Count different numbers possible using all the digits their frequency times

Given an array arr[] which contains the frequency of the digits (0-9), the task is to find the count of numbers possible using each digit its frequency times. That is, the final numbers generated should contain all of the digits their frequency times. Since, the count can be very large return the answer modulo 10^9+7.

Prerequisites: Factorial of a number, Modular multiplicative inverse
Examples:

Input : arr[] = {0, 1, 1, 0, 0, 0, 0, 0, 0, 0}
Output : 2
Frequency of 1 and 2 is 1 and all the rest is 0. 
Therefore, 2 possible numbers are 12 and 21.

Input : arr[] = {0, 5, 2, 0, 0, 0, 4, 0, 1, 1}
Output : 1081080

Approach: The problem can be easily solved using Permutation and Combinations. The ith index of array arr[] gives the frequency of ith digit. The problem can be thought of as finding the total permutation of X objects where X0 objects are of one type, X1 objects are of another type and so on. Then the possible count of numbers is given by:



Total Count = X! / (X0! * X1! *…..* X9!)

where X = Sum of frequencies of all the digits and Xi = the frequency of ith digit.

Below is the implementation of the above approach

C++

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// C++ implementation of the above approach
  
#include <bits/stdc++.h>
using namespace std;
  
#define ll long long
#define MAXN 100000
#define MOD 1000000007
  
// Initialize an array to store factorial values
ll fact[MAXN];
  
// Function to calculate and store X! values
void factorial()
{
    fact[0] = 1;
    for (int i = 1; i < MAXN; i++)
        fact[i] = (fact[i - 1] * i) % MOD;
}
  
// Iterative Function to calculate (x^y)%p in O(log y)
ll power(ll x, ll y, ll p)
{
    ll 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 that return modular
// inverse of x under modulo p
ll modInverse(ll x, ll p)
{
    return power(x, p - 2, p);
}
  
// Function that returns the count of
// different number possible by using
// all the digits its frequency times
ll countDifferentNumbers(ll arr[], ll P)
{
    // Preprocess factorial values
    factorial();
  
    // Initialize the result and sum
    // of aint the frequencies
    ll res = 0, X = 0;
  
    // Calculate the sum of frequencies
    for (int i = 0; i < 10; i++)
        X += arr[i];
  
    // Putting res equal to x!
    res = fact[X];
  
    // Multiplying res with modular
    // inverse of X0!, X1!, .., X9!
    for (int i = 0; i < 10; i++) {
        if (arr[i] > 1)
            res = (res * modInverse(fact[arr[i]], P)) % P;
    }
  
    return res;
}
  
// Driver Code
int main()
{
    ll arr[] = { 1, 0, 2, 0, 0, 7, 4, 0, 0, 3 };
    cout << countDifferentNumbers(arr, MOD);
  
    return 0;
}

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Java

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// Java program to implement
// the above approach
class GFG
{
  
static int MAXN = 100000;
static int MOD = 1000000007;
  
// Initialize an array to store factorial values
static long fact[] = new long[MAXN];
  
// Function to calculate and store X! values
static void factorial()
{
    fact[0] = 1;
    for (int i = 1; i < MAXN; 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== 1)
            res = (res * x) % p;
  
        // y must be even now
        y = y >> 1; // y = y/2
        x = (x * x) % p;
    }
    return res;
}
  
// Function that return modular
// inverse of x under modulo p
static long modInverse(long x, long p)
{
    return power(x, p - 2, p);
}
  
// Function that returns the count of
// different number possible by using
// along the digits its frequency times
static long countDifferentNumbers(long arr[], long P)
{
    // Preprocess factorial values
    factorial();
  
    // Initialize the result and sum
    // of aint the frequencies
    long res = 0, X = 0;
  
    // Calculate the sum of frequencies
    for (int i = 0; i < 10; i++)
        X += arr[i];
  
    // Putting res equal to x!
    res = fact[(int)X];
  
    // Multiplying res with modular
    // inverse of X0!, X1!, .., X9!
    for (int i = 0; i < 10; i++)
    {
        if (arr[i] > 1)
            res = (res * modInverse(fact[(int)arr[i]], P)) % P;
    }
  
    return res;
}
  
// Driver Code
public static void main(String[] args) 
{
    long arr[] = { 1, 0, 2, 0, 0, 7, 4, 0, 0, 3 };
    System.out.println(countDifferentNumbers(arr, MOD));
}
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

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# Python3 implementation of the above approach 
MAXN = 100000
MOD = 1000000007
  
# Initialize an array to store
# factorial values 
fact = [0] * MAXN; 
  
# Function to calculate and store X! values 
def factorial() :
    fact[0] = 1
    for i in range(1, MAXN) :
        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) %
  
        # y must be even now 
        y = y >> 1; # y = y/2 
        x = (x * x) % p
      
    return res
  
# Function that return modular 
# inverse of x under modulo p 
def modInverse(x, p) : 
    return power(x, p - 2, p) 
  
# Function that returns the count of 
# different number possible by using 
# all the digits its frequency times 
def countDifferentNumbers(arr, P) : 
  
    # Preprocess factorial values 
    factorial(); 
  
    # Initialize the result and sum 
    # of aint the frequencies 
    res = 0; X = 0
  
    # Calculate the sum of frequencies 
    for i in range(10) :
        X += arr[i] 
  
    # Putting res equal to x! 
    res = fact[X] 
  
    # Multiplying res with modular 
    # inverse of X0!, X1!, .., X9! 
    for i in range(10) :
        if (arr[i] > 1) :
            res = (res * modInverse(fact[arr[i]], P)) % P; 
  
    return res; 
  
# Driver Code 
if __name__ == "__main__"
  
    arr = [1, 0, 2, 0, 0, 7, 4, 0, 0, 3 ]
    print(countDifferentNumbers(arr, MOD)) 
      
# This code is contributed by Ryuga

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C#

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// C# program to implement
// the above approach
using System;
  
class GFG
{
  
    static int MAXN = 100000;
    static int MOD = 1000000007;
      
    // Initialize an array to store factorial values
    static long []fact = new long[MAXN];
      
    // Function to calculate and store X! values
    static void factorial()
    {
        fact[0] = 1;
        for (int i = 1; i < MAXN; 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== 1)
                res = (res * x) % p;
      
            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }
        return res;
    }
      
    // Function that return modular
    // inverse of x under modulo p
    static long modInverse(long x, long p)
    {
        return power(x, p - 2, p);
    }
      
    // Function that returns the count of
    // different number possible by using
    // along the digits its frequency times
    static long countDifferentNumbers(long []arr, long P)
    {
        // Preprocess factorial values
        factorial();
      
        // Initialize the result and sum
        // of aint the frequencies
        long res = 0, X = 0;
      
        // Calculate the sum of frequencies
        for (int i = 0; i < 10; i++)
            X += arr[i];
      
        // Putting res equal to x!
        res = fact[(int)X];
      
        // Multiplying res with modular
        // inverse of X0!, X1!, .., X9!
        for (int i = 0; i < 10; i++)
        {
            if (arr[i] > 1)
                res = (res * modInverse(fact[(int)arr[i]], P)) % P;
        }
      
        return res;
    }
      
    // Driver Code
    public static void Main(String[] args) 
    {
        long []arr = { 1, 0, 2, 0, 0, 7, 4, 0, 0, 3 };
        Console.WriteLine(countDifferentNumbers(arr, MOD));
    }
}
  
// This code has been contributed by 29AjayKumar

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Output:

245044800


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