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All possible binary numbers of length n with equal sum in both halves
• Difficulty Level : Hard
• Last Updated : 30 Sep, 2020

Given a number n, we need to print all n-digit binary numbers with equal sum in left and right halves. If n is odd, then mid element can be either 0 or 1.

Examples:

```Input  : n = 4
Output : 0000 0101 0110 1001 1010 1111
Input : n = 5
Output : 00000 00100 01001 01101 01010 01110 10001 10101 10010 10110 11011 11111

```

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

The idea is to recursively build left and right halves and keep track of difference between counts of 1s in them. We print a string when difference becomes 0 and there are no more characters to add.

## C++

 `// C++ program to generate all binary strings with``// equal sums in left and right halves.``#include ``using` `namespace` `std;`` ` `/* Default values are used only in initial call.``   ``n is number of bits remaining to be filled``   ``di is current difference between sums of``      ``left and right halves.``   ``left and right are current half substrings. */``void` `equal(``int` `n, string left=``""``, string right=``""``,``                                        ``int` `di=0)``{``    ``// TWO BASE CASES``    ``// If there are no more characters to add (n is 0)``    ``if` `(n == 0)``    ``{``        ``// If difference between counts of 1s and``        ``// 0s is 0 (di is 0)``        ``if` `(di == 0)``            ``cout << left + right << ``" "``;``        ``return``;``    ``}`` ` `    ``/* If 1 remains than string length was odd */``    ``if` `(n == 1)``    ``{``        ``// If difference is 0, we can put remaining``        ``// bit in middle.``        ``if` `(di == 0)``        ``{``            ``cout << left + ``"0"` `+ right << ``" "``;``            ``cout << left + ``"1"` `+ right << ``" "``;``        ``}``        ``return``;``    ``}`` ` `    ``/* If difference is more than what can be``       ``be covered with remaining n digits``       ``(Note that lengths of left and right``        ``must be same) */``    ``if` `((2 * ``abs``(di) <= n))``    ``{  ``         ` `         ``/* add 0 to end in both left and right``         ``half. Sum in both half will not``               ``change*/``         ``equal(n-2, left+``"0"``, right+``"0"``, di);`` ` `         ``/* add 0 to end in both left and right``         ``half* subtract 1 from di as right``         ``sum is increased by 1*/``         ``equal(n-2, left+``"0"``, right+``"1"``, di-1);`` ` `        ``/* add 1  in end in left half and 0 to the``        ``right half. Add 1 to di as left sum is``        ``increased by 1*/``        ``equal(n-2, left+``"1"``, right+``"0"``, di+1);`` ` `        ``/* add 1 in end to both left and right``          ``half the sum will not change*/``        ``equal(n-2, left+``"1"``, right+``"1"``, di);``    ``}``}`` ` `/* driver function */``int` `main()``{``    ``int` `n = 5; ``// n-bits``    ``equal(n);``    ``return` `0;``}`

## Java

 `// Java program to generate all binary strings ``// with equal sums in left and right halves.``import` `java.util.*;`` ` `class` `GFG``{`` ` `// Default values are used only in initial call.``// n is number of bits remaining to be filled``// di is current difference between sums of``// left and right halves.``// left and right are current half substrings. ``static` `void` `equal(``int` `n, String left, ``                         ``String right, ``int` `di)``{``    ``// TWO BASE CASES``    ``// If there are no more characters to add (n is 0)``    ``if` `(n == ``0``)``    ``{``        ``// If difference between counts of 1s and``        ``// 0s is 0 (di is 0)``        ``if` `(di == ``0``)``            ``System.out.print(left + right + ``" "``);``        ``return``;``    ``}`` ` `    ``/* If 1 remains than string length was odd */``    ``if` `(n == ``1``)``    ``{``        ``// If difference is 0, we can put ``        ``// remaining bit in middle.``        ``if` `(di == ``0``)``        ``{``            ``System.out.print(left + ``"0"` `+ right + ``" "``);``            ``System.out.print(left + ``"1"` `+ right + ``" "``);``        ``}``        ``return``;``    ``}`` ` `    ``/* If difference is more than what can be``    ``be covered with remaining n digits``    ``(Note that lengths of left and right``     ``must be same) */``    ``if` `((``2` `* Math.abs(di) <= n))``    ``{``         ` `            ``// add 0 to end in both left and right``            ``// half. Sum in both half will not``            ``// change``            ``equal(n - ``2``, left + ``"0"``, right + ``"0"``, di);`` ` `            ``// add 0 to end in both left and right``            ``// half* subtract 1 from di as right``            ``// sum is increased by 1``            ``equal(n - ``2``, left + ``"0"``, right + ``"1"``, di - ``1``);``         ` ` ` `        ``// add 1 in end in left half and 0 to the``        ``// right half. Add 1 to di as left sum is``        ``// increased by 1*``        ``equal(n - ``2``, left + ``"1"``, right + ``"0"``, di + ``1``);`` ` `        ``// add 1 in end to both left and right``        ``// half the sum will not change``        ``equal(n - ``2``, left + ``"1"``, right + ``"1"``, di);``    ``}``}`` ` `// Driver Code``public` `static` `void` `main(String args[])``{``    ``int` `n = ``5``; ``     ` `    ``// n-bits``    ``equal(n, ``""``, ``""``, ``0``);``}``}`` ` `// This code is contributed ``// by SURENDRA_GANGWAR`

## Python3

 `# Python program to generate all binary strings with``# equal sums in left and right halves.`` ` `# Default values are used only in initial call.``# n is number of bits remaining to be filled``# di is current difference between sums of``# left and right halves.``# left and right are current half substrings.``def` `equal(n: ``int``, left ``=` `"``", right = "``", di ``=` `0``):`` ` `    ``# TWO BASE CASES``    ``# If there are no more characters to add (n is 0)``    ``if` `n ``=``=` `0``:`` ` `        ``# If difference between counts of 1s and``        ``# 0s is 0 (di is 0)``        ``if` `di ``=``=` `0``:``            ``print``(left ``+` `right, end ``=` `" "``)``        ``return`` ` `    ``# If 1 remains than string length was odd``    ``if` `n ``=``=` `1``:`` ` `        ``# If difference is 0, we can put remaining``        ``# bit in middle.``        ``if` `di ``=``=` `0``:``            ``print``(left ``+` `"0"` `+` `right, end ``=` `" "``)``            ``print``(left ``+` `"1"` `+` `right, end ``=` `" "``)``        ``return`` ` `    ``# If difference is more than what can be``    ``# be covered with remaining n digits``    ``# (Note that lengths of left and right``    ``# must be same)``    ``if` `2` `*` `abs``(di) <``=` `n:``      ` `        ``# add 0 to end in both left and right``        ``# half. Sum in both half will not``        ``# change``        ``equal(n ``-` `2``, left ``+` `"0"``, right ``+` `"0"``, di)`` ` `        ``# add 0 to end in both left and right``        ``# half* subtract 1 from di as right``        ``# sum is increased by 1``        ``equal(n ``-` `2``, left ``+` `"0"``, right ``+` `"1"``, di ``-` `1``)`` ` `        ``# add 1 in end in left half and 0 to the``        ``# right half. Add 1 to di as left sum is``        ``# increased by 1``        ``equal(n ``-` `2``, left ``+` `"1"``, right ``+` `"0"``, di ``+` `1``)`` ` `        ``# add 1 in end to both left and right``        ``# half the sum will not change``        ``equal(n ``-` `2``, left ``+` `"1"``, right ``+` `"1"``, di)`` ` `# Driver Code``if` `__name__ ``=``=` `"__main__"``:``    ``n ``=` `5` `# n-bits``    ``equal(``5``)`` ` `# This code is contributed by``# sanjeev2552`

## C#

 `// C# program to generate all binary strings ``// with equal sums in left and right halves.``using` `System;`` ` `class` `GFG``{`` ` `// Default values are used only in initial call.``// n is number of bits remaining to be filled``// di is current difference between sums of``// left and right halves.``// left and right are current half substrings. ``static` `void` `equal(``int` `n, String left, ``                         ``String right, ``int` `di)``{``    ``// TWO BASE CASES``    ``// If there are no more characters``    ``// to add (n is 0)``    ``if` `(n == 0)``    ``{``        ``// If difference between counts of 1s ``        ``// and 0s is 0 (di is 0)``        ``if` `(di == 0)``            ``Console.Write(left + right + ``" "``);``        ``return``;``    ``}`` ` `    ``/* If 1 remains than string length was odd */``    ``if` `(n == 1)``    ``{``        ``// If difference is 0, we can put ``        ``// remaining bit in middle.``        ``if` `(di == 0)``        ``{``            ``Console.Write(left + ``"0"` `+ ``                          ``right + ``" "``);``            ``Console.Write(left + ``"1"` `+ ``                          ``right + ``" "``);``        ``}``        ``return``;``    ``}`` ` `    ``/* If difference is more than what can be``    ``be covered with remaining n digits``    ``(Note that lengths of left and right``    ``must be same) */``    ``if` `((2 * Math.Abs(di) <= n))``    ``{`` ` `         ` `            ``// add 0 to end in both left and right``            ``// half. Sum in both half will not``            ``// change``            ``equal(n - 2, left + ``"0"``, right + ``"0"``, di);`` ` `            ``// add 0 to end in both left and right``            ``// half* subtract 1 from di as right``            ``// sum is increased by 1``            ``equal(n - 2, left + ``"0"``, ``                  ``right + ``"1"``, di - 1);``         ` ` ` `        ``// add 1 in end in left half and 0 to the``        ``// right half. Add 1 to di as left sum is``        ``// increased by 1*``        ``equal(n - 2, left + ``"1"``, ``              ``right + ``"0"``, di + 1);`` ` `        ``// add 1 in end to both left and right``        ``// half the sum will not change``        ``equal(n - 2, left + ``"1"``, right + ``"1"``, di);``    ``}``}`` ` `// Driver Code``public` `static` `void` `Main(String []args)``{``    ``int` `n = 5; ``     ` `    ``// n-bits``    ``equal(n, ``""``, ``""``, 0);``}``}`` ` `// This code is contributed by 29AjayKumar`

Output:

```10001 10101 10010 10110 11011 11111
```

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