### 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 . 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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