All possible binary numbers of length n with equal sum in both halves

• Difficulty Level : Hard
• Last Updated : 07 Jul, 2021

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

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 Codepublic 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 Codeif __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 Codepublic static void Main(String []args){    int n = 5;         // n-bits    equal(n, "", "", 0);}} // This code is contributed by 29AjayKumar

Javascript



Output:

10001 10101 10010 10110 11011 11111

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