# Count of distinct numbers formed by shuffling the digits of a large number N

Given a large number N in the form of a string, the task is to determine the count of distinct numbers that can be formed by shuffling the digits of number N.

Note:

• N may contain leading zeros.
• The number itself is also taken into count.
• Since the answer could be very large, print result modulo 109+7.

Examples:

Input: N = “23”
Output: 2
Explanation:
23 can be shuffled as {23, 32}

Input: N = “0223”
Output: 12
Explanation:
0223 can be shuffled as {2230, 2203, 2023, 3220, 3202, 3022, 2320, 2302, 2032, 0232, 0322, 0223 }

Naive Approach: The naive idea is to find all the permutations of the given number and print the count of unique numbers generated. But since the given number N is very large, it cannot be used.

Time Complexity: O(N * N!)
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is to use the concept of Permutation and Combination and Fermat’s little theorem. Below are the steps:

1. Use Fermat’s Little Theorem to find Modulo Multiplicative Inverse under modulo M where M is 109+7.
2. Instead of finding all the permutations, the result will be factorial of the length of given number N divided by the product of factorial of the count of a number as: where,
K is the number of digits in N
C[i] is the count of digits(from 0 to 9) in N.

3. Create an array of in which at each index store the factorial of that index.
4. In order to store the count of each digit, create an array of size 10 and initialize it with 0.
5. Initialize a variable answer with a value factorial of the length of N. For each count of a digit, find it’s a modular multiplicative inverse under modulo m and multiple with the result as:

Since the count is According to fermat little theorm: Therefore the count is given by: Below is the implementation of the above approach:

## C++

 // C++ program for the above approach  #include  using namespace std;  #define ll long long int     // Recursive function to return the value  // of (x ^ n) % m  ll modexp(ll x, ll n, ll m)  {      // Base Case      if (n == 0) {          return 1;      }         // If N is even      else if (n % 2 == 0) {          return modexp((x * x) % m,                        n / 2, m);      }         // Else N is odd      else {          return (x * modexp((x * x) % m,                             (n - 1) / 2, m)                  % m);      }  }     // Function to find modular inverse  // of a number x under modulo m  ll modInverse(ll x, ll m)  {      // Using Fermat's little theorem      return modexp(x, m - 2, m);  }     // Function to count of numbers formed by  // shuffling the digits of a large number N  void countNumbers(string N)  {      // Modulo value      ll m = 1000000007;         // Array to store the factorials      // upto the maximum value of N      ll factorial;         // Store factorial of i at index i      factorial = 1;      for (ll i = 1; i < 100001; i++) {             factorial[i] = (factorial[i - 1] * i) % m;      }         // To store count of occurrence      // of a digit      ll count;         for (ll i = 0; i < 10; i++) {          count[i] = 0;      }         ll length = N.length();         for (ll i = 0; i < length; i++)             // Increment the count of          // digit occured          count[N[i] - '0']++;         // Assign the factorial of      // length of input      ll result = factorial[length];         // Multiplying result with the      // modulo multiplicative inverse of      // factorial of count of i      for (ll i = 0; i < 10; i++) {             result = (result                    * modInverse(factorial[count[i]], m))                   % m;      }         // Print the result      cout << result;  }     // Driver Code  int main()  {      // Given Number as string      string N = "0223";         // Function call      countNumbers(N);      return 0;  }

## Java

 // Java program for the above approach   import java.util.*;      class GFG{         // Recursive function to return the value   // of (x ^ n) % m   static long modexp(long x, long n, long m)   {              // Base Case       if (n == 0)       {           return 1;       }          // If N is even       else if (n % 2 == 0)      {           return modexp((x * x) % m,                          n / 2, m);       }          // Else N is odd       else     {           return (x * modexp((x * x) % m,                      (n - 1) / 2, m) % m);       }   }      // Function to find modular inverse   // of a number x under modulo m   static long modInverse(long x, long m)   {              // Using Fermat's little theorem       return modexp(x, m - 2, m);   }      // Function to count of numbers formed by   // shuffling the digits of a large number N   static void countNumbers(String N)   {              // Modulo value       long m = 1000000007;          // Array to store the factorials       // upto the maximum value of N       long factorial[] = new long ;          // Store factorial of i at index i       factorial = 1;       for(int i = 1; i < 100001; i++)      {           factorial[i] = (factorial[i - 1] * i) % m;       }          // To store count of occurrence       // of a digit       long count[] = new long ;          for(int i = 0; i < 10; i++)       {           count[i] = 0;       }          long length = N.length();          for(int i = 0; i < length; i++)                  // Increment the count of           // digit occured           count[N.charAt(i) - '0']++;          // Assign the factorial of       // length of input       long result = factorial[(int)length];          // Multiplying result with the       // modulo multiplicative inverse of       // factorial of count of i       for(int i = 0; i < 10; i++)      {           result = (result *                     modInverse(                        factorial[(int)count[i]], m)) % m;       }          // Print the result       System.out.println(result);  }      // Driver code  public static void main(String args[])   {             // Given number as string       String N = "0223";          // Function call      countNumbers(N);  }  }     // This code is contributed by Stream_Cipher

## Python3

 # Python3 program for the above approach      # Recursive function to return the value   # of (x ^ n) % m   def modexp(x, n, m):             # Base Case       if (n == 0):          return 1            # If N is even       else:          if (n % 2 == 0):               return modexp((x * x) % m,                             n / 2, m);                  # Else N is odd           else:              return (x * modexp((x * x) % m,                         (n - 1) / 2, m) % m)     # Function to find modular inverse   # of a number x under modulo m   def modInverse(x, m):             # Using Fermat's little theorem       return modexp(x, m - 2, m)     # Function to count of numbers formed by   # shuffling the digits of a large number N   def countNumbers(N):              # Modulo value       m = 1000000007            # Array to store the factorials       # upto the maximum value of N       factorial = [0 for x in range(100001)]             # Store factorial of i at index i       factorial = 1;              for i in range(1, 100001):           factorial[i] = (factorial[i - 1] * i) % m             # To store count of occurrence       # of a digit       count = [0 for x in range(10)]             for i in range(0, 10):          count[i] = 0                length = len(N)              for i in range(0, length):                 # Increment the count of           # digit occured           count[int(N[i])] += 1            # Assign the factorial of       # length of input       result = factorial[int(length)]             # Multiplying result with the       # modulo multiplicative inverse of       # factorial of count of i       for i in range(0, 10):          result = (result *                    modInverse(                        factorial[int(count[i])], m)) % m             # Print the result       print(result)     # Driver code     # Given number as string   N = "0223";      # Function call  countNumbers(N)     # This code is contributed by Stream_Cipher

## C#

 // C# program for the above approach   using System.Collections.Generic;   using System;     class GFG{         // Recursive function to return the value   // of (x ^ n) % m   static long modexp(long x, long n, long m)   {              // Base Case       if (n == 0)      {           return 1;       }          // If N is even       else if (n % 2 == 0)      {           return modexp((x * x) % m,                          n / 2, m);       }          // Else N is odd       else      {           return (x * modexp((x * x) % m,                      (n - 1) / 2, m) % m);       }   }      // Function to find modular inverse   // of a number x under modulo m   static long modInverse(long x, long m)   {              // Using Fermat's little theorem       return modexp(x, m - 2, m);   }      // Function to count of numbers formed by   // shuffling the digits of a large number N   static void countNumbers(string N)   {              // Modulo value       long m = 1000000007;          // Array to store the factorials       // upto the maximum value of N       long []factorial = new long ;          // Store factorial of i at index i       factorial = 1;       for(int i = 1; i < 100001; i++)       {           factorial[i] = (factorial[i - 1] * i) % m;       }          // To store count of occurrence       // of a digit       long []count = new long ;          for(int i = 0; i < 10; i++)      {           count[i] = 0;       }          long length = N.Length;          for(int i = 0; i < length; i++)              // Increment the count of           // digit occured           count[N[i] - '0']++;          // Assign the factorial of       // length of input       long result = factorial[(int)length];          // Multiplying result with the       // modulo multiplicative inverse of       // factorial of count of i       for(int i = 0; i < 10; i++)      {           result = (result *                     modInverse(                        factorial[(int)count[i]], m)) % m;       }              // Print the result       Console.WriteLine(result);  }      // Driver code  public static void Main()   {             // Given number as string       string N = "0223";          // Function call      countNumbers(N);  }  }     // This code is contributed by Stream_Cipher

Output:

12


Time Complexity: O(K + log(M)). O(K) is used to calculate the factorial of the number N and according to Fermat’s Little Theorem, it takes O(log(M)) to calculate the modulo multiplicative inverse of any number x under modulo m.
Auxiliary Space: O(log10N), where N is the given number N.

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