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

Last Updated : 11 Jul, 2025

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

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