我们得到一个整数N。我们需要计算这样的长度为N的二进制字符串的总数,以使长度为N / 2的第一个字符串中的“ 0”的数目是第二个字符串中的“ 0”的数目的两倍。长度N / 2。
注意:N始终是偶数正整数。
例子:
Input : N = 4
Output : 2
Explanation: 0010 and 0001 are two binary string such that the number of zero in the first half is double the number of zero in second half.
Input : N = 6
Output : 9
幼稚的方法:我们可以生成所有长度为N的二进制字符串,然后可以一一检查是否所选字符串遵循给定的方案。但是这种方法的时间复杂度是指数的,并且是O(N * 2 N )。
高效方法:此方法基于对该问题的一些数学分析。
此方法的先决条件:具有模函数的阶乘和组合式。
注意:令n = N / 2。
如果我们逐步执行一些分析:
- 含有字符串的数量在第一半字符串和在第二半字符串的一个“0”两“0”,等于n的C 2 * n个C1。
- 这在第二半字符串上半年字符串和两个“0”包含四个“0”字符串的数量,等于n的C 4 * n个C2。
- 这在第一半字符串和在第二半字符串三个“0”包含6“0”字符串的数量,等于n C 6 * N的C 3。
我们重复上述过程,直到如果n为偶数,前半部分的“ 0”数变为n,或者如果n为奇数,则为(n-1)。
因此,对于2 <= 2 * i <= n,我们的最终答案是Σ( n C (2 * i) * n C i ) 。
现在,我们只需使用置换和组合即可计算所需的字符串数。
算法 :
int n = N/2;
for(int i=2;i<=n;i=i+2)
ans += ( nCr(n, i) * nCr(n, i/2);
注意:您可以使用动态编程技术来预先计算CoeFunc(N,i)即n C i 。
如果我们预先计算O(N * N)中的CoeFunc(N,i),则时间复杂度为O(N)。
C++
// CPP for finding number of binary strings
// number of '0' in first half is double the
// number of '0' in second half of string
#include
// pre define some constant
#define mod 1000000007
#define max 1001
using namespace std;
// global values for pre computation
long long int nCr[1003][1003];
void preComputeCoeff()
{
for (int i = 0; i < max; i++) {
for (int j = 0; j <= i; j++) {
if (j == 0 || j == i)
nCr[i][j] = 1;
else
nCr[i][j] = (nCr[i - 1][j - 1] +
nCr[i - 1][j]) % mod;
}
}
}
// function to print number of required string
long long int computeStringCount(int N)
{
int n = N / 2;
long long int ans = 0;
// calculate answer using proposed algorithm
for (int i = 2; i <= n; i += 2)
ans = (ans + ((nCr[n][i] * nCr[n][i / 2])
% mod)) % mod;
return ans;
}
// Driver code
int main()
{
preComputeCoeff();
int N = 3;
cout << computeStringCount(N) << endl;
return 0;
}
Java
// Java program for finding number of binary strings
// number of '0' in first half is double the
// number of '0' in second half of string
class GFG {
// pre define some constant
static final long mod = 1000000007;
static final long max = 1001;
// global values for pre computation
static long nCr[][] = new long[1003][1003];
static void preComputeCoeff()
{
for (int i = 0; i < max; i++) {
for (int j = 0; j <= i; j++) {
if (j == 0 || j == i)
nCr[i][j] = 1;
else
nCr[i][j] = (nCr[i - 1][j - 1] +
nCr[i - 1][j]) % mod;
}
}
}
// function to print number of required string
static long computeStringCount(int N)
{
int n = N / 2;
long ans = 0;
// calculate answer using proposed algorithm
for (int i = 2; i <= n; i += 2)
ans = (ans + ((nCr[n][i] * nCr[n][i / 2])
% mod)) % mod;
return ans;
}
// main function
public static void main(String[] args)
{
preComputeCoeff();
int N = 3;
System.out.println( computeStringCount(N) );
}
}
// This code is contributed by Arnab Kundu.
Python3
# Python3 for finding number of binary strings
# number of '0' in first half is double the
# number of '0' in second half of string
# pre define some constant
mod = 1000000007
Max = 1001
# global values for pre computation
nCr = [[0 for _ in range(1003)]
for i in range(1003)]
def preComputeCoeff():
for i in range(Max):
for j in range(i + 1):
if (j == 0 or j == i):
nCr[i][j] = 1
else:
nCr[i][j] = (nCr[i - 1][j - 1] +
nCr[i - 1][j]) % mod
# function to print number of required string
def computeStringCount(N):
n = N // 2
ans = 0
# calculate answer using proposed algorithm
for i in range(2, n + 1, 2):
ans = (ans + ((nCr[n][i] *
nCr[n][i // 2]) % mod)) % mod
return ans
# Driver code
preComputeCoeff()
N = 3
print(computeStringCount(N))
# This code is contributed by mohit kumar
C#
// C# program for finding number of binary
// strings number of '0' in first half is
// double the number of '0' in second half
// of string
using System;
class GFG {
// pre define some constant
static long mod = 1000000007;
static long max = 1001;
// global values for pre computation
static long [,]nCr = new long[1003,1003];
static void preComputeCoeff()
{
for (int i = 0; i < max; i++)
{
for (int j = 0; j <= i; j++)
{
if (j == 0 || j == i)
nCr[i,j] = 1;
else
nCr[i,j] = (nCr[i - 1,j - 1]
+ nCr[i - 1,j]) % mod;
}
}
}
// function to print number of required
// string
static long computeStringCount(int N)
{
int n = N / 2;
long ans = 0;
// calculate answer using proposed
// algorithm
for (int i = 2; i <= n; i += 2)
ans = (ans + ((nCr[n,i]
* nCr[n,i / 2]) % mod)) % mod;
return ans;
}
// main function
public static void Main()
{
preComputeCoeff();
int N = 3;
Console.Write( computeStringCount(N) );
}
}
// This code is contributed by nitin mittal.
Javascript
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