Binary Strings With K Ones in C++ (Time Complexity: O(N*K))

## Understanding the Problem The problem requires us to find the number of binary strings of length `N` that contain exactly `K` ones. This is a combinatorial problem where we need to count the number of ways to place `K` ones in a string of length `N`. ### Core Challenge The core challenge is to efficiently compute the number of valid binary strings without generating all possible strings, which would be computationally expensive for larger values of `N`. ### Significance and Applications This problem is significant in fields like coding theory, cryptography, and combinatorial optimization. It helps in understanding how to efficiently count combinations and can be applied in scenarios where binary representations are used. ### Potential Pitfalls - Misunderstanding the constraints and generating all possible strings. - Not considering edge cases like `K = 0` or `K = N`. ## Approach ### Naive Solution A naive solution would involve generating all possible binary strings of length `N` and counting those with exactly `K` ones. This approach is not feasible due to its exponential time complexity, `O(2^N)`. ### Optimized Solution A more efficient approach uses combinatorial mathematics. The number of ways to choose `K` positions out of `N` for placing ones is given by the binomial coefficient `C(N, K)`, which can be computed using dynamic programming. ### Dynamic Programming Approach We can use a dynamic programming table `dp` where `dp[i][j]` represents the number of binary strings of length `i` with exactly `j` ones. #### Steps: 1. Initialize a 2D array `dp` of size `(N+1) x (K+1)` with all zeros. 2. Set `dp[i][0] = 1` for all `i` because there is exactly one way to have zero ones in any length string (all zeros). 3. Use the recurrence relation: - `dp[i][j] = dp[i-1][j] + dp[i-1][j-1]` - This relation states that the number of strings of length `i` with `j` ones is the sum of: - The number of strings of length `i-1` with `j` ones (adding a zero at the end). - The number of strings of length `i-1` with `j-1` ones (adding a one at the end). ## Algorithm ### Step-by-Step Breakdown 1. **Initialization**: - Create a 2D array `dp` of size `(N+1) x (K+1)` and initialize all elements to zero. - Set `dp[i][0] = 1` for all `i`. 2. **Filling the DP Table**: - Iterate over the lengths from `1` to `N`. - For each length, iterate over the number of ones from `1` to `K`. - Update the `dp` table using the recurrence relation. 3. **Result**: - The value `dp[N][K]` will contain the number of binary strings of length `N` with exactly `K` ones. ## Code Implementation

#include <iostream>
#include <vector>
using namespace std;

// Function to calculate the number of binary strings of length N with exactly K ones
int countBinaryStrings(int N, int K) {
    // Create a 2D DP array
    vector<vector<int>> dp(N + 1, vector<int>(K + 1, 0));

    // Base case: There is one way to have zero ones in any length string (all zeros)
    for (int i = 0; i <= N; ++i) {
        dp[i][0] = 1;
    }

    // Fill the DP table
    for (int i = 1; i <= N; ++i) {
        for (int j = 1; j <= K; ++j) {
            dp[i][j] = dp[i - 1][j]; // Adding a zero at the end
            if (j > 0) {
                dp[i][j] += dp[i - 1][j - 1]; // Adding a one at the end
            }
        }
    }

    // The result is the number of binary strings of length N with exactly K ones
    return dp[N][K];
}

int main() {
    int N = 4, K = 2;
    cout << "Number of binary strings of length " << N << " with " << K << " ones: " << countBinaryStrings(N, K) << endl;
    return 0;
}

## Complexity Analysis - **Time Complexity**: `O(N * K)` because we fill an `N x K` table. - **Space Complexity**: `O(N * K)` for the DP table. ### Comparison The optimized solution is significantly better than the naive approach, reducing the time complexity from exponential to polynomial. ## Edge Cases - **K = 0**: The result should be `1` (all zeros). - **K = N**: The result should be `1` (all ones). - **K > N**: The result should be `0` (impossible to have more ones than the length). ## Testing To test the solution comprehensively: 1. **Simple Cases**: Test with small values of `N` and `K`. 2. **Edge Cases**: Test with `K = 0`, `K = N`, and `K > N`. 3. **Large Cases**: Test with the maximum values of `N` and `K` within the constraints. ## Conclusion Understanding and solving combinatorial problems like this one is crucial for developing efficient algorithms. The dynamic programming approach provides a clear and efficient solution to count binary strings with a specific number of ones. ## Additional Resources - [Combinatorics and Binomial Coefficients](https://en.wikipedia.org/wiki/Binomial_coefficient) - [Dynamic Programming Concepts](https://www.geeksforgeeks.org/dynamic-programming/) - [Practice Problems on LeetCode](https://leetcode.com/) By practicing similar problems and understanding the underlying principles, you can improve your problem-solving skills and tackle more complex challenges.