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Count even length binary sequences with same sum of first and second half bits

  • Difficulty Level : Medium
  • Last Updated : 25 Aug, 2021

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

Examples: 

Input:  n = 1
Output: 2
There are 2 sequences of length 2*n, the
sequences are 00 and 11

Input:  n = 2
Output: 6
There are 6 sequences of length 2*n, the
sequences are 0101, 0110, 1010, 1001, 0000
and 1111

The idea is to fix first and last bits and then recur for n-1, i.e., remaining 2(n-1) bits. There are following possibilities when we fix first and last bits. 
1) First and last bits are same, remaining n-1 bits on both sides should also have the same sum. 
2) 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. 
2) 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.

Based on above facts, we get below recurrence formula.
diff is the expected difference between sum of first half digits and last half digits. Initially diff is 0. 

                  // When first and last bits are same
                  // there are two cases, 00 and 11
count(n, diff) =  2*count(n-1, diff) +    
    
                 // When first bit is 1 and last bit is 0
                 count(n-1, diff-1) +

                 // When first bit is 0 and last bit is 1
                 count(n-1, diff+1)

What should be base cases?
// When n == 1 (2 bit sequences)
1) If n == 1 and diff == 0, return 2
2) If n == 1 and |diff| == 1, return 1 

// We can't cover difference of more than n with 2n bits
3) If |diff| > n, return 0

Below is the implementation based of above Naive Recursive Solution



C++




// A Naive Recursive C++ program to count even
// length binary sequences such that the sum of
// first and second half bits is same
#include<bits/stdc++.h>
using namespace std;
 
// diff is difference between sums first n bits
// and last n bits respectively
int countSeq(int n, int diff)
{
    // We can't cover difference of more
    // than n with 2n bits
    if (abs(diff) > n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if (n == 1 && diff == 0)
        return 2;
    if (n == 1 && abs(diff) == 1)
        return 1;
 
    int res = // First bit is 0 & last bit is 1
              countSeq(n-1, diff+1) +
 
              // First and last bits are same
              2*countSeq(n-1, diff) +
 
              // First bit is 1 & last bit is 0
              countSeq(n-1, diff-1);
 
    return res;
}
 
// Driver program
int main()
{
    int n = 2;
    cout << "Count of sequences is "
         << countSeq(2, 0);
    return 0;
}

Java




// A Naive Recursive Java program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
import java.io.*;
 
class GFG {
 
// diff is difference between sums
// first n bits and last n bits respectively
static int countSeq(int n, int diff)
{
    // We can't cover difference of more
    // than n with 2n bits
    if (Math.abs(diff) > n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if (n == 1 && diff == 0)
        return 2;
    if (n == 1 && Math.abs(diff) == 1)
        return 1;
 
    int res = // First bit is 0 & last bit is 1
            countSeq(n-1, diff+1) +
 
            // First and last bits are same
            2*countSeq(n-1, diff) +
 
            // First bit is 1 & last bit is 0
            countSeq(n-1, diff-1);
 
    return res;
}
 
// Driver program
public static void main(String[] args)
{
    int n = 2;
    System.out.println("Count of sequences is "
                       + countSeq(2, 0));
}
}
 
// This code is contributed by Prerna Saini

Python3




# A Naive Recursive Python
# program to count even length
# binary sequences such that
# the sum of first and second
# half bits is same
 
# diff is difference between
# sums first n bits and last
# n bits respectively
def countSeq(n, diff):
 
    # We can't cover difference
    # of more than n with 2n bits
    if (abs(diff) > n):
        return 0
 
    # n == 1, i.e., 2
    # bit long sequences
    if (n == 1 and diff == 0):
        return 2
    if (n == 1 and abs(diff) == 1):
        return 1
 
    # First bit is 0 & last bit is 1
    # First and last bits are same
    # First bit is 1 & last bit is 0
    res = (countSeq(n - 1, diff + 1) +
           2 * countSeq(n - 1, diff) +
            countSeq(n - 1, diff - 1))    
             
    return res
 
# Driver Code
n = 2;
print("Count of sequences is %d " %
                  (countSeq(2, 0)))
     
# This code is contributed
# by Shivi_Aggarwal

C#




// A Naive Recursive C# program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
using System;
 
class GFG {
 
    // diff is difference between sums
    // first n bits and last n bits
    // respectively
    static int countSeq(int n, int diff)
    {
        // We can't cover difference
        // of more than n with 2n bits
        if (Math.Abs(diff) > n)
            return 0;
     
        // n == 1, i.e., 2 bit long
        // sequences
        if (n == 1 && diff == 0)
            return 2;
        if (n == 1 && Math.Abs(diff) == 1)
            return 1;
     
        // 1. First bit is 0 & last bit is 1
        // 2. First and last bits are same
        // 3. First bit is 1 & last bit is 0
        int res = countSeq(n-1, diff+1) +
                2 * countSeq(n-1, diff) +
                   countSeq(n-1, diff-1);
     
        return res;
    }
     
    // Driver program
    public static void Main()
    {
        Console.Write("Count of sequences is "
                             + countSeq(2, 0));
    }
}
 
// This code is contributed by nitin mittal.

PHP




<?php
// A Naive Recursive PHP program
// to count even length binary
// sequences such that the sum of
// first and second half bits is same
 
// diff is difference between
// sums first n bits and last
// n bits respectively
function countSeq($n, $diff)
{
    // We can't cover difference of
    // more than n with 2n bits
    if (abs($diff) > $n)
        return 0;
 
    // n == 1, i.e., 2
    // bit long sequences
    if ($n == 1 && $diff == 0)
        return 2;
         
    if ($n == 1 && abs($diff) == 1)
        return 1;
 
    $res = // First bit is 0 & last bit is 1
            countSeq($n - 1, $diff + 1) +
 
            // First and last bits are same
            2 * countSeq($n - 1, $diff) +
 
            // First bit is 1 & last bit is 0
            countSeq($n - 1, $diff - 1);
 
    return $res;
}
 
// Driver Code
$n = 2;
echo "Count of sequences is ",
              countSeq($n, 0);
 
// This code is contributed
// by shiv_bhakt.
?>

Javascript




<script>
// A Naive Recursive Javascript program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
 
    // diff is difference between sums
    // first n bits and last n bits respectively
    function countSeq(n,diff)
    {
     
        // We can't cover difference of more
        // than n with 2n bits
        if (Math.abs(diff) > n)
            return 0;
       
        // n == 1, i.e., 2 bit long sequences
        if (n == 1 && diff == 0)
            return 2;
        if (n == 1 && Math.abs(diff) == 1)
            return 1;
       
        let res = // First bit is 0 & last bit is 1
                countSeq(n-1, diff+1) +
       
                // First and last bits are same
                2*countSeq(n-1, diff) +
       
                // First bit is 1 & last bit is 0
                countSeq(n-1, diff-1);
       
        return res;
    }
     
    // Driver program
    let n = 2;
    document.write("Count of sequences is "
                       + countSeq(2, 0));
     
    // This code is contributed by avanitrachhadiya2155
</script>

Output: 

Count of sequences is 6

The time complexity of above solution is exponential. If we draw the complete recursion tree, we can observer that many subproblems are solved again and again. For example, when we start from n = 4 and diff = 0, we can reach (3, 0) through multiple paths. Since same subproblems are called again, this problem has Overlapping subproblems property. So min square sum problem has both properties (see this and this) of a Dynamic Programming problem. 

Below is a memoization based solution that uses a lookup table to compute the result. 

C++




// A memoization based C++ program to count even
// length binary sequences such that the sum of
// first and second half bits is same
#include<bits/stdc++.h>
using namespace std;
#define MAX 1000
 
// A lookup table to store the results of subproblems
int lookup[MAX][MAX];
 
// dif is difference between sums of first n bits
// and last n bits i.e., dif = (Sum of first n bits) -
//                              (Sum of last n bits)
int countSeqUtil(int n, int dif)
{
    // We can't cover difference of more
    // than n with 2n bits
    if (abs(dif) > n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if (n == 1 && dif == 0)
        return 2;
    if (n == 1 && abs(dif) == 1)
        return 1;
 
    // Check if this subproblem is already solved
    // n is added to dif to make sure index becomes
    // positive
    if (lookup[n][n+dif] != -1)
        return lookup[n][n+dif];
 
    int res = // First bit is 0 & last bit is 1
              countSeqUtil(n-1, dif+1) +
 
              // First and last bits are same
              2*countSeqUtil(n-1, dif) +
 
              // First bit is 1 & last bit is 0
              countSeqUtil(n-1, dif-1);
 
    // Store result in lookup table and return the result
    return lookup[n][n+dif] = res;
}
 
// A Wrapper over countSeqUtil().  It mainly initializes lookup
// table, then calls countSeqUtil()
int countSeq(int n)
{
    // Initialize all entries of lookup table as not filled
    memset(lookup, -1, sizeof(lookup));
 
    // call countSeqUtil()
    return countSeqUtil(n, 0);
}
 
// Driver program
int main()
{
    int n = 2;
    cout << "Count of sequences is "
         << countSeq(2);
    return 0;
}

Java




// A memoization based Java program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
import java.io.*;
 
class GFG {
     
// A lookup table to store the results of
// subproblems
static int lookup[][] = new int[1000][1000];
 
// dif is difference between sums of first
// n bits and last n bits i.e.,
// dif = (Sum of first n bits) - (Sum of last n bits)
static int countSeqUtil(int n, int dif)
{
    // We can't cover difference of
    // more than n with 2n bits
    if (Math.abs(dif) > n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if (n == 1 && dif == 0)
        return 2;
    if (n == 1 && Math.abs(dif) == 1)
        return 1;
 
    // Check if this subproblem is already
    // solved n is added to dif to make
    // sure index becomes positive
    if (lookup[n][n+dif] != -1)
        return lookup[n][n+dif];
 
    int res = // First bit is 0 & last bit is 1
            countSeqUtil(n-1, dif+1) +
 
            // First and last bits are same
            2*countSeqUtil(n-1, dif) +
 
            // First bit is 1 & last bit is 0
            countSeqUtil(n-1, dif-1);
 
    // Store result in lookup table
    // and return the result
    return lookup[n][n+dif] = res;
}
 
// A Wrapper over countSeqUtil(). It mainly
// initializes lookup table, then calls
// countSeqUtil()
static int countSeq(int n)
{
    // Initialize all entries of lookup
    // table as not filled
    // memset(lookup, -1, sizeof(lookup));
    for(int k = 0; k < lookup.length; k++)
    {
        for(int j = 0; j < lookup.length; j++)
        {
        lookup[k][j] = -1;
    }
    }
     
    // call countSeqUtil()
    return countSeqUtil(n, 0);
}
 
// Driver program
public static void main(String[] args)
{
    int n = 2;
    System.out.println("Count of sequences is "
                       + countSeq(2));
}
}
 
// This code is contributed by Prerna Saini

Python3




#A memoization based python program to count even
#length binary sequences such that the sum of
#first and second half bits is same
  
MAX=1000
 
#A lookup table to store the results of subproblems
lookup=[[0 for i in range(MAX)] for i in range(MAX)]
#dif is difference between sums of first n bits
#and last n bits i.e., dif = (Sum of first n bits) -
#                             (Sum of last n bits)
def countSeqUtil(n,dif):
 
    #We can't cover difference of more
    #than n with 2n bits
    if abs(dif)>n:
        return 0
    #n == 1, i.e., 2 bit long sequences
    if n==1 and dif==0:
        return 2
    if n==1 and abs(dif)==1:
        return 1
 
    #Check if this subproblem is already solved
    #n is added to dif to make sure index becomes
    #positive
    if lookup[n][n+dif]!=-1:
        return lookup[n][n+dif]
 
    #First bit is 0 & last bit is 1
    #+First and last bits are same
    #+First bit is 1 & last bit is 0
    res= (countSeqUtil(n-1, dif+1)+
          2*countSeqUtil(n-1, dif)+
          countSeqUtil(n-1, dif-1))
          
         
    #Store result in lookup table and return the result
    lookup[n][n+dif]=res
    return res
 
#A Wrapper over countSeqUtil(). It mainly initializes lookup
#table, then calls countSeqUtil()
def countSeq(n):
    #Initialize all entries of lookup table as not filled
    global lookup
    lookup=[[-1 for i in range(MAX)] for i in range(MAX)]
    #call countSeqUtil()
    res=countSeqUtil(n,0)
    return res
 
#Driver Code
if __name__=='__main__':
    n=2
    print('Count of Sequences is ',countSeq(n))
     
#This Code is contributed by sahilshelangia

C#




// A memoization based C# program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
 
using System;
class GFG {
     
// A lookup table to store the results of
// subproblems
static int [,]lookup = new int[1000,1000];
 
// dif is difference between sums of first
// n bits and last n bits i.e.,
// dif = (Sum of first n bits) - (Sum of last n bits)
static int countSeqUtil(int n, int dif)
{
    // We can't cover difference of
    // more than n with 2n bits
    if (Math.Abs(dif) > n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if (n == 1 && dif == 0)
        return 2;
    if (n == 1 && Math.Abs(dif) == 1)
        return 1;
 
    // Check if this subproblem is already
    // solved n is added to dif to make
    // sure index becomes positive
    if (lookup[n,n+dif] != -1)
        return lookup[n,n+dif];
 
    int res = // First bit is 0 & last bit is 1
            countSeqUtil(n-1, dif+1) +
 
            // First and last bits are same
            2*countSeqUtil(n-1, dif) +
 
            // First bit is 1 & last bit is 0
            countSeqUtil(n-1, dif-1);
 
    // Store result in lookup table
    // and return the result
    return lookup[n,n+dif] = res;
}
 
// A Wrapper over countSeqUtil(). It mainly
// initializes lookup table, then calls
// countSeqUtil()
static int countSeq(int n)
{
    // Initialize all entries of lookup
    // table as not filled
    // memset(lookup, -1, sizeof(lookup));
    for(int k = 0; k < lookup.GetLength(0); k++)
    {
        for(int j = 0; j < lookup.GetLength(1); j++)
        {
        lookup[k,j] = -1;
    }
    }
     
    // call countSeqUtil()
    return countSeqUtil(n, 0);
}
 
// Driver program
public static void Main()
{
    int n = 2;
    Console.WriteLine("Count of sequences is "
                    + countSeq(n));
}
}
 
// This code is contributed by Ryuga

PHP




<?php
// A memoization based PHP program to count even
// length binary sequences such that the sum of
// first and second half bits is same
$MAX = 1000;
 
// A lookup table to store the results
// of subproblems
$lookup = array_fill(0, $MAX,
          array_fill(0, $MAX, -1));
 
// dif is difference between sums of first n bits
// and last n bits i.e., dif = (Sum of first n bits) -
//                               (Sum of last n bits)
function countSeqUtil($n, $dif)
{
    global $lookup;
     
    // We can't cover difference of more
    // than n with 2n bits
    if (abs($dif) > $n)
        return 0;
 
    // n == 1, i.e., 2 bit long sequences
    if ($n == 1 && $dif == 0)
        return 2;
    if ($n == 1 && abs($dif) == 1)
        return 1;
 
    // Check if this subproblem is already solved
    // n is added to dif to make sure index becomes
    // positive
    if ($lookup[$n][$n + $dif] != -1)
        return $lookup[$n][$n + $dif];
 
    $res = // First bit is 0 & last bit is 1
            countSeqUtil($n - 1, $dif + 1) +
 
            // First and last bits are same
            2 * countSeqUtil($n - 1, $dif) +
 
            // First bit is 1 & last bit is 0
            countSeqUtil($n - 1, $dif - 1);
 
    // Store result in lookup table and return the result
    return $lookup[$n][$n + $dif] = $res;
}
 
// A Wrapper over countSeqUtil(). It mainly
// initializes lookup table, then calls countSeqUtil()
function countSeq($n)
{
    // Initialize all entries of
    // lookup table as not filled
 
    // call countSeqUtil()
    return countSeqUtil($n, 0);
}
 
// Driver Code
$n = 2;
echo "Count of sequences is " . countSeq($n);
 
// This code is contributed by mits
?>

Javascript




<script>
// A memoization based Javascript program to
// count even length binary sequences
// such that the sum of first and
// second half bits is same
     
    // A lookup table to store the results of
    // subproblems
    let lookup = new Array(1000);
    for(let i = 0; i < 1000; i++)
    {
        lookup[i] = new Array(1000);
    }
     
    // dif is difference between sums of first
    // n bits and last n bits i.e.,
    // dif = (Sum of first n bits) - (Sum of last n bits)
    function countSeqUtil(n, dif)
    {
     
        // We can't cover difference of
        // more than n with 2n bits
        if (Math.abs(dif) > n)
            return 0;
      
        // n == 1, i.e., 2 bit long sequences
        if (n == 1 && dif == 0)
            return 2;
        if (n == 1 && Math.abs(dif) == 1)
            return 1;
      
        // Check if this subproblem is already
        // solved n is added to dif to make
        // sure index becomes positive
        if (lookup[n][n + dif] != -1)
            return lookup[n][n + dif];
      
        let res = // First bit is 0 & last bit is 1
                countSeqUtil(n - 1, dif + 1) +
      
                // First and last bits are same
                2*countSeqUtil(n - 1, dif) +
      
                // First bit is 1 & last bit is 0
                countSeqUtil(n - 1, dif - 1);
      
        // Store result in lookup table
        // and return the result
        return lookup[n][n + dif] = res;
    }
     
    // A Wrapper over countSeqUtil(). It mainly
    // initializes lookup table, then calls
    // countSeqUtil()
    function countSeq(n)
    {
     
        // Initialize all entries of lookup
        // table as not filled
        // memset(lookup, -1, sizeof(lookup));
        for(let k = 0; k < lookup.length; k++)
        {
            for(let j = 0; j < lookup.length; j++)
            {
                lookup[k][j] = -1;
            }
        }
          
        // call countSeqUtil()
        return countSeqUtil(n, 0);
    }
     
    // Driver program
    let n = 2;
    document.write("Count of sequences is "
                       + countSeq(2));
     
    // This code is contributed by rag2127
</script>

Output: 

Count of sequences is 6

Worst case time complexity of this solution is O(n2) as diff can be maximum n.

Below is O(n) solution for the same. 

Number of n-bit strings with 0 ones = nC0
Number of n-bit strings with 1 ones = nC1
...
Number of n-bit strings with k ones = nCk
...
Number of n-bit strings with n ones = nCn 

So, we can get required result using below 

No. of 2*n bit strings such that first n bits have 0 ones & 
last n bits have 0 ones = nC0 * nC0

No. of 2*n bit strings such that first n bits have 1 ones & 
last n bits have 1 ones = nC1 * nC1

....

and so on.

Result = nC0*nC0 + nC1*nC1 + ... + nCn*nCn
       = ∑(nCk)2 
        0 <= k <= n 

Below is the implementation based on above idea. 

C++




// A O(n) C++ program to count even length binary sequences
// such that the sum of first and second half bits is same
#include<iostream>
using namespace std;
 
// Returns the count of even length sequences
int countSeq(int n)
{
    int nCr=1, res = 1;
 
    // Calculate SUM ((nCr)^2)
    for (int r = 1; r<=n ; r++)
    {
        // Compute nCr using nC(r-1)
        // nCr/nC(r-1) = (n+1-r)/r;
        nCr = (nCr * (n+1-r))/r;  
 
        res += nCr*nCr;
    }
 
    return res;
}
 
// Driver program
int main()
{
    int n = 2;
    cout << "Count of sequences is "
         << countSeq(n);
    return 0;
}

Java




// Java program to find remaining
// chocolates after k iterations
class GFG
{
// A O(n) C++ program to count
// even length binary sequences
// such that the sum of first
// and second half bits is same
 
// Returns the count of
// even length sequences
static int countSeq(int n)
{
    int nCr = 1, res = 1;
 
    // Calculate SUM ((nCr)^2)
    for (int r = 1; r <= n ; r++)
    {
        // Compute nCr using nC(r-1)
        // nCr/nC(r-1) = (n+1-r)/r;
        nCr = (nCr * (n + 1 - r)) / r;
 
        res += nCr * nCr;
    }
 
    return res;
}
 
// Driver code
public static void main(String args[])
{
    int n = 2;
    System.out.print("Count of sequences is ");
    System.out.println(countSeq(n));
}
}
 
// This code is contributed
// by Shivi_Aggarwal

Python




# A Python program to count
# even length binary sequences
# such that the sum of first
# and second half bits is same
 
# Returns the count of
# even length sequences
def countSeq(n):
 
    nCr = 1
    res = 1
 
    # Calculate SUM ((nCr)^2)
    for r in range(1, n + 1):
     
        # Compute nCr using nC(r-1)
        # nCr/nC(r-1) = (n+1-r)/r;
        nCr = (nCr * (n + 1 - r)) / r;
 
        res += nCr * nCr;
 
    return res;
 
# Driver Code
n = 2
print("Count of sequences is"),
print (int(countSeq(n)))
     
# This code is contributed
# by Shivi_Aggarwal

C#




// C# program to find remaining
// chocolates after k iteration
using System;
 
class GFG {
     
// A O(n) C# program to count
// even length binary sequences
// such that the sum of first
// and second half bits is same
 
// Returns the count of
// even length sequences
static int countSeq(int n)
{
    int nCr = 1, res = 1;
 
    // Calculate SUM ((nCr)^2)
    for (int r = 1; r <= n ; r++)
    {
        // Compute nCr using nC(r-1)
        // nCr/nC(r-1) = (n+1-r)/r;
        nCr = (nCr * (n + 1 - r)) / r;
 
        res += nCr * nCr;
    }
 
    return res;
}
 
// Driver code
public static void Main()
{
    int n = 2;
    Console.Write("Count of sequences is ");
    Console.Write(countSeq(n));
}
}
 
// This code is contributed
// by ChitraNayal

PHP




<?php
// A Php program to count even
// length binary sequences
// such that the sum of first
// and second half bits is same
 
// Returns the count of
// even length sequences
function countSeq($n)
{
    $nCr = 1;
    $res = 1;
 
    // Calculate SUM ((nCr)^2)
    for ($r = 1; $r <= $n; $r++)
    {
        // Compute nCr using nC(r-1)
        // nCr/nC(r-1) = (n+1-r)/r;
        $nCr = ($nCr * ($n + 1 - $r)) / $r;
 
        $res = $res + ($nCr * $nCr);
    }
 
    return $res;
}
 
// Driver Code
$n = 2;
echo ("Count of sequences is ");
echo countSeq($n);
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
    // Javascript program to find remaining chocolates after k iteration
     
    // A O(n) C# program to count
    // even length binary sequences
    // such that the sum of first
    // and second half bits is same
 
    // Returns the count of
    // even length sequences
    function countSeq(n)
    {
        let nCr = 1, res = 1;
 
        // Calculate SUM ((nCr)^2)
        for (let r = 1; r <= n ; r++)
        {
            // Compute nCr using nC(r-1)
            // nCr/nC(r-1) = (n+1-r)/r;
            nCr = (nCr * (n + 1 - r)) / r;
 
            res += nCr * nCr;
        }
 
        return res;
    }
     
    let n = 2;
    document.write("Count of sequences is ");
    document.write(countSeq(n));
     
    // This code is contributed by mukesh07.
</script>

Output: 

Count of sequences is 6

Thanks to d_geeks, Saurabh Jain and Mysterious Mind for suggesting above O(n) solution.
This article is contributed by Pawan. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
 

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