Find all even length binary sequences with same sum of first and second half bits | Iterative

Given a number N, find all binary sequences of length 2*N such that sum of first N bits is same as the sum of last N bits.
Examples: 

Input: N = 2 
Output: 
0000 
0101 
0110 
1001 
1010 
1111

Input: N = 1 
Output: 
00 
11 

Note: The recursive approach to this problem can be found here. 
Approach: 
A simple approach to run a loop from 0 to 2^2*N and convert into the binary form and check whether the sum of first half is equal to the sum of the second half.
If the above condition is true, then print that number, else check for the next one.
Below is the implementation of the above approach: 

C++

// C++ implementation
#include <iostream>
#include <malloc.h>
#include <math.h>
using namespace std;

// Function to convert the
// number into binary and
// store the number into
// an array
void convertToBinary(int num, int a[], int n)
{

    int pointer = n - 1;
    while (num > 0) {
        a[pointer] = num % 2;
        num = num / 2;
        pointer--;
    }
}

// Function to check if the
// sum of the digits till
// the mid of the array and
// the sum of the digits
// from mid till n is the
// same, if they are same
// then print that binary
void checkforsum(int a[], int n)
{

    int sum1 = 0;
    int sum2 = 0;
    int mid = n / 2;

    // Calculating the sum from
    // 0 till mid and store
    // in sum1
    for (int i = 0; i < mid; i++)
        sum1 = sum1 + a[i];

    // Calculating the sum
    // from mid till n and
    // store in sum2
    for (int j = mid; j < n; j++)
        sum2 = sum2 + a[j];

    // If sum1 is same as
    // sum2 print the binary
    if (sum1 == sum2) {
        for (int i = 0; i < n; i++)
            cout << a[i];
        cout << "\n";
    }
}

// Function to print sequence
void print_seq(int m)
{

    int n = (2 * m);

    // Creating the array
    int a[n];

    // Initialize the array
    // with 0 to store the
    // binary nnumbers
    for (int j = 0; j < n; j++) {
        a[j] = 0;
    }

    for (int i = 0; i < (int)pow(2, n); i++) {

        // Converting the number
        // into binary first
        convertToBinary(i, a, n);

        // Checking if the sum of
        // the first half of the
        // array is same as the
        // sum of the next half
        checkforsum(a, n);
    }
}

// Driver Code
int main()
{
    int m = 2;

    print_seq(m);

    return 0;
}

Java

// Java code implementation for above approach
class GFG
{

    // Function to convert the
    // number into binary and
    // store the number into
    // an array
    static void convertToBinary(int num,
                                int a[], int n)
    {
        int pointer = n - 1;
        while (num > 0)
        {
            a[pointer] = num % 2;
            num = num / 2;
            pointer--;
        }
    }

    // Function to check if the
    // sum of the digits till
    // the mid of the array and
    // the sum of the digits
    // from mid till n is the
    // same, if they are same
    // then print that binary
    static void checkforsum(int a[], int n)
    {
        int sum1 = 0;
        int sum2 = 0;
        int mid = n / 2;

        // Calculating the sum from
        // 0 till mid and store
        // in sum1
        for (int i = 0; i < mid; i++)
            sum1 = sum1 + a[i];

        // Calculating the sum
        // from mid till n and
        // store in sum2
        for (int j = mid; j < n; j++)
            sum2 = sum2 + a[j];

        // If sum1 is same as
        // sum2 print the binary
        if (sum1 == sum2)
        {
            for (int i = 0; i < n; i++)
                System.out.print(a[i]);
            System.out.println();
        }
    }

    // Function to print sequence
    static void print_seq(int m)
    {

        int n = (2 * m);

        // Creating the array
        int a[] = new int[n];

        // Initialize the array
        // with 0 to store the
        // binary nnumbers
        for (int j = 0; j < n; j++)
        {
            a[j] = 0;
        }

        for (int i = 0; i < (int)Math.pow(2, n); i++)
        {

            // Converting the number
            // into binary first
            convertToBinary(i, a, n);

            // Checking if the sum of
            // the first half of the
            // array is same as the
            // sum of the next half
            checkforsum(a, n);
        }
    }

    // Driver Code
    public static void main (String[] args)
    {
        int m = 2;

        print_seq(m);
    }
}

// This code is contributed by AnkitRai01

Python3

# Python3 implementation of above approach

# Function to convert the number into binary
# and store the number into an array
def convertToBinary(num, a, n):

    pointer = n - 1
    while (num > 0):
        a[pointer] = num % 2
        num = num // 2
        pointer -= 1

# Function to check if the sum of the digits till
# the mid of the array and the sum of the digits
# from mid till n is the same, if they are same
# then print that binary
def checkforsum(a, n):

    sum1 = 0
    sum2 = 0
    mid = n // 2

    # Calculating the sum from 0 till mid
    # and store in sum1
    for i in range(mid):
        sum1 = sum1 + a[i]

    # Calculating the sum from mid till n
    # and store in sum2
    for j in range(mid, n):
        sum2 = sum2 + a[j]

    # If sum1 is same as sum2 print the binary
    if (sum1 == sum2):
        for i in range(n):
            print(a[i], end = "")
        print()

# Function to print sequence
def print_seq(m):

    n = (2 * m)

    # Creating the array
    a = [0 for i in range(n)]


    for i in range(pow(2, n)):

        # Converting the number
        # into binary first
        convertToBinary(i, a, n)

        # Checking if the sum of the first half
        # of the array is same as the sum of
        # the next half
        checkforsum(a, n)

# Driver Code
m = 2

print_seq(m)

# This code is contributed by mohit kumar

C#

// C# code implementation for above approach
using System;

class GFG
{

    // Function to convert the
    // number into binary and
    // store the number into
    // an array
    static void convertToBinary(int num,
                                int []a, int n)
    {
        int pointer = n - 1;
        while (num > 0)
        {
            a[pointer] = num % 2;
            num = num / 2;
            pointer--;
        }
    }

    // Function to check if the
    // sum of the digits till
    // the mid of the array and
    // the sum of the digits
    // from mid till n is the
    // same, if they are same
    // then print that binary
    static void checkforsum(int []a, int n)
    {
        int sum1 = 0;
        int sum2 = 0;
        int mid = n / 2;

        // Calculating the sum from
        // 0 till mid and store
        // in sum1
        for (int i = 0; i < mid; i++)
            sum1 = sum1 + a[i];

        // Calculating the sum
        // from mid till n and
        // store in sum2
        for (int j = mid; j < n; j++)
            sum2 = sum2 + a[j];

        // If sum1 is same as
        // sum2 print the binary
        if (sum1 == sum2)
        {
            for (int i = 0; i < n; i++)
                Console.Write(a[i]);
            Console.WriteLine();
        }
    }

    // Function to print sequence
    static void print_seq(int m)
    {

        int n = (2 * m);

        // Creating the array
        int []a = new int[n];

        // Initialize the array
        // with 0 to store the
        // binary nnumbers
        for (int j = 0; j < n; j++)
        {
            a[j] = 0;
        }

        for (int i = 0; i < (int)Math.Pow(2, n); i++)
        {

            // Converting the number
            // into binary first
            convertToBinary(i, a, n);

            // Checking if the sum of
            // the first half of the
            // array is same as the
            // sum of the next half
            checkforsum(a, n);
        }
    }

    // Driver Code
    public static void Main (String[] args)
    {
        int m = 2;

        print_seq(m);
    }
}

// This code is contributed by PrinciRaj1992

Javascript

<script>
// Javascript implementation

// Function to convert the
// number into binary and
// store the number into
// an array
function convertToBinary(num, a, n)
{

    let pointer = n - 1;
    while (num > 0) {
        a[pointer] = num % 2;
        num = parseInt(num / 2);
        pointer--;
    }
}

// Function to check if the
// sum of the digits till
// the mid of the array and
// the sum of the digits
// from mid till n is the
// same, if they are same
// then print that binary
function checkforsum(a, n)
{

    let sum1 = 0;
    let sum2 = 0;
    let mid = parseInt(n / 2);

    // Calculating the sum from
    // 0 till mid and store
    // in sum1
    for (let i = 0; i < mid; i++)
        sum1 = sum1 + a[i];

    // Calculating the sum
    // from mid till n and
    // store in sum2
    for (let j = mid; j < n; j++)
        sum2 = sum2 + a[j];

    // If sum1 is same as
    // sum2 print the binary
    if (sum1 == sum2) {
        for (let i = 0; i < n; i++)
            document.write(a[i]);
        document.write("<br>");
    }
}

// Function to print sequence
function print_seq(m)
{

    let n = (2 * m);

    // Creating the array
    let a = new Array(n);

    // Initialize the array
    // with 0 to store the
    // binary nnumbers
    for (let j = 0; j < n; j++) {
        a[j] = 0;
    }

    for (let i = 0; i < Math.pow(2, n); i++) {

        // Converting the number
        // into binary first
        convertToBinary(i, a, n);

        // Checking if the sum of
        // the first half of the
        // array is same as the
        // sum of the next half
        checkforsum(a, n);
    }
}

// Driver Code
    let m = 2;

    print_seq(m);

</script>

Output:

0000
0101
0110
1001
1010
1111

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