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Count of factors of combination of N and K (nCk)

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  • Last Updated : 07 Apr, 2022
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Given integers N and K, the task is to find the number of factors of NCK. Since the answer can be very large return the count of factors modulo 998244353.

Example: 

Input: N = 5, K = 2
Output: 4
Explanation: 5C2 = 10 which have {1, 2, 5, 10} as its divisors. So answer would be 4.

Input: N = 10, K = 3
Output: 16

 

Approach:  The problem can be solved based on the following mathematical fact:

^{N}C_{K} = (N * (N-1)* \cdots * (N-K+1))/(K*(K-1)* \cdots *2*1)

This value is always a whole number (Let’s say M).

M = \prod_{i}^{}p_{i}^{e_{i}}

where pi is a prime number. 
Therefore, number of factors of M is

M = \prod_{i}^{}{(e_{i}+1)}

Based on the above fact, the problem can be solved by finding the prime factors of M and their exponents. To find the prime factors of M, find the prime factors of the numerator and prime factors of the denominator

Follow the steps mentioned below to solve the problem:

  • Do prime factorization of first K natural numbers using Sieve of Eratosthenes to find prime factors of denominator.
  • Similarly, do prime factorization of natural numbers in the range [N – K +1, N] to find prime factors of numerator.
  • After finding the prime factorization of all the terms in the numerator and denominator of NCK, use the formula shown above to find the number of factors of NCK.

Below is the implementation of the above approach:

C++




// C++ program to implement the approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to count the number of factors
int divisorsOfNchooseK(long long N, long long K)
{
 
    // Parameter in time and space complexity
    long long M = max((long long)sqrt(N), K);
 
    // Sieve of eratosthenes for finding prime upto M
    vector<bool> prime(M + 1, true);
    prime[0] = prime[1] = false;
    for (long long p = 2; p * p < prime.size(); p++) {
        if (prime[p]) {
            for (long long i = 2 * p; i < prime.size();
                 i += p) {
                prime[i] = false;
            }
        }
    }
 
    // Store the denominator values in deno vector
    vector<long long> deno(K + 1);
 
    for (long long i = 1; i <= K; i++) {
        deno[i] = i;
    }
 
    // Store the numerator values in nume vector
    vector<long long> nume(K);
 
    long long offset = N - K + 1;
 
    for (long long i = 0; i < K; i++) {
        nume[i] = offset + i;
    }
 
    // Variable for storing answer
    long long ans = 1;
 
    // Iterate through all prime upto M
    for (long long p = 2; p < prime.size(); p++) {
        if (prime[p]) {
            // Store the power of p in
            // prime factorization of C(N, K)
            long long int power = 0;
 
            // Do prime factorization of deno terms
            for (long long i = p; i <= K; i += p) {
                while (deno[i] % p == 0) {
                    power--;
                    deno[i] /= p;
                }
            }
 
            // Do prime factorization of nume terms
            for (long long i
                 = ((N - K + 1) + p - 1) / p * p;
                 i <= N; i += p) {
                while (nume[i - offset] % p == 0) {
                    power++;
                    nume[i - offset] /= p;
                }
            }
            ans *= (power + 1);
            ans %= 998244353;
        }
    }
 
    // Find whether any term in
    // numerator is divisible by some prime
    // greater than √N
    for (long long i = N - K + 1; i <= N; i++) {
        if (nume[i - offset] != 1) {
            ans *= 2; // Coefficient of this prime will be 1
            ans %= 998244353;
        }
    }
 
    return ans;
}
// Driver code
int main()
{
    long long N = 10, K = 3;
 
    // Function call
    int ans = divisorsOfNchooseK(N, K);
    cout << ans;
    return 0;
}


Java




// Java program to implement the approach
import java.util.*;
 
class GFG {
 
  // Function to count the number of factors
  static long divisorsOfNchooseK(long N, long K)
  {
 
    // Parameter in time and space complexity
    long M = Math.max((long)Math.sqrt(N), K);
 
    // Sieve of eratosthenes for finding prime upto M
    boolean[] prime = new boolean[(int)M + 1];
    for (long x = 0; x < M + 1; x++) {
      prime[(int)x] = true;
    }
 
    prime[0] = prime[1] = false;
    for (long p = 2; p * p < prime.length; p++) {
      if (prime[(int)p] == true) {
        for (long i = 2 * p; i < prime.length;
             i += p) {
          prime[(int)i] = false;
        }
      }
    }
 
    // Store the denominator values in deno vector
    long[] deno = new long[(int)K + 1];
 
    for (long i = 1; i <= K; i++) {
      deno[(int)i] = i;
    }
 
    // Store the numerator values in nume vector
    long[] nume = new long[(int)K];
 
    long offset = N - K + 1;
 
    for (long i = 0; i < K; i++) {
      nume[(int)i] = offset + i;
    }
 
    // Variable for storing answer
    long ans = 1;
 
    // Iterate through all prime upto M
    for (long p = 2; p < prime.length; p++) {
      if (prime[(int)p] == true) {
        // Store the power of p in
        // prime factorization of C(N, K)
        long power = 0;
 
        // Do prime factorization of deno terms
        for (long i = p; i <= K; i += p) {
          while (deno[(int)i] % p == 0) {
            power--;
            deno[(int)i] /= p;
          }
        }
 
        // Do prime factorization of nume terms
        for (long i = ((N - K + 1) + p - 1) / p * p;
             i <= N; i += p) {
          while (nume[(int)(i - offset)] % p == 0) {
            power++;
            nume[(int)(i - offset)] /= p;
          }
        }
        ans *= (power + 1);
        ans %= 998244353;
      }
    }
 
    // Find whether any term in
    // numerator is divisible by some prime
    // greater than √N
    for (long i = N - K + 1; i <= N; i++) {
      if (nume[(int)(i - offset)] != 1) {
        ans *= 2; // Coefficient of this prime will
        // be 1
        ans %= 998244353;
      }
    }
 
    return ans;
  }
 
  // Driver code
  public static void main (String[] args) {
    long N = 10, K = 3;
 
    // Function call
    long ans = divisorsOfNchooseK(N, K);
    System.out.print(ans);
  }
}
 
// This code is contributed by hrithikgarg03188.


Python3




# python3 program to implement the approach
 
import math
 
# Function to count the number of factors
def divisorsOfNchooseK(N, K) :
 
 
    # Parameter in time and space complexity
    M = max(int(math.sqrt(N)), K)
 
    # Sieve of eratosthenes for finding prime upto M
    prime = [ True for _ in range(M + 1) ]
    prime[0] = prime[1] = False
    for p in range(2, int(math.sqrt(len(prime)))) :
        if (prime[p]) :
            for i in range(2*p, len(prime), p) :
                prime[i] = False
         
 
    # Store the denominator values in deno vector
    deno = [ 0 for _ in range(K + 1) ]
 
    for i in range(1, K+1) :
        deno[i] = i
     
 
    # Store the numerator values in nume vector
    nume = [ 0 for _ in range(K) ]
 
    offset = N - K + 1
  
 
    for i in range(0, K) :
        nume[i] = offset + i
      
     
    # Variable for storing answer
    ans = 1
 
    # Iterate through all prime upto M
    for p in range(2, len(prime)) :
        if (prime[p]) :
            # Store the power of p in
            # prime factorization of C(N, K)
            power = 0
 
            # Do prime factorization of deno terms
            for i in range(p, K+1, p) :
                while (deno[i] % p == 0) :
                    power -= 1
                    deno[i] //= p
                 
             
            # Do prime factorization of nume terms
            for i in range(((N - K + 1) + p - 1) // p * p, N+1, p) :
                
                while (nume[i - offset] % p == 0) :
                    power += 1
                    nume[i - offset] //= p
                 
             
            ans *= (power + 1)
            ans %= 998244353
     
 
    # Find whether any term in
    # numerator is divisible by some prime
    # greater than √N
    for i in range(N - K + 1, N+1) :
        if (nume[i - offset] != 1) :
            ans *= 2 # Coefficient of this prime will be 1
            ans %= 998244353
 
    return ans
 
# Driver code
if __name__ == "__main__" :
 
    N, K = 103
 
    # Function call
    ans = divisorsOfNchooseK(N, K)
    print(ans)
     
    # This code is contributed by rakeshsahni


C#




// C# program to implement the approach
using System;
class GFG {
 
    // Function to count the number of factors
    static long divisorsOfNchooseK(long N, long K)
    {
 
        // Parameter in time and space complexity
        long M = Math.Max((long)Math.Sqrt(N), K);
 
        // Sieve of eratosthenes for finding prime upto M
        bool[] prime = new bool[M + 1];
        for (long x = 0; x < M + 1; x++) {
            prime[x] = true;
        }
 
        prime[0] = prime[1] = false;
        for (long p = 2; p * p < prime.Length; p++) {
            if (prime[p] == true) {
                for (long i = 2 * p; i < prime.Length;
                     i += p) {
                    prime[i] = false;
                }
            }
        }
 
        // Store the denominator values in deno vector
        long[] deno = new long[K + 1];
 
        for (long i = 1; i <= K; i++) {
            deno[i] = i;
        }
 
        // Store the numerator values in nume vector
        long[] nume = new long[K];
 
        long offset = N - K + 1;
 
        for (long i = 0; i < K; i++) {
            nume[i] = offset + i;
        }
 
        // Variable for storing answer
        long ans = 1;
 
        // Iterate through all prime upto M
        for (long p = 2; p < prime.Length; p++) {
            if (prime[p] == true) {
                // Store the power of p in
                // prime factorization of C(N, K)
                long power = 0;
 
                // Do prime factorization of deno terms
                for (long i = p; i <= K; i += p) {
                    while (deno[i] % p == 0) {
                        power--;
                        deno[i] /= p;
                    }
                }
 
                // Do prime factorization of nume terms
                for (long i = ((N - K + 1) + p - 1) / p * p;
                     i <= N; i += p) {
                    while (nume[i - offset] % p == 0) {
                        power++;
                        nume[i - offset] /= p;
                    }
                }
                ans *= (power + 1);
                ans %= 998244353;
            }
        }
 
        // Find whether any term in
        // numerator is divisible by some prime
        // greater than √N
        for (long i = N - K + 1; i <= N; i++) {
            if (nume[i - offset] != 1) {
                ans *= 2; // Coefficient of this prime will
                          // be 1
                ans %= 998244353;
            }
        }
 
        return ans;
    }
   
    // Driver code
    public static void Main()
    {
        long N = 10, K = 3;
 
        // Function call
        long ans = divisorsOfNchooseK(N, K);
        Console.Write(ans);
    }
}
 
// This code is contributed by Samim Hossain Mondal.


Javascript




<script>
 
// JavaScript program to implement the approach
 
// Function to count the number of factors
function divisorsOfNchooseK(N, K){
 
    // Parameter in time and space complexity
    let M = Math.max(Math.floor(Math.sqrt(N)), K);
 
    // Sieve of eratosthenes for finding prime upto M
    let prime = new Array(M + 1).fill(true);
    prime[0] = prime[1] = false;
    for (let p = 2; p * p < prime.length; p++) {
        if (prime[p]) {
            for (let i = 2 * p; i < prime.length;
                 i += p) {
                prime[i] = false;
            }
        }
    }
 
    // Store the denominator values in deno vector
    let deno = new Array(K+1).fill(0);
 
    for (let i = 1; i <= K; i++) {
        deno[i] = i;
    }
 
    // Store the numerator values in nume vector
    let nume = new Array(K).fill(0);
 
    let offset = N - K + 1;
    for (let i = 0; i < K; i++) {
        nume[i] = offset + i;
    }
 
    // Variable for storing answer
    let ans = 1;
 
    // Iterate through all prime upto M
    for (let p = 2; p < prime.length; p++) {
        if (prime[p]) {
            // Store the power of p in
            // prime factorization of C(N, K)
            let power = 0;
 
            // Do prime factorization of deno terms
            for (let i = p; i <= K; i += p) {
                while (deno[i] % p == 0) {
                    power--;
                    deno[i] = Math.floor(deno[i] / p);
                }
            }
             
            // Do prime factorization of nume terms
            for (let i = Math.floor(((N - K + 1) + p - 1) / p) * p; i <= N; i += p) {
                 
                while (nume[i - offset] % p == 0) {
                    power++;
                    nume[i - offset] = Math.floor(nume[i - offset]/p);
                }
            }
            ans *= (power + 1);
            ans %= 998244353;
        }
    }
 
    // Find whether any term in
    // numerator is divisible by some prime
    // greater than √N
    for (let i = N - K + 1; i <= N; i++) {
        if (nume[i - offset] != 1) {
            ans *= 2; // Coefficient of this prime will be 1
            ans %= 998244353;
        }
    }
 
    return ans;
}
 
// Driver code
let N = 10, K = 3;
 
// Function call
let ans = divisorsOfNchooseK(N, K);
document.write(ans);
 
// This code is contributed by  shinjanpatra
 
</script>


Output

16

Time Complexity: O(M * loglogM)
Auxiliary Space:  O(M) where M is max(√N, K)


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