## Saturday, April 30, 2016

### Count number of binary strings without consecutive 1's - GeeksforGeeks

Count number of binary strings without consecutive 1's - GeeksforGeeks
Given a positive integer N, count all possible distinct binary strings of length N such that there are no consecutive 1's.
Examples:
Input:  N = 2  Output: 3  // The 3 strings are 00, 01, 10

1) Compute number of binary strings without consecutive 1’s using the approach discussed here. Let this count be c.
2) Count of all possible binary strings with consecutive 1’s is 2^n where n is input length.
3) Total binary strings with consecutive 1 is 2^n – c.
This problem can be solved using Dynamic Programming. Let a[i] be the number of binary strings of length i which do not contain any two consecutive 1’s and which end in 0. Similarly, let b[i] be the number of such strings which end in 1. We can append either 0 or 1 to a string ending in 0, but we can only append 0 to a string ending in 1. This yields the recurrence relation:
a[i] = a[i - 1] + b[i - 1]
b[i] = a[i - 1]
The base cases of above recurrence are a[1] = b[1] = 1. The total number of strings of length i is just a[i] + b[i].
int countStrings(int n)
{
int a[n], b[n];
a[0] = b[0] = 1;
for (int i = 1; i < n; i++)
{
a[i] = a[i-1] + b[i-1];
b[i] = a[i-1];
}
return a[n-1] + b[n-1]
}

If we take a closer look at the pattern, we can observe that the count is actually (n+2)’th Fibonacci number for n >= 1. The Fibonacci Numbers are 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 141, ….
n = 1, count = 2  = fib(3)
n = 2, count = 3  = fib(4)
n = 3, count = 5  = fib(5)
n = 4, count = 8  = fib(6)
n = 5, count = 13 = fib(7)
................
Therefore we can count the strings in O(Log n) time also using the method 5 here.