# 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 <= 10^{6}R <= 10^{6}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: 114015!/(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 quereies 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 10^{6}, precomputing values till 10^{6} 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; ` `} ` |

*chevron_right*

*filter_none*

## Python3

# Python 3 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

**Output:**

1365 1140

**Time Complexity:** O(10^{6}) for precomputing and O(1) for every query.

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