# List of all binary combinations for a number in Java

I am working on a project involving "Dynamic Programming" and am struck on this trivial thing, please help.

Suppose I take 4 as an input, I want to display something like: 0000 to 1111

But, if I input 5, I want to display like: 00000 to 11111 and so on.

EDIT: Please don't post asking me for the code. This is not a homework problem and I don't need any code, just tell me the logic for it and I would be happy.

EDIT2: WTH is happening with Stackoverflow, did I ask any of you to write code for me? I want the person who downvoted to upvote it. What is a point of this forum if I can't for help?

Share the logic with me. We can discuss and I do not require the code for this.

EDIT3: Here I am posting the code which I tried. I hope this "SATISFIES" all the people who were thinking I have not tried anything.

```import java.util.ArrayList;
```

public class RegularInvestigator {

public ArrayList createCombinations(ArrayList listOfFlightNumbers) {

```ArrayList<String> result = new ArrayList<String>();

for(int i = 1; i < listOfFlightNumbers.size(); i++) {

String binaryEqvivalent = Integer.toBinaryString(i);System.out.println(binaryEqvivalent);
String element = "";

for(int j = 0; j < binaryEqvivalent.length(); j++)
if(binaryEqvivalent.charAt(j) == '1')
element += listOfFlightNumbers + " ";

result.add(element.substring(0, element.length() - 1));
}

return result;
```

}

```private String getContent(ArrayList<String> flight) {
String temp = "";

for(int i = 0; i < flight.size() - 1; i++)  temp += flight.get(i) + " ";

temp += flight.get(flight.size() - 1);

return temp;
```

}

private ArrayList removeElementAtIndex(ArrayList flight, int position) {

```ArrayList<String> res = new ArrayList<String>();

for(int i = 0; i < flight.size(); i++) {
if(i != position) res.add(flight.get(i));
}

return res;
```

} }

EDIT4: Thank you phoxis, PengOne, Jerry Coffin and oliholz for your valuable answers :)

• Get input n
• Count from i=0 to (2^n) - 1
• for each value of i bitmask each bit of i and display.

```public void outBinary(int value){
for (int i = 0; i < Math.pow(2, value); i++) {
System.out.println(Integer.toBinaryString(i));
}
}
```

with leading zeros something like that

```    for (int i = 0; i < Math.pow(2, value); i++) {
StringBuilder binary = new StringBuilder(Integer.toBinaryString(i));
for(int j = binary.length(); j < value; j++) {
binary.insert( 0, '0' );
}
System.out.println(binary);
}
```

Either use phoxis's very nice solution, or just iterate them lexicographically (this is really the same solution!): Given a binary string of a given length, get the next lexicographic string by finding the rightmost zero entry, change it to a 1, and change everything to the right of it back to a 0, e.g.

```0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
```

I'm a bit lost as to how you'd apply dynamic programming to this. It's just a matter of counting from 0 to one less than the specified maximum value (where the maximum value is 1 shifted left the specified number of bits).

Edit: I should add that there are other possibilities (e.g., gray codes) but absent some reason to do otherwise, simple binary counting is probably the simplest to implement.

```int x = 5;

for(int i = 0; i < (1 << x); i++){
System.out.println(Integer.toBinaryString(i));
}
```

here is the code is to find the combination

```/*
* To change this template file, choose Tools | Templates
* and open the template in the editor.
*/
package rotateimage;

/**
*
* @author ANGEL
*/
public class BinaryPermutaion {

public static void main(String[] args) {
//object creation
BinaryPermutaion binaryDigit=new BinaryPermutaion();
//Recursive call of the function to print the binary string combinations
binaryDigit.printBinary("", 4);
}

/**
*
* @param soFar String to be printed
* @param iterations number of combinations
*/
public void printBinary(String soFar, int iterations) {
if(iterations == 0) {
System.out.println(soFar);
}
else {
printBinary(soFar + "0", iterations - 1);
printBinary(soFar + "1", iterations - 1);
}
}
}
```

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