从二进制字符串中的1开始的唯一排列数

给定一个由0和1组成的二进制字符串。任务是找到以1开头的字符串的唯一排列数。

注意：由于答案可能非常大，请以模数10 ⁹ + 7打印答案。

例子：

Input : str ="10101001001"
Output : 210

Input : str ="101110011"
Output : 56

这个想法是首先在给定的字符串找到1的计数和0的计数。现在让我们考虑一下字符串的长度$L$并且该字符串包含至少一个1。设1为$n$和0的数字是$m$ 。在n个1中，我们必须在字符串的开头放置一个1，因此我们要在n-1 1的左边，而m 0的w必须排列这些(n-1) 1和m 0的长度(L-1)的字符串。

因此，排列数将是：

(L-1)! / ((n-1)!*(m)!)

下面是上述想法的实现：

C++

// C++ program to find number of unique permutations
// of a binary string starting with 1
  
#include 
using namespace std;
  
#define MAX 1000003
#define mod 1000000007
  
// Array to store factorial of i at
// i-th index
long long fact[MAX];
  
// Precompute factorials under modulo mod
// upto MAX
void factorial()
{
    // factorial of 0 is 1
    fact[0] = 1;
  
    for (int i = 1; i < MAX; i++) {
        fact[i] = (fact[i - 1] * i) % mod;
    }
}
  
// Iterative Function to calculate (x^y)%p in O(log y)
long long power(long long x, long long y, long long p)
{
    long long res = 1; // Initialize result
  
    x = x % p; // Update x if it is more than or
    // equal to p
  
    while (y > 0) {
        // If y is odd, multiply x with result
        if (y & 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function to find the modular inverse
long long inverse(int n)
{
  
    return power(n, mod - 2, mod);
}
  
// Function to find the number of permutation
// starting with 1 of a binary string
int countPermutation(string s)
{
    // Generate factorials upto MAX
    factorial();
  
    int length = s.length(), num1 = 0, num0;
    long long count = 0;
  
    // find number of 1's
    for (int i = 0; i < length; i++) {
        if (s[i] == '1')
            num1++;
    }
  
    // number of 0's
    num0 = length - num1;
  
    // Find the number of permuattion of
    // string starting with 1 using the formulae:
    // (L-1)! / ((n-1)!*(m)!)
    count = (fact[length - 1] * 
            inverse((fact[num1 - 1] * 
                    fact[num0]) % mod)) % mod;
  
    return count;
}
  
// Driver code
int main()
{
    string str = "10101001001";
  
    cout << countPermutation(str);
  
    return 0;
}

Java

// Java program to find number of unique permutations 
// of a binary string starting with 1 
    
public class Improve {
      
    final static int MAX = 1000003 ;
    final static int mod = 1000000007 ;
      
    // Array to store factorial of i at 
    // i-th index 
    static long fact[] = new long [MAX]; 
      
    // Pre-compute factorials under modulo mod 
    // upto MAX 
    static void factorial() 
    { 
        // factorial of 0 is 1 
        fact[0] = 1; 
        
        for (int i = 1; i < MAX; i++) { 
            fact[i] = (fact[i - 1] * i) % mod; 
        } 
    } 
        
    // Iterative Function to calculate (x^y)%p in O(log y) 
    static long power(long x, long y, long p) 
    { 
        long res = 1; // Initialize result 
        
        x = x % p; // Update x if it is more than or 
        // equal to p 
        
        while (y > 0) { 
            // If y is odd, multiply x with result 
            if (y % 2 != 0) 
                res = (res * x) % p; 
        
            // y must be even now 
            y = y >> 1; // y = y/2 
            x = (x * x) % p; 
        } 
        return res; 
    } 
        
    // Function to find the modular inverse 
    static long inverse(int n) 
    { 
        
        return power(n, mod - 2, mod); 
    } 
        
    // Function to find the number of permutation 
    // starting with 1 of a binary string 
    static int countPermutation(String s) 
    { 
        // Generate factorials upto MAX 
        factorial(); 
        
        int length = s.length(), num1 = 0, num0; 
        long count = 0; 
        
        // find number of 1's 
        for (int i = 0; i < length; i++) { 
            if (s.charAt(i) == '1') 
                num1++; 
        } 
        
        // number of 0's 
        num0 = length - num1; 
        
        // Find the number of permuattion of 
        // string starting with 1 using the formulae: 
        // (L-1)! / ((n-1)!*(m)!) 
        count = (fact[length - 1] *  
                inverse((int) ((fact[num1 - 1] *  
                        fact[num0]) % mod))) % mod; 
        
        return (int) count; 
    } 
      
    // Driver code
    public static void main(String args[])
    {
           String str = "10101001001"; 
             
           System.out.println(countPermutation(str)); 
    }
    // This Code is contributed by ANKITRAI1
}

Python 3

# Python 3 program to find number 
# of unique permutations of a
# binary string starting with 1
MAX = 1000003
mod = 1000000007
  
# Array to store factorial of 
# i at i-th index
fact = [0] * MAX
  
# Precompute factorials under 
# modulo mod upto MAX
def factorial():
      
    # factorial of 0 is 1
    fact[0] = 1
  
    for i in range(1, MAX):
        fact[i] = (fact[i - 1] * i) % mod
  
# Iterative Function to calculate
# (x^y)%p in O(log y)
def power(x, y, p):
      
    res = 1 # Initialize result
  
    x = x % p # Update x if it is 
              # more than or equal to p
  
    while (y > 0) :
          
        # If y is odd, multiply
        # x with result
        if (y & 1):
            res = (res * x) % p
  
        # y must be even now
        y = y >> 1 # y = y/2
        x = (x * x) % p
      
    return res
  
# Function to find the modular inverse
def inverse( n):
    return power(n, mod - 2, mod)
  
# Function to find the number of 
# permutation starting with 1 of 
# a binary string
def countPermutation(s):
      
    # Generate factorials upto MAX
    factorial()
  
    length = len(s) 
    num1 = 0
    count = 0
  
    # find number of 1's
    for i in range(length) :
        if (s[i] == '1'):
            num1 += 1
              
    # number of 0's
    num0 = length - num1
  
    # Find the number of permuattion 
    # of string starting with 1 using 
    # the formulae: (L-1)! / ((n-1)!*(m)!)
    count = (fact[length - 1] *
            inverse((fact[num1 - 1] *
                     fact[num0]) % mod)) % mod
  
    return count
  
# Driver code
if __name__ =="__main__":
    s = "10101001001"
  
    print(countPermutation(s))
  
# This code is contributed 
# by ChitraNayal

C#

// C# program to find number of 
// unique permutations of a
// binary string starting with 1 
using System;
      
class GFG
{ 
static int MAX = 1000003 ; 
static int mod = 1000000007 ; 
  
// Array to store factorial 
// of i at i-th index 
static long []fact = new long [MAX]; 
  
// Pre-compute factorials under 
// modulo mod upto MAX 
static void factorial() 
{ 
    // factorial of 0 is 1 
    fact[0] = 1; 
      
    for (int i = 1; i < MAX; i++) 
    { 
        fact[i] = (fact[i - 1] * i) % mod; 
    } 
} 
      
// Iterative Function to calculate 
// (x^y)%p in O(log y) 
static long power(long x, 
                  long y, long p) 
{ 
    long res = 1; // Initialize result 
      
    x = x % p; // Update x if it is more 
               // than or equal to p 
      
    while (y > 0) 
    { 
        // If y is odd, multiply 
        // x with result 
        if (y % 2 != 0) 
            res = (res * x) % p; 
      
        // y must be even now 
        y = y >> 1; // y = y/2 
        x = (x * x) % p; 
    } 
    return res; 
} 
      
// Function to find the modular inverse 
static long inverse(int n) 
{ 
    return power(n, mod - 2, mod); 
} 
      
// Function to find the number of 
// permutation starting with 1 of
// a binary string 
static int countPermutation(string s) 
{ 
    // Generate factorials upto MAX 
    factorial(); 
      
    int length = s.Length, num1 = 0, num0; 
    long count = 0; 
      
    // find number of 1's 
    for (int i = 0; i < length; i++) 
    { 
        if (s[i] == '1') 
            num1++; 
    } 
      
    // number of 0's 
    num0 = length - num1; 
      
    // Find the number of permuattion 
    // of string starting with 1 using 
    // the formulae: (L-1)! / ((n-1)!*(m)!) 
    count = (fact[length - 1] * 
             inverse((int) ((fact[num1 - 1] * 
                    fact[num0]) % mod))) % mod; 
      
    return (int) count; 
} 
  
// Driver code 
public static void Main() 
{ 
    string str = "10101001001"; 
          
    Console.WriteLine(countPermutation(str)); 
} 
} 
  
// This code is contributed
// by anuj_67

输出：

210

时间复杂度： O(n)，其中n是字符串的长度。