Binary Strings With At Most K Consecutive Ones in Java

Understanding the Problem

The core challenge of this problem is to generate all possible binary strings of length N and count those that have at most K consecutive ones. This problem is significant in various fields such as coding theory, data compression, and error detection.

Potential pitfalls include misunderstanding the constraint of "at most K consecutive ones" and not efficiently generating or counting the valid strings.

Approach

To solve this problem, we can use dynamic programming (DP). The idea is to build the solution incrementally and use previously computed results to find the current result efficiently.

We can define a DP table where dp[i][j] represents the number of valid binary strings of length i that end with exactly j consecutive ones.

We will initialize the DP table and then fill it using the following rules:

If the current string ends with a '0', it can be appended to any valid string of length i-1.
If the current string ends with '1', it can be appended to any valid string of length i-1 that ends with j-1 consecutive ones, provided j is less than or equal to K.

Algorithm

1. Initialize a DP table dp with dimensions (N+1) x (K+1) to store the number of valid strings.

2. Set the base case: dp[0][0] = 1 (an empty string is a valid string).

3. Iterate over the length of the string from 1 to N.

4. For each length, iterate over the number of consecutive ones from 0 to K.

5. Update the DP table based on the rules mentioned above.

6. Sum up all the values in the last row of the DP table to get the total number of valid strings.

Code Implementation

public class BinaryStrings {
    public static int countBinaryStrings(int N, int K) {
        // Initialize the DP table
        int[][] dp = new int[N + 1][K + 1];
        
        // Base case: An empty string is a valid string
        dp[0][0] = 1;
        
        // Fill the DP table
        for (int i = 1; i <= N; i++) {
            for (int j = 0; j <= K; j++) {
                // If the current string ends with '0'
                for (int l = 0; l <= K; l++) {
                    dp[i][0] += dp[i - 1][l];
                }
                // If the current string ends with '1'
                if (j > 0) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }
        
        // Sum up all the valid strings of length N
        int result = 0;
        for (int j = 0; j <= K; j++) {
            result += dp[N][j];
        }
        
        return result;
    }

    public static void main(String[] args) {
        int N = 4;
        int K = 2;
        System.out.println(countBinaryStrings(N, K)); // Output: 13
    }
}

Complexity Analysis

The time complexity of this approach is O(N * K) because we are filling a DP table of size (N+1) x (K+1). The space complexity is also O(N * K) due to the storage required for the DP table.

Edge Cases

Some potential edge cases include:

N = 0: The output should be 1 (an empty string).
K = 0: The output should be 1 (only the string of all zeros is valid).
K >= N: All binary strings of length N are valid.

Testing

To test the solution comprehensively, consider the following test cases:

N = 4, K = 2: Expected output is 13.
N = 5, K = 1: Expected output is 8.
N = 3, K = 0: Expected output is 1.

Thinking and Problem-Solving Tips

When approaching such problems, it is crucial to break down the problem into smaller subproblems and use dynamic programming to solve them efficiently. Practice similar problems to develop a deeper understanding of DP techniques.

Conclusion

In this blog post, we discussed how to count binary strings of length N with at most K consecutive ones using dynamic programming. We covered the problem definition, approach, algorithm, code implementation, complexity analysis, edge cases, and testing. Understanding and solving such problems is essential for improving problem-solving skills and preparing for coding interviews.

Additional Resources

For further reading and practice, consider the following resources:

Binary Strings with at most K Consecutive Ones in Java (Time Complexity: O(N*K))