counting


Binary Strings of the form *111*


Given an integer N, we have to count the number of possible binary strings(string is madeup of only 0s and 1s) of length N, and matches the regular expression pattern [111]. For example if N is 4, then 1110 , 0111 , 1111 are the possibilities.
I have worked on it, and have got the following recurrence:
count(a,N) = 4*count(0,N-3) + 2*count(1,N-3)+ count(2,N-3); if(a == 0)
count(a,N) = 2*count(0,N-2)+ count(1,N-1); if(a == 1)
count(a,N) = count(0,N-1); if(a == 2)
Answer is [(2^N) - Count(0,N)].
By using memorization, we can compute the the Count(0,N) in O(N).
but is there any better algorithm other than O(N)
Without completely answering your homework, try combinatorics,
For length N,
There are (N-2) ways to place a 111 in the block (Starting at the 0 index, 1 index, 2 index).. And then 2^(n-3) ways to fill the remaining space.
You just need to be careful about dupes..

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Binary Strings of the form *111*

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