计数具有相同的前半部分和后半部分的偶数长度二进制序列

给定数字n，请找到长度为2n的所有二进制序列的计数，以使前n位的总和与后n位的总和相同。
例子：

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

这个想法是固定第一个和最后一个比特，然后递归n-1个，即剩余的2(n-1)个比特。当我们固定第一位和最后一位时，存在以下可能性。
1)第一位和最后一位相同，两边剩下的n-1位也应具有相同的总和。
2)第一位为1，最后一位为0，左侧剩余的n-1位之和应比右侧剩余的n-1位之和少1。
2)第1位为0和最后的位为1，其余的上左侧n-1个比特的总和应为1比右侧的总和n-1个位的更多。
基于以上事实，我们得到以下递推公式。
diff是前半个数字和后半个数字的总和之间的期望差。最初diff为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

下面是基于以上朴素递归解决方案的实现。

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
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

 $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

// 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

输出：

Count of sequences is 6

上述解决方案的时间复杂度是指数的。如果绘制完整的递归树，则可以观察到许多子问题一次又一次地得到解决。例如，当我们从n = 4开始并且diff = 0时，我们可以通过多条路径达到(3，0)。由于再次调用了相同的问题，因此此问题具有“重叠子问题”属性。因此，最小平方和问题具有动态规划问题的两个属性(请参阅此内容)。
下面是一个基于备忘录的解决方案，它使用查找表来计算结果。

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
using namespace std;
#define MAX 1000
 
// A lookup table to store the results of subproblems
int lookup[MAX][MAX];
 
// dif is diference 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 diference 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 subbproblem 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 diference 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 diference 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 subbproblem 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 diference 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 diference 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 subbproblem 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 diference 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 diference 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 subbproblem 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

的PHP

 $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 subbproblem 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
?>

Java脚本

// 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 diference 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 diference 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 subbproblem 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

输出：

Count of sequences is 6

最坏的情况是此解决方案的时间复杂度为O(n ² )，因为diff可以最大为n。
下面是相同的O(n)解决方案。

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 

因此，我们可以使用以下方法获得所需的结果

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 

下面是基于以上思想的实现。

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
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

输出：

Count of sequences is 6