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Find all even length binary sequences with same sum of first and second half bits | Iterative
• Difficulty Level : Medium
• Last Updated : 24 Oct, 2019

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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Note: The recursive approach to this problem can be found here.
Approach:
A simple approach to run a loop from 0 to 22*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 ` `#include ` `#include ` `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 prsequence ` `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 `

Output:

```0000
0101
0110
1001
1010
1111
```

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