Binary Numbers without Consecutive 1's

Find all n-digit binary numbers without any consecutive 1’s

Great, you're tackling a classic dynamic programming problem! Let's start breaking it down with a Socratic approach.

1. First, how would you define the problem in terms of smaller sub-problems? Think about a smaller version of the same problem: if you had to find all binary numbers of length ( n-1 ) without any consecutive 1’s, how would that help you solve for length ( n )?

2. Consider a binary number of length ( n ). What are the possible endings for this number, and how does each possibility affect your next step?

If I had all binary numbers of length n-1 then I can append 0 or 1 to the end

Excellent observation! However, there's a crucial detail to consider: appending a '1' to the end of a binary number is only valid if the last digit of the ( n-1 ) length number isn't already a '1'. This is because the problem restricts us from having consecutive '1's.

So, with that in mind:

1. If you have a binary number of length ( n-1 ) that ends in '0', what are your options for extending it to length ( n )?

2. Conversely, if your ( n-1 ) length number ends in '1', what options do you have for the last digit of the ( n ) length number?

Ah I missed that. Okay so if it ends in 0 we extend it with 1 or 0 if it ends with 1 we extend it with 0 only .

Exactly right! You've correctly identified how to extend the binary numbers based on their last digit. This insight is crucial for solving the problem using dynamic programming.

Now, let's move on to formulating our dynamic programming approach:

1. How would you define your state? Remember, in dynamic programming, a state is typically a set of parameters that uniquely defines the subproblems. In this case, what parameters do you think are necessary to define the state?

2. Once you've defined the state, think about the recurrence relation. This is the formula that expresses the solution to a problem in terms of the solutions to its subproblems. Given your observation on how to extend binary numbers, how would you express the number of valid ( n )-digit binary strings in terms of ( n-1 )-digit strings?

What is a state is it like the parameters of my recursive call ?

Yes, exactly! In dynamic programming, a "state" refers to the parameters of your recursive call that uniquely define each subproblem. For each state, you want to store the result of the subproblem so that you don't have to compute it again. This is known as memoization.

In the context of your problem, finding all n-digit binary numbers without any consecutive 1’s, think about what parameters would uniquely define each subproblem. Here are a couple of questions to guide your thinking:

1. What information do you need to know about your current position to decide how to form valid n-digit binary numbers?

2. Does the choice of the next digit (0 or 1) depend on the previous digit? How might this affect the definition of your state?

Okay so i will define my function is binaryNumbersWithoutConsecutiveOnes(K) which will return all binary numbers without consecutive ones of length K or less

That's a good start for defining your function! However, we need to refine it a bit to fit the dynamic programming paradigm more closely.

Since your goal is to find all binary numbers of a certain length without consecutive 1s, it's helpful to consider not just the length of the binary number (K) but also the last digit of the current number you're constructing. This is because the choice of your next digit (0 or 1) depends on what the last digit was, to avoid consecutive 1s.

So, you might consider a function like binaryNumbersWithoutConsecutiveOnes(K, lastDigit), where K is the remaining length of the number you need to construct, and lastDigit is the last digit of the current number.

1. With this function, what would be the base case or the simplest instance of the problem?

2. How would you use the results of binaryNumbersWithoutConsecutiveOnes(K-1, 0) and binaryNumbersWithoutConsecutiveOnes(K-1, 1) to construct the result for binaryNumbersWithoutConsecutiveOnes(K, lastDigit)?

The base case will be a single digit. And for last digit 0 its just 0 and for last digit 1 its just 1. To construct the solution we take the solution and based on last digit if its zero append 0 or 1 otherwise if its one we only append zero

You've got the base case right and a good understanding of how to construct the solution based on the last digit. Let's summarize and refine this approach:

1. Base Case: When K = 1, the function binaryNumbersWithoutConsecutiveOnes(K, lastDigit) returns just '0' and '1', as single-digit binary numbers are always valid.

2. Recursive Case: For K > 1, the function needs to consider the value of lastDigit.

   • If lastDigit is '0', you can append both '0' and '1' to each of the binary numbers obtained from binaryNumbersWithoutConsecutiveOnes(K-1, 0) and binaryNumbersWithoutConsecutiveOnes(K-1, 1).
   • If lastDigit is '1', you can only append '0' to each of the binary numbers obtained from binaryNumbersWithoutConsecutiveOnes(K-1, 1).

3. Combining Results: Concatenate the results from the above two steps to get all valid binary numbers of length K without consecutive 1s.

Now, let's think about how you would implement this. You can use memoization to store the results of subproblems to avoid redundant calculations.

A question to consider:

• How would you implement memoization in this context? What would be the key for your memoization table, and what value would it store?

key will be K and last digit

That's correct! Using ( K ) and lastDigit as keys for memoization is a great approach. This ensures that for each combination of the length of the binary number and its last digit, you compute the list of valid numbers only once.

Here's how you can structure your memoization table and implement this:

1. Memoization Table: Create a table (like a dictionary in Python) where the key is a tuple consisting of ( K ) and lastDigit, and the value is the list of all valid binary numbers of length ( K ) that end with lastDigit.

2. Function Implementation:

   • In your binaryNumbersWithoutConsecutiveOnes(K, lastDigit) function, first check if the key (K, lastDigit) is already in your memoization table. If it is, return the stored value.
   • If not, compute the value using the recursive approach you've outlined:
     • If K = 1, handle the base case by returning ['0', '1'] if lastDigit is not specified, or the corresponding single-digit number if it is.
     • For K > 1, concatenate '0' and '1' (if lastDigit is '0') or just '0' (if lastDigit is '1') to the results of the recursive calls, and store this list in your memoization table before returning it.

Would you like an example of how to code this function, or do you have any other questions on how to proceed?