Queries of nCr%p in O(1) time complexity
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 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 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# 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
<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>
Time Complexity: O(N) for precomputing and O(1) for every query.
Auxiliary Space: O(N)
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