To solve this problem, we need to count the number of valid binary sequences (composed of 0s and 1s) of length ( n ) where no two consecutive 1s appear. This is a classic problem that can be efficiently solved using dynamic programming.

Approach

The key insight here is to recognize the pattern in valid sequences:

• Let ( f(n) ) be the number of valid sequences of length ( n ).
• For a sequence of length ( n ):
  • If the last element is 0, the first ( n-1 ) elements can be any valid sequence of length ( n-1 ) (count = ( f(n-1) )).
  • If the last element is 1, the previous element must be 0, so the first ( n-2 ) elements can be any valid sequence of length ( n-2 ) (count = ( f(n-2) )).
• Thus, the recurrence relation is: ( f(n) = f(n-1) + f(n-2) ).
• Base cases:
  • ( f(1) = 2 ) (sequences: "0", "1").
  • ( f(2) = 3 ) (sequences: "00", "01", "10").

Solution Code

def count_valid_sequences(n):
    if n == 1:
        return 2
    elif n == 2:
        return 3
    a, b = 2, 3  # f(1), f(2)
    for _ in range(3, n+1):
        c = a + b
        a = b
        b = c
    return b

# Example usage:
n = 5
print(count_valid_sequences(n))  # Output: 13

Explanation

• The code uses an iterative approach to compute the result, which is efficient with a time complexity of ( O(n) ) and space complexity of ( O(1) ) (since we only keep track of the last two values).
• For each length from 3 to ( n ), we compute the number of valid sequences by adding the counts of valid sequences of the previous two lengths.
• This approach avoids recursion and handles large values of ( n ) efficiently.

This solution correctly computes the number of valid binary sequences with no consecutive 1s for any given length ( n ).

Answer: The function count_valid_sequences(n) returns the number of valid sequences. For example, if ( n=5 ), the answer is 13. For a general ( n ), the result is given by the function. If you need the answer for a specific ( n ), replace the example value and run the code.

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