Generate all Binary Strings of length N with equal count of 0s and 1s

Given an integer N, the task is to generate all the binary strings with equal 0s and 1s. If no strings are possible, print -1

Examples: 

Input: N = 2  
Output: “01”, “10”
Explanation: All possible binary strings of length 2 are: 01, 10, 11, 00. Out of these, only 2 have equal number of 0s and 1s  

Input: 4  
Output:  “0011”, “0101”, “0110”, “1100”, “1010”, “1001” 

Approach: The task can be solved by using recursion. If N is odd, then the answer is -1, else, we can use recursion to generate all the binary strings with equal 0s and 1s. Follow the below steps to solve the problem:

• Variable ones keep track of the number of 1’s and variable zeros keeps a track of the number of 0’s in the string.
• Both ones and zeros should have frequency N/2.
• Base condition: The string s stores the output string. So, when the length of s reaches N we stop recursive calls and print the output string s.
• If the frequency of 1’s is less than N/2 then add 1 to the string and increment ones.
• If the frequency of 0’s is less than N/2 then add 0 to the string and increment zeros.

Below is the implementation of the above code:

C++

// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;

// Recursive function that prints
// all strings of N length with equal 1's and 0's
void binaryNum(int n, string s, int ones,
               int zeros)
{

    // String s contains the output to be printed
    // ones stores the frequency of 1's
    // zeros stores the frequency of 0's
    // Base Condition: When the length of string s
    // becomes N
    if (s.length() == n)
    {
        cout << (s) << endl;
        return;
    }

    // If frequency of 1's is less than N/2 then
    // add 1 to the string and increment ones
    if (ones < n / 2)
        binaryNum(n, s + "1", ones + 1, zeros);

    // If frequency of 0's is less than N/2 then
    // add 0 to the string and increment zeros
    if (zeros < n / 2)
        binaryNum(n, s + "0", ones, zeros + 1);
}

// Driver Code
int main()
{

    string s = "";
    binaryNum(4, s, 0, 0);
    return 0;
}

// This code is contributed by Potta Lokesh

Java

// Java program for the above approach
import java.io.*;

class GFG {

    // Recursive function that prints
    // all strings of N length with equal 1's and 0's
    static void binaryNum(int n, String s, int ones,
                          int zeros)
    {
        // String s contains the output to be printed
        // ones stores the frequency of 1's
        // zeros stores the frequency of 0's
        // Base Condition: When the length of string s
        // becomes N
        if (s.length() == n) {
            System.out.println(s);
            return;
        }

        // If frequency of 1's is less than N/2 then
        // add 1 to the string and increment ones
        if (ones < n / 2)
            binaryNum(n, s + "1", ones + 1, zeros);

        // If frequency of 0's is less than N/2 then
        // add 0 to the string and increment zeros
        if (zeros < n / 2)
            binaryNum(n, s + "0", ones, zeros + 1);
    }

    // Driver Code
    public static void main(String[] args)
    {
        String s = "";
        binaryNum(4, s, 0, 0);
    }
}

Python3

# python code for the above approach

# Recursive function that prints
# all strings of N length with equal 1's and 0's
def binaryNum(n, s, ones, zeros):

    # String s contains the output to be printed
    # ones stores the frequency of 1's
    # zeros stores the frequency of 0's
    # Base Condition: When the length of string s
    # becomes N
    if (len(s) == n):

        print(s)
        return

    # If frequency of 1's is less than N/2 then
    # add 1 to the string and increment ones
    if (ones < n / 2):
        binaryNum(n, s + "1", ones + 1, zeros)

    # If frequency of 0's is less than N/2 then
    # add 0 to the string and increment zeros
    if (zeros < n / 2):
        binaryNum(n, s + "0", ones, zeros + 1)

# Driver Code
if __name__ == "__main__":

    s = ""
    binaryNum(4, s, 0, 0)

# This code is contributed by rakeshsahni

C#

// C# program for the above approach
using System;

class GFG {

    // Recursive function that prints
    // all strings of N length with equal 1's and 0's
    static void binaryNum(int n, string s, int ones,
                          int zeros)
    {
        // String s contains the output to be printed
        // ones stores the frequency of 1's
        // zeros stores the frequency of 0's
        // Base Condition: When the length of string s
        // becomes N
        if (s.Length == n) {
            Console.WriteLine(s);
            return;
        }

        // If frequency of 1's is less than N/2 then
        // add 1 to the string and increment ones
        if (ones < n / 2)
            binaryNum(n, s + "1", ones + 1, zeros);

        // If frequency of 0's is less than N/2 then
        // add 0 to the string and increment zeros
        if (zeros < n / 2)
            binaryNum(n, s + "0", ones, zeros + 1);
    }

    // Driver Code
    public static void Main(string[] args)
    {
        string s = "";
        binaryNum(4, s, 0, 0);
    }
}

// This code is contributed by ukasp.

Javascript

<script>
// javascript program for the above approach

   // Recursive function that prints
    // all strings of N length with equal 1's and 0's
    function binaryNum(n, s, ones, zeros)
    {

        // String s contains the output to be printed
        // ones stores the frequency of 1's
        // zeros stores the frequency of 0's
        // Base Condition: When the length of string s
        // becomes N
        if (s.length == n) {
            document.write(s+"<br>");
            return;
        }

        // If frequency of 1's is less than N/2 then
        // add 1 to the string and increment ones
        if (ones < n / 2)
            binaryNum(n, s + "1", ones + 1, zeros);

        // If frequency of 0's is less than N/2 then
        // add 0 to the string and increment zeros
        if (zeros < n / 2)
            binaryNum(n, s + "0", ones, zeros + 1);
    }

// Driver Code
var s = "";
binaryNum(4, s, 0, 0);

// This code is contributed by 29AjayKumar
</script>

Output

1100
1010
1001
0110
0101
0011

Time Complexity: O(2^N)
Auxiliary Space: O(1)