide and lexicographical ordering.
n=4, there will be 15 [ we need to print all of them too ]
My Finding;
1. I found out that, this is just a bell number ( see the pattern 1,2,5,15.... )
2. I found, this is Dynamic programming question, we can generate the next n+1 from n; How
n=2 has
AA
AB
n=3
Take AA; there are three possiblilties to append a character is A,B,C
But since the last character is A; so lexicographically A and B can be appended at the most, since appending C could conflict with B.
AA(A)
AA(B)
Take AB; there are three possibilities to append a character is A,B,C
But since the last character is B; which is lexicographically smaller than A, so lexicographically A, B,C can be appended easily,
AB(A)
AB(B)
AB(C) <- this is possible since its not colliding with any thing
So ans; 5
AA(A)
AA(B)
AB(A)
AB(B)
AB(C)
Now lets try n=4 [ now its become complicated ;A,B,C,D]
Take AAA; possibilty A,B, not possible C/D conflict with B
AAA(A)
AAA(B)
Take AAB, Possibility is A, B C but what about D; is it possible ? Yes/No
AAB(A)
AAB(B)
AAB(C)
AAB(D) <- it collide with last C so discard ;
Hence with AAB
AAB(A)
AAB(B)
AAB(C)
Take ;
ABA(A)
ABA(B)
ABA(C)
ABA(D) Not possible ; collide with C
If you observe there is a pattern with last character to first character from right to left;
Solution: Brute force is obvious solution; and generating number is also fine since you can use Bell number [ i was not able to come up with this solution though, found later]
Any one can help me over here?
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