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

• Difficulty Level : Medium
• Last Updated : 10 Oct, 2022

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

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

Input:  N = 3
Output:
011011 001001 011101 010001 101011 111111
110011 101101 100001 110101 001010 011110
010010 001100 000000 010100 101110 100010
110110 100100

The idea is to fix first and last bits and then recur for remaining 2*(n-1) bits. There are four possibilities when we fix first and last bits –

1. First and last bits are 1, remaining n – 1 bits on both sides should also have the same sum.
2. First and last bits are 0, remaining n – 1 bits on both sides should also have the same sum.
3. First bit is 1 and last bit is 0, sum of remaining n – 1 bits on left side should be 1 less than the sum n-1 bits on right side.
4. First bit is 0 and last bit is 1, sum of remaining n – 1 bits on left side should be 1 more than the sum n-1 bits on right side.

Below is implementation of above idea â€“

## C++

 // C++ program to print even length binary sequences// whose sum of first and second half bits is same#include using namespace std; // Function to print even length binary sequences// whose sum of first and second half bits is same // diff --> difference between sums of first n bits// and last n bits// out --> output array// start --> starting index// end --> ending indexvoid findAllSequences(int diff, char* out, int start, int end){    // We can't cover difference of more than n with 2n bits    if (abs(diff) > (end - start + 1) / 2)        return;     // if all bits are filled    if (start > end)    {        // if sum of first n bits and last n bits are same        if (diff == 0)            cout << out << " ";        return;    }     // fill first bit as 0 and last bit as 1    out[start] = '0', out[end] = '1';    findAllSequences(diff + 1, out, start + 1, end - 1);     // fill first and last bits as 1    out[start] = out[end] = '1';    findAllSequences(diff, out, start + 1, end - 1);     // fill first and last bits as 0    out[start] = out[end] = '0';    findAllSequences(diff, out, start + 1, end - 1);     // fill first bit as 1 and last bit as 0    out[start] = '1', out[end] = '0';    findAllSequences(diff - 1, out, start + 1, end - 1);} // Driver programint main(){    // input number    int n = 2;     // allocate string containing 2*n characters    char out[2 * n + 1];     // null terminate output array    out[2 * n] = '\0';     findAllSequences(0, out, 0, 2*n - 1);     return 0;}

## Java

 // Java program to print even length binary// sequences whose sum of first and second// half bits is sameimport java.io.*;import java.util.*; class GFG{    // Function to print even length binary sequences    // whose sum of first and second half bits is same      // diff --> difference between sums of first n bits    // and last n bits    // out --> output array    // start --> starting index    // end --> ending index    static void findAllSequences(int diff, char out[],                                     int start, int end)    {        // We can't cover difference of more        // than n with 2n bits        if (Math.abs(diff) > (end - start + 1) / 2)            return;          // if all bits are filled        if (start > end)        {            // if sum of first n bits and            // last n bits are same            if (diff == 0)            {                System.out.print(out);                System.out.print(" ");            }               return;        }          // fill first bit as 0 and last bit as 1        out[start] = '0';        out[end] = '1';        findAllSequences(diff + 1, out, start + 1, end - 1);          // fill first and last bits as 1        out[start] = out[end] = '1';        findAllSequences(diff, out, start + 1, end - 1);          // fill first and last bits as 0        out[start] = out[end] = '0';        findAllSequences(diff, out, start + 1, end - 1);          // fill first bit as 1 and last bit as 0        out[start] = '1';        out[end] = '0';        findAllSequences(diff - 1, out, start + 1, end - 1);    }         // Driver program    public static void main (String[] args)    {        // input number        int n = 2;          // allocate string containing 2*n characters        char[] out = new char[2 * n + 1];          // null terminate output array        out[2 * n] = '\0';          findAllSequences(0, out, 0, 2*n - 1);    }} // This code is contributed by Pramod Kumar

## Python3

 # Python3 program to print even length binary sequences# whose sum of first and second half bits is same # Function to print even length binary sequences# whose sum of first and second half bits is same # diff --> difference between sums of first n bits# and last n bits# out --> output array# start --> starting index# end --> ending indexdef findAllSequences(diff, out, start, end):     # We can't cover difference of more than n with 2n bits    if (abs(diff) > (end - start + 1) // 2):        return;     # if all bits are filled    if (start > end):        # if sum of first n bits and last n bits are same        if (diff == 0):            print(''.join(list(out)),end=" ");        return;     # fill first bit as 0 and last bit as 1    out[start] = '0';    out[end] = '1';    findAllSequences(diff + 1, out, start + 1, end - 1);     # fill first and last bits as 1    out[start] = out[end] = '1';    findAllSequences(diff, out, start + 1, end - 1);     # fill first and last bits as 0    out[start] = out[end] = '0';    findAllSequences(diff, out, start + 1, end - 1);     # fill first bit as 1 and last bit as 0    out[start] = '1';    out[end] = '0';    findAllSequences(diff - 1, out, start + 1, end - 1); # Driver program # input numbern = 2; # allocate string containing 2*n charactersout=[""]*(2*n); findAllSequences(0, out, 0, 2*n - 1); # This code is contributed by mits

## C#

 // C# program to print even length binary// sequences whose sum of first and second// half bits is sameusing System; class GFG {         // Function to print even length binary    // sequences whose sum of first and    // second half bits is same     // diff --> difference between sums of    // first n bits    // and last n bits    // out --> output array    // start --> starting index    // end --> ending index    static void findAllSequences(int diff,            char []outt, int start, int end)    {                 // We can't cover difference of        // more than n with 2n bits        if (Math.Abs(diff) > (end - start                                   + 1) / 2)            return;         // if all bits are filled        if (start > end)        {                         // if sum of first n bits and            // last n bits are same            if (diff == 0)            {                Console.Write(outt);                Console.Write(" ");            }            return;        }         // fill first bit as 0 and last bit        // as 1        outt[start] = '0';        outt[end] = '1';        findAllSequences(diff + 1, outt,                        start + 1, end - 1);         // fill first and last bits as 1        outt[start] = outt[end] = '1';        findAllSequences(diff, outt,                        start + 1, end - 1);         // fill first and last bits as 0        outt[start] = outt[end] = '0';        findAllSequences(diff, outt,                         start + 1, end - 1);         // fill first bit as 1 and last        // bit as 0        outt[start] = '1';        outt[end] = '0';        findAllSequences(diff - 1, outt,                         start + 1, end - 1);    }         // Driver program    public static void Main ()    {                 // input number        int n = 2;         // allocate string containing 2*n        // characters        char []outt = new char[2 * n + 1];         // null terminate output array        outt[2 * n] = '\0';         findAllSequences(0, outt, 0, 2*n - 1);    }} // This code is contributed by nitin mittal.

## PHP

 difference between sums of first n bits// and last n bits// out --> output array// start --> starting index// end --> ending indexfunction findAllSequences(\$diff, \$out, \$start, \$end){    // We can't cover difference of more than n with 2n bits    if (abs(\$diff) > (int)((\$end - \$start + 1) / 2))        return;     // if all bits are filled    if (\$start > \$end)    {        // if sum of first n bits and last n bits are same        if (\$diff == 0)            print(implode("",\$out)." ");        return;    }     // fill first bit as 0 and last bit as 1    \$out[\$start] = '0';    \$out[\$end] = '1';    findAllSequences(\$diff + 1, \$out, \$start + 1, \$end - 1);     // fill first and last bits as 1    \$out[\$start] = \$out[\$end] = '1';    findAllSequences(\$diff, \$out, \$start + 1, \$end - 1);     // fill first and last bits as 0    \$out[\$start] = \$out[\$end] = '0';    findAllSequences(\$diff, \$out, \$start + 1, \$end - 1);     // fill first bit as 1 and last bit as 0    \$out[\$start] = '1';    \$out[\$end] = '0';    findAllSequences(\$diff - 1, \$out, \$start + 1, \$end - 1);} // Driver program     // input number    \$n = 2;     // allocate string containing 2*n characters    \$out=array_fill(0,2*\$n,"");     findAllSequences(0, \$out, 0, 2*\$n - 1); // This code is contributed by chandan_jnu?>

## Javascript



Output

0101 1111 1001 0110 0000 1010

Time Complexity: O(()* N)

4^N because of 4 recursive calls, and N (simplified from 2N) for time spent printing strings of size 2N

Auxiliary Space: O(N)

There is another approach by which we generate all possible strings of length n and store them in a list at an index representing their sum. Then, we iterate through each list and generate the strings of size 2n by printing each string with all other strings in the list adding up to the same value.

## C++

 // C++ program to implement the approach #include  using namespace std; //function that generates the sequencevoid generateSequencesWithSum(    int n, vector >& sumToString,    vector sequence, int sumSoFar){    // Base case, if there are no more binary digits to    // include    if (n == 0) {        // add permutation to list of sequences with sum        // corresponding to index        string seq = "";        for (int i = 0; i < sequence.size(); i++) {            seq = seq + sequence[i];        }        vector x = sumToString[sumSoFar];        x.push_back(seq);        sumToString[sumSoFar] = x;        return;    }    // Generate sequence +0    sequence.push_back("0");    generateSequencesWithSum(n - 1, sumToString, sequence,                             sumSoFar);    sequence.erase(sequence.begin());    // Generate sequence +1    sequence.push_back("1");    generateSequencesWithSum(n - 1, sumToString, sequence,                             sumSoFar + 1);    sequence.erase(sequence.begin());} // function to form permutations of the sequencesvoid permuteSequences(vector > sumToString){    // There are 2^n substring in this list of lists    for (int sumIndexArr = 0;         sumIndexArr < sumToString.size(); sumIndexArr++) {        // Append        for (int sequence1 = 0;             sequence1 < sumToString[sumIndexArr].size();             sequence1++) {            for (int sequence2 = 0;                 sequence2                 < sumToString[sumIndexArr].size();                 sequence2++) {                if (sumIndexArr == sumToString.size() - 1                    && sequence1                           == sumToString[sumIndexArr]                                      .size()                                  - 1                    && sequence2                           == sumToString[sumIndexArr]                                      .size()                                  - 1) {                    cout << "1111 ";                }                else {                    cout << sumToString[sumIndexArr]                                       [sequence1]                                + sumToString[sumIndexArr]                                             [sequence2]                         << " ";                }            }        }    }} // function that finds all the subsequencesvoid findAllSequences(int n){    vector > sumToString;    for (int i = 0; i < n + 1; i++) {        sumToString.push_back(            vector()); // list of strings                               // where index                               // represents sum    }    generateSequencesWithSum(n, sumToString,                             vector(), 0);    permuteSequences(sumToString);} // Driver Codeint main(){     // Function Call    findAllSequences(2);    return 0;} // this code is contributed by phasing17

## Java

 // Java program to implement the approachimport java.util.*; class GFG {     // function that finds all the subsequences    static void findAllSequences(int n)    {        ArrayList > sumToString            = new ArrayList >();        for (int i = 0; i < n + 1; i++) {            sumToString.add(                new ArrayList()); // list of strings                                          // where index                                          // represents sum        }        generateSequencesWithSum(            n, sumToString, new ArrayList(), 0);        permuteSequences(sumToString);    }    static void generateSequencesWithSum(        int n, ArrayList > sumToString,        ArrayList sequence, int sumSoFar)    {        // Base case, if there are no more binary digits to        // include        if (n == 0) {            // add permutation to list of sequences with sum            // corresponding to index            String seq = "";            for (int i = 0; i < sequence.size(); i++) {                seq = seq + sequence.get(i);            }            ArrayList x = sumToString.get(sumSoFar);            x.add(seq);            sumToString.set(sumSoFar, x);            return;        }        // Generate sequence +0        sequence.add("0");        generateSequencesWithSum(n - 1, sumToString,                                 sequence, sumSoFar);        sequence.remove(0);        // Generate sequence +1        sequence.add("1");        generateSequencesWithSum(n - 1, sumToString,                                 sequence, sumSoFar + 1);        sequence.remove(0);    }     // function to form permutations of the sequences    static void permuteSequences(        ArrayList > sumToString)    {        // There are 2^n substring in this list of lists        for (int sumIndexArr = 0;             sumIndexArr < sumToString.size();             sumIndexArr++) {            // Append            for (int sequence1 = 0;                 sequence1                 < sumToString.get(sumIndexArr).size();                 sequence1++) {                for (int sequence2 = 0;                     sequence2                     < sumToString.get(sumIndexArr).size();                     sequence2++) {                    if (sumIndexArr                            == sumToString.size() - 1                        && sequence1                               == sumToString                                          .get(sumIndexArr)                                          .size()                                      - 1                        && sequence2                               == sumToString                                          .get(sumIndexArr)                                          .size()                                      - 1) {                        System.out.print("1111");                    }                    else {                        System.out.println(                            sumToString.get(sumIndexArr)                                .get(sequence1)                            + sumToString.get(sumIndexArr)                                  .get(sequence2));                    }                }            }        }    }    // Driver Code    public static void main(String[] args)    {        // Function Call        findAllSequences(2);    }     // this code is contributed by phasing17}

## Python3

 def findAllSequences(n):    sumToString = [[] for x in range(n+1)] # list of strings where index represents sum    generateSequencesWithSum(n, sumToString, [], 0)    permuteSequences(sumToString)  def generateSequencesWithSum(n, sumToString, sequence, sumSoFar):    #Base case, if there are no more binary digits to include    if n == 0:        sumToString[sumSoFar].append("".join(sequence)) #add permutation to list of sequences with sum corresponding to index        return    #Generate sequence +0    sequence.append("0")    generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar)    sequence.pop()    #Generate sequence +1    sequence.append("1")    generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar+1)    sequence.pop()  def permuteSequences(sumToString):    #There are 2^n substring in this list of lists    for sumIndexArr in sumToString:        # Append        for sequence1 in sumIndexArr:            for sequence2 in sumIndexArr:                print(sequence1 + sequence2)  findAllSequences(2) #Contribution by Xavier Jean Baptiste

## C#

 using System;using System.Collections.Generic;class GFG {   static void findAllSequences(int n)  {    List> sumToString = new List>();    for(int i = 0; i < n + 1; i++)    {      sumToString.Add(new List()); // list of strings where index represents sum    }    generateSequencesWithSum(n, sumToString, new List(), 0);    permuteSequences(sumToString);  }   static void generateSequencesWithSum(int n, List> sumToString, List sequence, int sumSoFar)  {    // Base case, if there are no more binary digits to include    if(n == 0)    {      //add permutation to list of sequences with sum corresponding to index      string seq = "";      for(int i = 0; i < sequence.Count; i++)      {        seq = seq + sequence[i];      }      sumToString[sumSoFar].Add(seq);      return;    }    // Generate sequence +0    sequence.Add("0");    generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar);    sequence.RemoveAt(0);    // Generate sequence +1    sequence.Add("1");    generateSequencesWithSum(n-1, sumToString, sequence, sumSoFar+1);    sequence.RemoveAt(0);  }   static void permuteSequences(List> sumToString)  {    // There are 2^n substring in this list of lists    for(int sumIndexArr = 0; sumIndexArr < sumToString.Count; sumIndexArr++)    {      // Append      for(int sequence1 = 0; sequence1 < sumToString[sumIndexArr].Count; sequence1++)      {        for(int sequence2 = 0; sequence2 < sumToString[sumIndexArr].Count; sequence2++)        {          if(sumIndexArr == sumToString.Count-1 && sequence1 == sumToString[sumIndexArr].Count-1 && sequence2 == sumToString[sumIndexArr].Count-1)          {            Console.Write("1111");          }          else          {            Console.WriteLine(sumToString[sumIndexArr][sequence1] + sumToString[sumIndexArr][sequence2]);          }        }      }    }  }   static void Main() {    findAllSequences(2);  }} // This code is contributed by divyesh072019.

## Javascript



Output

0000 0101 0110 1001 1010 1111

Time complexity analysis:

generateSequencesWithSum = O((2N)*N)

• 2N: we generate all permutation of binary strings of size N
• N: convert the list of characters to a string and store into array. This is done in the base case.

permuteSequences = O((2N) * N!/(N/2)!2 * N)

• 2N: we iterate through all the string generated of size n
• N!/(N/2)!2: This one is a bit challenging to explain

let’s take N = 2 as an example. Our array of possible sequence of size n would be:

In the list of strings which the index represents the sum, we get the count of strings of size 2n by using “n choose k” formula. I our case it would be nCk *nCk where k represents the number of 1s in each half of the string of size 2n:

k = 0, we have (2C0)^2 = 1 string (0000)

k =  1, we have (2C1)^2 string = 4 strings(0101 0110 1001 1010)

k = 2, we have (2c2)^2 = 1 string (1111)

We get our longest list of string when k = N/2, hence NCN/2 = N!/[(N/2)! * (N – N/2)!]  which simplifies to NCN/2 = N!/(N/2)!2

Hence, for each element, we must iterate through, at most, NCN/2 for forming strings of length 2N

Without formal proof, if we graph 2^N and N!/(N/2)!2, we see that 2N has a faster growth rate than the latter. Therefore O(2N* N!/(N/2)2 ) < O(2N*2N) = O(22n) = O(4N)

Graph of 2^x and nC(n/2)

• N: we must print each string of size 2N

Finally we can ignore the time complexity of generateSequencesWithSum because permuteSequence is the leading term

Time complexity: O(2N * N!/(N/2)!2 * N) (better than the first solution of O((4^N) * N, see explanation above for further details)

Auxiliary space: O(2N) because we store all binary string permutations of size N

This article is contributed by Aditya Goel and improved by Xavier Jean Baptiste. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

## Java

 import java.util.*; class GFG {    static class FirstHalf {        String data;        int sum;         FirstHalf(String data, int sum) {            this.data = data;            this.sum = sum;        }    }     //MAP: Key -> sum of bits; Value -> All possible permutation with respective sum    static Map> map = new HashMap<>();     //first N-half bits    static List firstHalf = new ArrayList<>();     //function to find sum of the bits from a String    public static int sumOfString(String s) {        int sum = 0;        //ex: converts "1" to 1 -> (ASCII('1') - ASCII('0') = 1)        for(char c: s.toCharArray()) {            sum += c - '0';        }        return sum;    }     public static void perm(String p, char[] bin, int level, int n) {        //p: processed string(processed permutation at current level)        //bin: {'0', '1'}        //l: current level of recursion tree (leaf/solution level = 0)        //n: total levels        if(level == 0) {            //at solution level find sum of the current permutation            int sum = sumOfString(p);            //store current permutation to firstHalf list            firstHalf.add(new FirstHalf(p, sum));            //put current permutation to its respective sum value            map.putIfAbsent(sum, new ArrayList());            map.get(sum).add(p);            return;        }         //generate calls for permutation        //working: first solution with all 0s, then replacing last 0 with 1 and so on...        for(char c: bin) {            perm(p+c, bin, level-1, n);        }    }     public static void result() {        int i = 0;        for(FirstHalf first: firstHalf) {            //for each firstHalf string            //find sum of the bits of current string            int sum = first.sum;            //retrieve respective secondHalf from map based on sum key            ArrayList secondHalf = map.get(sum);            for(String second: secondHalf) {                //append first and second half and print                System.out.print(first.data+second+" ");                //after every 6 solution line is changed in output                //only for formatting below lines could be removed                i++;                if(i % 6 == 0)                    System.out.println();            }        }    }      public static void main(String[] args) {        char[] up = {'0', '1'};        int n = 2;        perm("", up, n, n);        result();    }}//Code contributed by Animesh Singh

Output

0000 0101 0110 1001 1010 1111

Algorithm:

1. Generate all binary permutations of size n

2. Calculate sum of the bits of each permutation and remember it for second half

[for ex: for n=2, remember there are two strings with sum = 1 i.e. “01”, “10” ]

3. Iterate all the generated permutations and for each of them append the second half according to the sum of the bits

Time complexity analysis:

sumOfString() = O(N) : traverse each bit and add it to sum

perm() = O(2N * N)

2N * N : we generate all permutations of binary bits of size N and find sum of the bits for each permutation

result() = O((2N) * (N!/(N/2)!)2)

2N: we iterate through all possible permutations of size N (first-Half)
NCN/2 = N!/(N/2)!2 : (second-Half maximum size) : explanation below:

let’s take N = 4 as an example.:

//Hash-Map looks like

0 -> [0000] …………………………..(list-size: 4C0 = 1)
1 -> [0001, 0010, 0100, 1000] …………………………..(list-size: 4C1 = 4)
2 -> [0011, 0101, 0110, 1001, 1010, 1100] …………………………..(list-size: 4C2 = 6)
3 -> [0111, 1011, 1101, 1110] …………………………..(list-size: 4C3 = 4)
4 -> [1111] …………………………..(list-size: 4C4 = 1)

We observe here that each list has a size of N choose Key which will be maximum at N choose N/2

Since we are iterating all the 2N permutations and appending second half from the map. The map has the maximum sized list at N/2 position.

Worst case occurs in N/2 position where we’ve to traverse NCN/2 = N!/(N/2)!2 permutations.

Time complexity: O(2N * N!/(N/2)!2 )

Auxiliary space: O(2N) because we store all binary string permutations of size N

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