Compute nCr % p | Set 1 (Introduction and Dynamic Programming Solution)

2.7

Given three numbers n, r and p, compute value of nCr mod p.

Example:

Input:  n = 10, r = 2, p = 13
Output: 6
Explanation: 10C2 is 45 and 45 % 13 is 6.


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A Simple Solution is to first compute nCr, then compute nCr % p. This solution works fine when the value of nCr is small.

What if the value of nCr is large?
The value of nCr%p is generally needed for large values of n when nCr cannot fit in a variable, and causes overflow. So computing nCr and then using modular operator is not a good idea as there will be overflow even for slightly larger values of n and r. For example the methods discussed here and here cause overflow for n = 50 and r = 40.

The idea is to compute nCr using below formula

   C(n, r) = C(n-1, r-1) + C(n-1, r)
   C(n, 0) = C(n, n) = 1

Working of Above formula and Pascal Triangle:
Let us see how above formula works for C(4,3)

1==========>> n = 0, C(0,0) = 1
1–1========>> n = 1, C(1,0) = 1, C(1,1) = 1
1–2–1======>> n = 2, C(2,0) = 1, C(2,1) = 2, C(2,2) = 1
1–3–3–1====>> n = 3, C(3,0) = 1, C(3,1) = 3, C(3,2) = 3, C(3,3)=1
1–4–6–4–1==>> n = 4, C(4,0) = 1, C(4,1) = 4, C(4,2) = 6, C(4,3)=4, C(4,4)=1
So here every loop on i, builds i’th row of pascal triangle, using (i-1)th row

Extension of above formula for modular arithmetic:
We can use distributive property of modulor operator to find nCr % p using above formula.

   C(n, r)%p = [ C(n-1, r-1)%p + C(n-1, r)%p ] % p
   C(n, 0) = C(n, n) = 1

The above formula can implemented using Dynamic Programming using a 2D array.

The 2D array based dynamic programming solution can be further optimized by constructing one row at a time. See Space optimized version in below post for details.

Binomial Coefficient using Dynamic Programming

Below is C++ implementation based on the space optimized version discussed in above post.

// A Dynamic Programming based solution to compute nCr % p
#include<bits/stdc++.h>
using namespace std;

// Returns nCr % p
int nCrModp(int n, int r, int p)
{
    // The array C is going to store last row of
    // pascal triangle at the end. And last entry
    // of last row is nCr
    int C[r+1];
    memset(C, 0, sizeof(C));

    C[0] = 1; // Top row of Pascal Triangle

    // One by constructs remaining rows of Pascal
    // Triangle from top to bottom
    for (int i = 1; i <= n; i++)
    {
        // Fill entries of current row using previous
        // row values
        for (int j = min(i, r); j > 0; j--)

            // nCj = (n-1)Cj + (n-1)C(j-1);
            C[j] = (C[j] + C[j-1])%p;
    }
    return C[r];
}

// Driver program
int main()
{
    int n = 10, r = 2, p = 13;
    cout << "Value of nCr % p is " << nCrModp(n, r, p);
    return 0;
}

Output:

Value of nCr % p is 6

Time complexity of above solution is O(n*r) and it requires O(n) space. There are more and better solutions to above problem.

Compute nCr % p | Set 2 (Lucas Theorem)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

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