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

• Difficulty Level : Medium
• Last Updated : 14 Jun, 2021

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 the 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 a 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

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

Since the count is

According to Fermat Little theorem:

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[100001];       // Store factorial of i at index i     factorial[0] = 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[10];       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 [100001];        // Store factorial of i at index i      factorial[0] = 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 [10];        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[0] = 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 [100001];        // Store factorial of i at index i      factorial[0] = 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 [10];        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

## Javascript

 

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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