Question: So i'm trying to generate all binaries of a size n but with the condition that only k 1s. i.e

for size n = 4, k=2, (there is 2 over 4 combinations)

```
1100
1010
1001
0110
0101
0011

```

I'm stuck and can't figure out how to generate this.

Answer: 
Here's my non-recursive take on this algorithm. Because there are `2^n` permutations of binary strings, we can use a for-loop to iterate through every possible string and check if the amount of "1"s is not equal to `k`:

```
private static void generate(int n, int k) {
    for (int i = 0; i < Math.pow(2, n); i++) {
        if (Integer.bitCount(i) != k) {
            continue;
        }

        String binary = Integer.toBinaryString(i);

        if (binary.length() < n) {
            System.out.format("%0" + (n - binary.length()) + "d%s\n", 0, binary);
        } else {
            System.out.println(binary);
        }
    }
}

```
**Below is the solution using Recursion as an approach in java**

```
public class NumberOfBinaryPatternsSpecificOnes {

    static int[] bitArray = new int[]{0,1}; // kept binary bits in array

    public static void main(String args[])
    {   
        System.out.println("Below are the patterns\n");
        int n = 4;
        int k = 2;
        drawBinaryPattern(n,"",k,0);
    }
    private static void drawBinaryPattern(int n,String seed,int numberOfOnes,int currentCount)
    {
        if(n==0)
        {
            if(currentCount==numberOfOnes){
                System.out.println(seed);
            }
            return;
        }   
        for(int j=0;j<bitArray.length;j++)
        {
            String temp = seed+bitArray[j];
            int currentcountTemp = bitArray[j]==1?(currentCount+1):(currentCount);

            if(currentcountTemp>numberOfOnes)
            {
                return;
            }
            drawBinaryPattern(n-1,temp,numberOfOnes,currentcountTemp);
        }
    }
}

```