Queries of nCr%p in O(1) time complexity

Given Q queries and P where P is a prime number, each query has two numbers N and R and the task is to calculate nCr mod p. 
Constraints: 

N <= 106
R <= 106
p is a prime number

Examples:

Input: 
Q = 2 p = 1000000007 
1st query: N = 15, R = 4 
2nd query: N = 20, R = 3 
Output:
1st query: 1365 
2nd query: 1140 
15!/(4!*(15-4)!)%1000000007 = 1365 
20!/(20!*(20-3)!)%1000000007 = 1140

A naive approach is to calculate nCr using formulae by applying modular operations at any time. Hence time complexity will be O(q*n).

A better approach is to use fermat little theorem. According to it nCr can also be written as (n!/(r!*(n-r)!) ) mod which is equivalent to (n!*inverse(r!)*inverse((n-r)!) ) mod p. So, precomputing factorial of numbers from 1 to n will allow queries to be answered in O(log n). The only calculation that needs to be done is calculating inverse of r! and (n-r)!. Hence overall complexity will be q*( log(n)) .

A efficient approach will be to reduce the better approach to an efficient one by precomputing the inverse of factorials. Precompute inverse of factorial in O(n) time and then queries can be answered in O(1) time. Inverse of 1 to N natural number can be computed in O(n) time using Modular multiplicative inverse. Using recursive definition of factorial, the following can be written:

n! = n * (n-1) !
taking inverse on both side 
inverse( n! ) = inverse( n ) * inverse( (n-1)! )

Since N’s maximum value is 10⁶, precomputing values till 10⁶ will do.

Below is the implementation of the above approach:

C++

// C++ program to answer queries
// of nCr in O(1) time.
#include <bits/stdc++.h>
#define ll long long
const int N = 1000001;
using namespace std;

// array to store inverse of 1 to N
ll factorialNumInverse[N + 1];

// array to precompute inverse of 1! to N!
ll naturalNumInverse[N + 1];

// array to store factorial of first N numbers
ll fact[N + 1];

// Function to precompute inverse of numbers
void InverseofNumber(ll p)
{
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
        naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
}
// Function to precompute inverse of factorials
void InverseofFactorial(ll p)
{
    factorialNumInverse[0] = factorialNumInverse[1] = 1;

    // precompute inverse of natural numbers
    for (int i = 2; i <= N; i++)
        factorialNumInverse[i] = (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
}

// Function to calculate factorial of 1 to N
void factorial(ll p)
{
    fact[0] = 1;

    // precompute factorials
    for (int i = 1; i <= N; i++) {
        fact[i] = (fact[i - 1] * i) % p;
    }
}

// Function to return nCr % p in O(1) time
ll Binomial(ll N, ll R, ll p)
{
    // n C r = n!*inverse(r!)*inverse((n-r)!)
    ll ans = ((fact[N] * factorialNumInverse[R])
              % p * factorialNumInverse[N - R])
             % p;
    return ans;
}

// Driver Code
int main()
{
    // Calling functions to precompute the
    // required arrays which will be required
    // to answer every query in O(1)
    ll p = 1000000007;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);

    // 1st query
    ll N = 15;
    ll R = 4;
    cout << Binomial(N, R, p) << endl;

    // 2nd query
    N = 20;
    R = 3;
    cout << Binomial(N, R, p) << endl;

    return 0;
}

Java

// Java program to answer queries
// of nCr in O(1) time
import java.io.*;

class GFG{

static int N = 1000001;

// Array to store inverse of 1 to N
static long[] factorialNumInverse = new long[N + 1];

// Array to precompute inverse of 1! to N!
static long[] naturalNumInverse = new long[N + 1];

// Array to store factorial of first N numbers
static long[] fact = new long[N + 1];

// Function to precompute inverse of numbers
public static void InverseofNumber(int p)
{
    naturalNumInverse[0] = naturalNumInverse[1] = 1;

    for(int i = 2; i <= N; i++)
        naturalNumInverse[i] = naturalNumInverse[p % i] *
                                 (long)(p - p / i) % p;
}

// Function to precompute inverse of factorials
public static void InverseofFactorial(int p)
{
    factorialNumInverse[0] = factorialNumInverse[1] = 1;

    // Precompute inverse of natural numbers
    for(int i = 2; i <= N; i++)
        factorialNumInverse[i] = (naturalNumInverse[i] *
                           factorialNumInverse[i - 1]) % p;
}

// Function to calculate factorial of 1 to N
public static void factorial(int p)
{
    fact[0] = 1;

    // Precompute factorials
    for(int i = 1; i <= N; i++)
    {
        fact[i] = (fact[i - 1] * (long)i) % p;
    }
}

// Function to return nCr % p in O(1) time
public static long Binomial(int N, int R, int p)
{

    // n C r = n!*inverse(r!)*inverse((n-r)!)
    long ans = ((fact[N] * factorialNumInverse[R]) %
                       p * factorialNumInverse[N - R]) % p;

    return ans;
}

// Driver code
public static void main (String[] args)
{

    // Calling functions to precompute the
    // required arrays which will be required
    // to answer every query in O(1)
    int p = 1000000007;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);

    // 1st query
    int n = 15;
    int R = 4;
    System.out.println(Binomial(n, R, p));

    // 2nd query
    n = 20;
    R = 3;
    System.out.println(Binomial(n, R, p));
}
}

// This code is contributed by RohitOberoi

Python3

# Python3 program to answer queries
# of nCr in O(1) time.
N = 1000001

# array to store inverse of 1 to N
factorialNumInverse = [None] * (N + 1)

# array to precompute inverse of 1! to N!
naturalNumInverse = [None] * (N + 1)

# array to store factorial of
# first N numbers
fact = [None] * (N + 1)

# Function to precompute inverse of numbers
def InverseofNumber(p):
    naturalNumInverse[0] = naturalNumInverse[1] = 1
    for i in range(2, N + 1, 1):
        naturalNumInverse[i] = (naturalNumInverse[p % i] *
                                   (p - int(p / i)) % p)

# Function to precompute inverse
# of factorials
def InverseofFactorial(p):
    factorialNumInverse[0] = factorialNumInverse[1] = 1

    # precompute inverse of natural numbers
    for i in range(2, N + 1, 1):
        factorialNumInverse[i] = (naturalNumInverse[i] *
                                  factorialNumInverse[i - 1]) % p

# Function to calculate factorial of 1 to N
def factorial(p):
    fact[0] = 1

    # precompute factorials
    for i in range(1, N + 1):
        fact[i] = (fact[i - 1] * i) % p

# Function to return nCr % p in O(1) time
def Binomial(N, R, p):

    # n C r = n!*inverse(r!)*inverse((n-r)!)
    ans = ((fact[N] * factorialNumInverse[R])% p *
                      factorialNumInverse[N - R])% p
    return ans

# Driver Code
if __name__ == '__main__':

    # Calling functions to precompute the
    # required arrays which will be required
    # to answer every query in O(1)
    p = 1000000007
    InverseofNumber(p)
    InverseofFactorial(p)
    factorial(p)

    # 1st query
    N = 15
    R = 4
    print(Binomial(N, R, p))

    # 2nd query
    N = 20
    R = 3
    print(Binomial(N, R, p))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to answer queries
// of nCr in O(1) time
using System;

class GFG{

static int N = 1000001;

// Array to store inverse of 1 to N
static long[] factorialNumInverse = new long[N + 1];

// Array to precompute inverse of 1! to N!
static long[] naturalNumInverse = new long[N + 1];

// Array to store factorial of first N numbers
static long[] fact = new long[N + 1];

// Function to precompute inverse of numbers
static void InverseofNumber(int p)
{
    naturalNumInverse[0] = naturalNumInverse[1] = 1;

    for(int i = 2; i <= N; i++)
        naturalNumInverse[i] = naturalNumInverse[p % i] *
                                 (long)(p - p / i) % p;
}

// Function to precompute inverse of factorials
static void InverseofFactorial(int p)
{
    factorialNumInverse[0] = factorialNumInverse[1] = 1;

    // Precompute inverse of natural numbers
    for(int i = 2; i <= N; i++)
        factorialNumInverse[i] = (naturalNumInverse[i] *
                            factorialNumInverse[i - 1]) % p;
}

// Function to calculate factorial of 1 to N
static void factorial(int p)
{
    fact[0] = 1;

    // Precompute factorials
    for(int i = 1; i <= N; i++)
    {
        fact[i] = (fact[i - 1] * (long)i) % p;
    }
}

// Function to return nCr % p in O(1) time
static long Binomial(int N, int R, int p)
{

    // n C r = n!*inverse(r!)*inverse((n-r)!)
    long ans = ((fact[N] * factorialNumInverse[R]) %
                       p * factorialNumInverse[N - R]) % p;

    return ans;
}

// Driver code
static void Main()
{

    // Calling functions to precompute the
    // required arrays which will be required
    // to answer every query in O(1)
    int p = 1000000007;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);

    // 1st query
    int n = 15;
    int R = 4;
    Console.WriteLine(Binomial(n, R, p));

    // 2nd query
    n = 20;
    R = 3;
    Console.WriteLine(Binomial(n, R, p));
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// Javascript program to answer queries
// of nCr in O(1) time.
var N = 1000001;

// array to store inverse of 1 to N
factorialNumInverse = Array(N+1).fill(0);

// array to precompute inverse of 1! to N!
naturalNumInverse = Array(N+1).fill(0);

// array to store factorial of first N numbers
fact = Array(N+1).fill(0);

// Function to precompute inverse of numbers
function InverseofNumber(p)
{
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (var i = 2; i <= N; i++)
        naturalNumInverse[i] = (naturalNumInverse[p % i] * (p - parseInt(p / i))) % p;
}

// Function to precompute inverse of factorials
function InverseofFactorial(p)
{
    factorialNumInverse[0] = factorialNumInverse[1] = 1;

    // precompute inverse of natural numbers
    for (var i = 2; i <= N; i++)
        factorialNumInverse[i] = ((naturalNumInverse[i] * factorialNumInverse[i - 1])) % p;
}

// Function to calculate factorial of 1 to N
function factorial(p)
{
    fact[0] = 1;

    // precompute factorials
    for (var i = 1; i <= N; i++) {
        fact[i] = (fact[i - 1] * i) % p;
    }
}

// Function to return nCr % p in O(1) time
function Binomial(N, R, p)
{

    // n C r = n!*inverse(r!)*inverse((n-r)!)
    var ans = ((((fact[N] * factorialNumInverse[R])% p) * factorialNumInverse[N - R]))% p;
    return ans;
}

// Driver Code

// Calling functions to precompute the
// required arrays which will be required
// to answer every query in O(1)
p = 100000007;
InverseofNumber(p);
InverseofFactorial(p);
factorial(p);

// 1st query
N = 15;
R = 4;
document.write(Binomial(N, R, p)+"<br>")

// 2nd query
N = 20;
R = 3;
document.write(Binomial(N, R, p));

// This code is contributed by noob2000.
</script>

Output:

1365
1140

Time Complexity: O(N) for precomputing and O(1) for every query.
Auxiliary Space: O(N)